Nuts for Numbers

Friday, 22nd Sep, 2023 by Krilling for Company in Arithmetic, Base Behavior, Integer Sequences, Mathematics and tagged 210, 2772, Arithmetic, base 10, base 8, base 9, integer sums, Mr Logic, Mr Logic and nuts, nuts, nutty numbers, nutty sums, palindromes, recreational math, recreational mathematics, recreational maths, summing integers, sums of consecutive integers, Viz, Viz comic | Leave a comment

I was looking at palindromes created by sums of consecutive integers. And I came across this beautiful result:

2772 = sum(22..77)

2772 = 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45 + 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55 + 56 + 57 + 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66 + 67 + 68 + 69 + 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77

You could call 2772 a nutty sum, because 77 is held inside 22 like a kernel inside a nutshell. Here some more nutty sums, sum(n₁..n₂), where n₂ is a kernel in the shell of n₁:

1599 = sum(19..59)
2772 = sum(22..77)
22113 = sum(23..211)
159999 = sum(199..599)
277103 = sum(203..771)
277722 = sum(222..777)
267786 = sum(266..778)
279777 = sum(277..797)
1152217 = sum(117..1522)
1152549 = sum(149..1525)
1152767 = sum(167..1527)
4296336 = sum(436..2963)
5330303 = sum(503..3303)
6235866 = sum(626..3586)
8418316 = sum(816..4183)
10470075 = sum(1075..4700)
11492217 = sum(1117..4922)
13052736 = sum(1306..5273)
13538277 = sum(1377..5382)
14557920 = sum(1420..5579)
15999999 = sum(1999..5999)
25175286 = sum(2516..7528)
26777425 = sum(2625..7774)
27777222 = sum(2222..7777)
37949065 = sum(3765..9490)
53103195 = sum(535..10319)
111497301 = sum(1101..14973)

Of course, you can go the other way and find nutty sums where sum(n₁..n₂) produces n₁ as a kernel inside the shell of n₂:

147 = sum(4..17)
210 = sum(1..20)
12056 = sum(20..156)
13467 = sum(34..167)
22797 = sum(79..227)
22849 = sum(84..229)
26136 = sum(61..236)
1145520 = sum(145..1520)
1208568 = sum(208..1568)
1334667 = sum(334..1667)
1540836 = sum(540..1836)
1931590 = sum(315..1990)
2041462 = sum(414..2062)
2041863 = sum(418..2063)
2158083 = sum(158..2083)
2244132 = sum(244..2132)
2135549 = sum(554..2139)
2349027 = sum(902..2347)
2883558 = sum(883..2558)
2989637 = sum(989..2637)

When you look at nutty sums in other bases, you’ll find that the number “210” is always triangular and always a nutty sum in bases > 2:

210 = sum(1..20) in b3 → 21 = sum(1..6) in b10
210 = sum(1..20) in b4 → 36 = sum(1..8) in b10
210 = sum(1..20) in b5 → 55 = sum(1..10) in b10
210 = sum(1..20) in b6 → 78 = sum(1..12) in b10
210 = sum(1..20) in b7 → 105 = sum(1..14) in b10
210 = sum(1..20) in b8 → 136 = sum(1..16) in b10
210 = sum(1..20) in b9 → 171 = sum(1..18) in b10
210 = sum(1..20) in b10
210 = sum(1..20) in b11 → 253 = sum(1..22) in b10
210 = sum(1..20) in b12 → 300 = sum(1..24) in b10
210 = sum(1..20) in b13 → 351 = sum(1..26) in b10
210 = sum(1..20) in b14 → 406 = sum(1..28) in b10
210 = sum(1..20) in b15 → 465 = sum(1..30) in b10
210 = sum(1..20) in b16 → 528 = sum(1..32) in b10
210 = sum(1..20) in b17 → 595 = sum(1..34) in b10
210 = sum(1..20) in b18 → 666 = sum(1..36) in b10
210 = sum(1..20) in b19 → 741 = sum(1..38) in b10
210 = sum(1..20) in b20 → 820 = sum(1..40) in b10
[…]

Why is 210 always a nutty sum like that? Because the formula for sum(n₁..n₂) is (n₁*n₂) * (n₂-n₁+1) / 2. In all bases > 2, the sum of 1 to 20 (where 20 = 2 * b) is therefore:

(1+20) * (20-1+1) / 2 = 21 * 20 / 2 = 21 * 10 = 210

And here are nutty sums of both kinds (n₁ inside n₂ and n₂ inside n₁) for base 8:

210 = sum(1..20) in b8 → 136 = sum(1..16) in b10
12653 = sum(26..153) → 5547 = sum(22..107)
23711 = sum(71..231) → 10185 = sum(57..153)
2022323 = sum(223..2023) → 533715 = sum(147..1043)
2032472 = sum(247..2032) → 537914 = sum(167..1050)
2271564 = sum(715..2264) → 619380 = sum(461..1204)
2307422 = sum(742..2302) → 626450 = sum(482..1218)
125265253 = sum(2526..15253) → 22375083 = sum(1366..6827)

3246710 = sum(310..2467) in b8 → 871880 = sum(200..1335)
in b10
5326512 = sum(512..3265) → 1420618 = sum(330..1717)
15540671 = sum(1571..5406) → 3588537 = sum(889..2822)
21625720 = sum(2120..6257) → 4664272 = sum(1104..3247)

And for base 9:

125 = sum(2..15) in b9 → 104 = sum(2..14) in b10
210 = sum(1..20) → 171 = sum(1..18)
12858 = sum(28..158) → 8720 = sum(26..134)
1128462 = sum(128..1462) → 609824 = sum(107..1109)
1288588 = sum(288..1588) → 708344 = sum(242..1214)
1475745 = sum(475..1745) → 817817 = sum(392..1337)
2010707 = sum(107..2007) → 1070017 = sum(88..1465)
2034446 = sum(344..2046) → 1085847 = sum(283..1500)
2040258 = sum(402..2058) → 1089341 = sum(326..1511)
2063410 = sum(341..2060) → 1104768 = sum(280..1512)
2215115 = sum(215..2115) → 1191281 = sum(176..1553)
2255505 = sum(555..2205) → 1217840 = sum(455..1625)
2475275 = sum(475..2275) → 1348880 = sum(392..1688)
2735455 = sum(735..2455) → 1499927 = sum(599..1832)

1555 = sum(15..55) in b9 → 1184 = sum(14..50) in b10
155858 = sum(158..558) → 96200 = sum(134..458)
1148181 = sum(181..1481) → 622720 = sum(154..1126)
2211313 = sum(213..2113) → 1188525 = sum(174..1551)
2211747 = sum(247..2117) → 1188880 = sum(205..1555)
6358585 = sum(685..3585) → 3404912 = sum(563..2669)
7037453 = sum(703..3745) → 3745245 = sum(570..2795)
7385484 = sum(784..3854) → 3953767 = sum(643..2884)
13518167 = sum(1367..5181) → 6685072 = sum(1033..3799)
15588588 = sum(1588..5588) → 7794224 = sum(1214..4130)
17603404 = sum(1704..6034) → 8859865 = sum(1300..4405)
26750767 = sum(2667..7507) → 13201360 = sum(2005..5515)

Post-Performative Post-Scriptum…

Viz ’s Mr Logic would be a fan of nutty sums. And unlike real nuts, they wouldn’t prove fatal:

Mr Logic Goes Nuts (strip from Viz comic)

(click for full-size)

Square Pairs

Wednesday, 13th Sep, 2023 by Krilling for Company in Integer Sequences, Mathematics, Powers, Prime numbers and tagged 13, 4n + 1, 4n + 3, Albert Girard, David Wells, facts about 13, mathematics, maths, Penguin Dictionary of Curious and Interesting Mathematics, Pierre de Fermat, primes, Pythagorean primes, sum of two squares, thirteen | Leave a comment

Girard knew and Fermat a few years later proved the beautiful theorem that every prime of the form 4n + 1; that is, the primes 5, 13, 17, 29, 37, 41, 53… is the sum of two squares in exactly one way. Primes of the form 4n + 3, such as 3, 7, 11, 19, 23, 31, 43, 47… are never the sum of two squares. — David Wells, The Penguin Dictionary of Curious and Interesting Numbers (1986), entry for “13”.

Elsewhere other-accessible…

• Fermat’s theorem on sums of two squares
• Pythagorean primes

Fine as Nine

Thursday, 20th Jul, 2023 by Krilling for Company in Geometry, Integer Sequences, Mathematics, Polygons, Reciprocals, Triangular Numbers and tagged enneagonal numbers, icosikaihenagonal, Integer Sequences, nonagons, pentadecagonal numbers, Polygonal numbers, reciprocals, recreational math, recreational mathematics, recreational maths, Triangular Numbers | Leave a comment

This is a regular nonagon (a polygon with nine sides):

A nonagon or enneagon (from Wikipedia)

And this is the endlessly repeating decimal of the reciprocal of 7:

1/7 = 0.142857142857142857142857…

What is the curious connection between 1/7 and nonagons? If I’d been asked that a week ago, I’d’ve had no answer. Then I found a curious connection when I was looking at the leading digits of polygonal numbers. A polygonal number is a number that can be represented in the form of a polygon. Triangular numbers look like this:

* = 1
* ** = 3 * ** *** = 6 * ** *** **** = 10
* ** *** **** ***** = 15

By looking at the shapes rather than the numbers, it’s easy to see that you generate the triangular numbers by simply summing the integers:

1 = 1 1+2=3 1+2+3=6 1+2+3+4=10 1+2+3+4+5=15

Now try the square numbers:

* = 1
** ** = 4 *** *** *** = 9 **** **** **** **** = 16 ***** ***** ***** ***** ***** = 25

You generate the square numbers by summing the odd integers:

1 = 1 1+3 = 4 1+3+5 = 9 1+3+7 = 16 1+3+7+9 = 25

Next come the pentagonal numbers, the hexagonal numbers, the heptagonal numbers, and so on. I was looking at the leading digits of these numbers and trying to find patterns. For example, when do the leading digits of the k-th triangular number, tri(k), match the digits of k? This is when:

tri(1) = 1 tri(19) = 190 tri(199) = 19900 tri(1999) = 1999000 tri(19999) = 199990000 tri(199999) = 19999900000 [...]

That pattern is easy to explain. The formula for the k-th polygonal number is k * ((pn-2)*k + (4-pn)) / 2, where pn = 3 for the triangular numbers, 4 for the square numbers, 5 for the pentagonal numbers, and so on. Therefore the k-th triangular number is k * (k + 1) / 2. When k = 19, the formula is 19 * (19 + 1) / 2 = 19 * 20 / 2 = 19 * 10 = 190. And so on. Now try the pol(k) = leaddig(pol(k)) for higher polygonal numbers. The patterns are easy to predict until you get to the nonagonal numbers:

square(10) = 100 square(100) = 10000 square(1000) = 1000000 square(10000) = 100000000 square(100000) = 10000000000 [...]
pentagonal(7) = 70 pentagonal(67) = 6700 pentagonal(667) = 667000 pentagonal(6667) = 66670000 pentagonal(66667) = 6666700000 [...] hexagonal(6) = 66 hexagonal(51) = 5151 hexagonal(501) = 501501 hexagonal(5001) = 50015001 hexagonal(50001) = 5000150001 [...] heptagonal(5) = 55 heptagonal(41) = 4141 heptagonal(401) = 401401 heptagonal(4001) = 40014001 heptagonal(40001) = 4000140001 [...] octagonal(4) = 40 octagonal(34) = 3400 octagonal(334) = 334000 octagonal(3334) = 33340000 octagonal(33334) = 3333400000 [...]
nonagonal(4) = 46 nonagonal(30) = 3075 nonagonal(287) = 287574 nonagonal(2858) = 28581429 nonagonal(28573) = 2857385719 nonagonal(285715) = 285715000000 nonagonal(2857144) = 28571444285716 nonagonal(28571430) = 2857143071428575 nonagonal(285714287) = 285714287571428574 nonagonal(2857142858) = 28571428581428571429 nonagonal(28571428573) = 2857142857385714285719 nonagonal(285714285715) = 285714285715000000000000 nonagonal(2857142857144) = 28571428571444285714285716 nonagonal(28571428571430) = 2857142857143071428571428575 nonagonal(285714285714287) = 285714285714287571428571428574 nonagonal(2857142857142858) = 28571428571428581428571428571429 nonagonal(28571428571428573) = 2857142857142857385714285714285719 nonagonal(285714285714285715) = 285714285714285715000000000000000000 nonagonal(2857142857142857144) = 28571428571428571444285714285714285716 nonagonal(28571428571428571430) = 2857142857142857143071428571428571428575 [...]

What’s going on with the leading digits of the nonagonals? Well, they’re generating a different reciprocal. Or rather, they’re generating the multiple of a different reciprocal:

1/7 * 2 = 2/7 = 0.285714285714285714285714285714...

And why does 1/7 have this curious connection with the nonagonal numbers? Because the nonagonal formula is k * (7k-5) / 2 = k * ((9-2) * k + (4-pn)) / 2. Now look at the pentadecagonal numbers, where pn = 15:

pentadecagonal(1538461538461538461540) = 15384615384615384615406923076923076923076930
2/13 = 0.153846153846153846153846153846...
pentadecagonal formula = k * (13k - 11) / 2 = k * ((15-2)*k + (4-15)) / 2

Penultimately, let’s look at the icosikaihenagonal numbers, where pn = 21:

icosikaihenagonal(2) = 21 icosikaihenagonal(12) = 1266 icosikaihenagonal(107) = 107856 icosikaihenagonal(1054) = 10544743 icosikaihenagonal(10528) = 1052878960 icosikaihenagonal(105265) = 105265947385 icosikaihenagonal(1052633) = 10526335263165 icosikaihenagonal(10526317) = 1052631731578951 icosikaihenagonal(105263159) = 105263159210526318 icosikaihenagonal(1052631580) = 10526315801578947370 icosikaihenagonal(10526315791) = 1052631579163157894746 icosikaihenagonal(105263157896) = 105263157896368421052636 icosikaihenagonal(1052631578949) = 10526315789497368421052643 icosikaihenagonal(10526315789475) = 1052631578947542105263157900 icosikaihenagonal(105263157894738) = 105263157894738263157894736845 icosikaihenagonal(1052631578947370) = 10526315789473706842105263157905 icosikaihenagonal(10526315789473686) = 1052631578947368689473684210526331 icosikaihenagonal(105263157894736843) = 105263157894736843000000000000000000 icosikaihenagonal(1052631578947368422) = 10526315789473684220526315789473684211 icosikaihenagonal(10526315789473684212) = 1052631578947368421257894736842105263166
2/19 = 0.1052631578947368421052631579
icosikaihenagonal formula = k * (19k - 17) / 2 = k * ((21-2)*k + (4-21)) / 2

And ultimately, let’s look at this other pattern in the leading digits of the triangular numbers, which I can’t yet explain at all:

tri(904) = 409060 tri(6191) = 19167336 tri(98984) = 4898965620 tri(996694) = 496699963165 tri(9989894) = 49898996060565 tri(99966994) = 4996699994681515 tri(999898994) = 499898999601055515 tri(9999669994) = 49996699999451815015 tri(99998989994) = 4999898999960055555015 tri(999996699994) = 499996699999945018150015 tri(9999989899994) = 49999898999996005055550015 tri(99999966999994) = 4999996699999994500181500015 tri(999999898999994) = 499999898999999600500555500015 [...]

A Walk on the Wide Side

Friday, 23rd Jun, 2023 by Krilling for Company in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Irrational numbers, φ, Mathematics, Phi and tagged Arithmetic, base 10, base 11, base 12, carries, carry, digit-sums of Fibonacci numbers, duodecimal, logarithms, powers of 2, recreational math, recreational mathematics, recreational maths, undecimal, undenary, unodecimal | Leave a comment

How wide is a number? The obvious answer is to count digits and say that 1 and 9 are one digit wide, 11 and 99 are two digits wide, 111 and 999 are three digits wide, and so on. But that isn’t a very good answer. 111 and 999 are both three digits wide, but 999 is nine larger times than 111. And although 111 and 999 are both one digit wider than 11 and 99, 111 is much closer to 99 than 999 is to 111.

So there’s got to be a better answer to the question. I came across it indirectly, when I started looking at carries in powers. I wanted to know how fast a number grew in digit-width as it was multiplied repeatedly by, say, 2. For example, 2^3 = 8 and 2^4 = 16, so there’s been a carry at the far left and 2^4 = 16 has increased in digit-width by 1 over 2^3 = 8. After that, 2^6 = 64 and 2^7 = 128, so there’s another carry and another increase in digit-width. I wrote a program to sum the carries and divide them by the power. If I were better at math, I would’ve known what the value of carries / power was going to be. Here’s the program beginning to find it (it begins with a carry of 1, to mark 2^0 = 1 as creating a digit ex nihilo, as it were):

8 = 2^3 16 = 2^4 → 2 / 4 = 0.5 64 = 2^6 128 = 2^7 → 3 / 7 = 0.4285714285714285714285714286 512 = 2^9 1024 = 2^10 → 4 / 10 = 0.4 8192 = 2^13 16384 = 2^14 → 5 / 14 = 0.3571428571428571428571428571 65536 = 2^16 131072 = 2^17 → 6 / 17 = 0.3529411764705882352941176471 524288 = 2^19 1048576 = 2^20 → 7 / 20 = 0.35 8388608 = 2^23 16777216 = 2^24 → 8 / 24 = 0.3... 67108864 = 2^26 134217728 = 2^27 → 9 / 27 = 0.3... 536870912 = 2^29 1073741824 = 2^30 → 10 / 30 = 0.3... 8589934592 = 2^33 17179869184 = 2^34 → 11 / 34 = 0.3235294117647058823529411765 68719476736 = 2^36 137438953472 = 2^37 → 12 / 37 = 0.3243243243243243243243243243 549755813888 = 2^39 1099511627776 = 2^40 → 13 / 40 = 0.325 8796093022208 = 2^43 17592186044416 = 2^44 → 14 / 44 = 0.318... 70368744177664 = 2^46 140737488355328 = 2^47 → 15 / 47 = 0.3191489361702127659574468085 562949953421312 = 2^49 1125899906842624 = 2^50 → 16 / 50 = 0.32 9007199254740992 = 2^53 18014398509481984 = 2^54 → 17 / 54 = 0.3148... 72057594037927936 = 2^56 144115188075855872 = 2^57 → 18 / 57 = 0.3157894736842105263157894737 576460752303423488 = 2^59 1152921504606846976 = 2^60 → 19 / 60 = 0.316... 9223372036854775808 = 2^63 18446744073709551616 = 2^64 → 20 / 64 = 0.3125 73786976294838206464 = 2^66 147573952589676412928 = 2^67 → 21 / 67 = 0.3134328358208955223880597015 590295810358705651712 = 2^69 1180591620717411303424 = 2^70 → 22 / 70 = 0.3142857... 9444732965739290427392 = 2^73 18889465931478580854784 = 2^74 → 23 / 74 = 0.3108... 75557863725914323419136 = 2^76 151115727451828646838272 = 2^77 → 24 / 77 = 0.3116883... 604462909807314587353088 = 2^79 1208925819614629174706176 = 2^80 → 25 / 80 = 0.3125 9671406556917033397649408 = 2^83 19342813113834066795298816 = 2^84 → 26 / 84 = 0.3095238095238095238095238095 77371252455336267181195264 = 2^86 154742504910672534362390528 = 2^87 → 27 / 87 = 0.3103448275862068965517241379 618970019642690137449562112 = 2^89 1237940039285380274899124224 = 2^90 → 28 / 90 = 0.31... 9903520314283042199192993792 = 2^93 19807040628566084398385987584 = 2^94 → 29 / 94 = 0.3085106382978723404255319149 79228162514264337593543950336 = 2^96 158456325028528675187087900672 = 2^97 → 30 / 97 = 0.3092783505154639175257731959 633825300114114700748351602688 = 2^99 1267650600228229401496703205376 = 2^100 → 31 / 100 = 0.31

After calculating 2^p higher and higher (I discarded trailing digits of 2^p), I realized that the answer — carries / power — was converging on a value of slightly less than 0.30103. In the end (doh!), I realized that what I was calculating was the logarithm of 2 in base 10:

log(2) = 0.3010299956639811952137388947... 10^0.301029995663981... = 2

You can use then same carries-and-powers method to approximate the values of other logarithms:

log(1) = 0 log(2) = 0.3010299956639811952137388947... log(3) = 0.4771212547196624372950279033... log(4) = 0.6020599913279623904274777894... log(5) = 0.6989700043360188047862611053... log(6) = 0.7781512503836436325087667980... log(7) = 0.8450980400142568307122162586... log(8) = 0.9030899869919435856412166842... log(9) = 0.9542425094393248745900558065...

I also realized logarithms are a good answer to the question I raised above: How wide is a number? The logs of the powers of 2 are multiples of log(2):

log(2^1) = log(2) = 0.301029995663981195213738894 log(2^2) = log(4) = 0.602059991327962390427477789 = 2 * log(2) log(2^3) = log(8) = 0.903089986991943585641216684 = 3 * log(2) log(2^4) = log(16) = 1.204119982655924780854955579 = 4 * log(2) log(2^5) = log(32) = 1.505149978319905976068694474 = 5 * log(2) log(2^6) = log(64) = 1.806179973983887171282433368 = 6 * log(2) log(2^7) = log(128) = 2.107209969647868366496172263 = 7 * log(2) log(2^8) = log(256) = 2.408239965311849561709911158 = 8 * log(2) log(2^9) = log(512) = 2.709269960975830756923650053 = 9 * log(2) log(2^10) = log(1024) = 3.010299956639811952137388947 = 10 * log(2)

4 is 2 times larger than 2 and, in a sense, the width of 4 is 0.301029995663981… greater than the width of 2. As you can see, when the integer part of the log-sum increases by 1, so does the digit-width of the power:

log(2^3) = log(8) = 0.903089986991943585641216684 = 3 * log(2) log(2^4) = log(16) = 1.204119982655924780854955579 = 4 * log(2)
[...] log(2^6) = log(64) = 1.806179973983887171282433368 = 6 * log(2) log(2^7) = log(128) = 2.107209969647868366496172263 = 7 * log(2) [...]
log(2^9) = log(512) = 2.709269960975830756923650053 = 9 * log(2) log(2^10) = log(1024) = 3.01029995663981195213738894 = 10 * log(2)

In other words, powers of 2 are increasing in width by 0.301029995663981… units. When the increase flips the integer part of the log-sum up by 1, the digit-width or digit-count also increases by 1. To find the digit-count of a number, n, in a particular base, you simply take the integer part of log(n,b) and add 1. In base 10, the log of 123456789 is 8.091514… The integer part is 8 and 8+1 = 9. But it also makes perfect sense that log(1) = 0. No matter how many times you multiply a number by 1, the number never changes. That is, its width stays the same. So you can say that 1 has a width of 0, while 2 has a width of 0.301029995663981…

Logarithms also answer a question pre-previously raised on Overlord of the Über-Feral: Why are the Fibonacci numbers so productive in base 11 for digsum(fib(k)) = k? In base 10, such numbers are quickly exhausted:

digsum(fib(1)) = 1 = digsum(1) digsum(fib(5)) = 5 = digsum(5) digsum(fib(10)) = 10 = digsum(55) digsum(fib(31)) = 31 = digsum(1346269) digsum(fib(35)) = 35 = digsum(9227465) digsum(fib(62)) = 62 = digsum(4052739537881) digsum(fib(72)) = 72 = digsum(498454011879264) digsum(fib(175)) = 175 = digsum(1672445759041379840132227567949787325) digsum(fib(180)) = 180 = digsum(18547707689471986212190138521399707760) digsum(fib(216)) = 216 = digsum(619220451666590135228675387863297874269396512) digsum(fib(251)) = 251 = digsum(12776523572924732586037033894655031898659556447352249) digsum(fib(252)) = 252 = digsum(20672849399056463095319772838289364792345825123228624) digsum(fib(360)) = 360 digsum(fib(494)) = 494 digsum(fib(540)) = 540 digsum(fib(946)) = 946 digsum(fib(1188)) = 1188 digsum(fib(2222)) = 2222

In base 11, such numbers go on and on:

digsum(fib(1),b=11) = 1 = digsum(1) (k=1) digsum(fib(5),b=11) = 5 = digsum(5) (k=5) digsum(fib(12),b=11) = 12 = digsum(1A2) (k=13) digsum(fib(38),b=11) = 38 = digsum(855138A1) (k=41) digsum(fib(49)) = 49 = digsum(2067A724762) (k=53) (c=5) digsum(fib(50)) = 50 = digsum(542194A6905) (k=55) digsum(fib(55)) = 55 = digsum(54756364A280) (k=60) digsum(fib(56)) = 56 = digsum(886283256841) (k=61) digsum(fib(82)) = 82 = digsum(57751318A9814A6410) (k=90) digsum(fib(89)) = 89 = digsum(140492673676A06482A2) (k=97) digsum(fib(144)) = 144 = digsum(401631365A48A784A09392136653457871) (k=169) (c=10) digsum(fib(159)) = 159 = digsum(67217257641069185100889658A1AA72A0805) (k=185) digsum(fib(166)) = 166 = digsum(26466A3A88237918577363A2390343388205432) (k=193) digsum(fib(186)) = 186 = digsum(6A963147A9599623A20A05390315140A21992A96005) (k=215) digsum(fib(221)) = 221 (k=265) (c=15) digsum(fib(225)) = 225 (k=269) digsum(fib(2A1)) = 2A1 (k=353) digsum(fib(2A3)) = 2A3 (k=355)
[...] digsum(fib(39409)) = 39409 (k=56395) digsum(fib(3958A)) = 3958A (k=56605) (c=295) digsum(fib(3965A)) = 3965A (k=56693) digsum(fib(3A106)) = 3A106 (k=57360) digsum(fib(3AA46)) = 3AA46 (k=58493) digsum(fib(40140)) = 40140 (k=58729) digsum(fib(4222A)) = 4222A (k=61500) (c=300) digsum(fib(42609)) = 42609 (k=61961) digsum(fib(42775)) = 42775 (k=62155) digsum(fib(4287A)) = 4287A (k=62281) digsum(fib(430A2)) = 430A2 (k=62669) digsum(fib(43499)) = 43499 (k=63149) (c=305) digsum(fib(435A9)) = 435A9 (k=63281) [...] digsum(fib(157476)) = 157476 (k=244140) (c=525) digsum(fib(158470)) = 158470 (k=245465) digsum(fib(159037)) = 159037 (k=246275) digsum(fib(159285)) = 159285 (k=246570) digsum(fib(159978)) = 159978 (k=247409) digsum(fib(162993)) = 162993 (k=252750) (c=530) digsum(fib(163A32)) = 163A32 (k=254135) digsum(fib(164918)) = 164918 (k=255329) digsum(fib(166985)) = 166985 (k=258065) digsum(fib(167234)) = 167234 (k=258493) digsum(fib(167371)) = 167371 (k=258655) (c=535) digsum(fib(1676A5)) = 1676A5 (k=259055) digsum(fib(16992A)) = 16992A (k=261997) [...]

When do these numbers run out in base 11? I don’t know, but I do know why there are so many of them. The answer involves the logarithm of a special number. The most famous aspect of Fibonacci numbers is that the ratio, fib(k) / fib(k-1), of successive numbers converges on an irrational constant known as Φ. Here are the first Fibonacci numbers, where fib(k) = fib(k-2) + fib(k-1) (in other words, 1+1 = 2, 1+2 = 3, 2+3 = 5, and so on):

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, ...

And here are the first ratios:

1 / 1 = 1 2 / 1 = 2 3 / 2 = 1.5 5 / 3 = 1.6... 8 / 5 = 1.6 13 / 8 = 1.625 21 / 13 = 1.6153846... 34 / 21 = 1.619047... 55 / 34 = 1.617647058823529411764705882 89 / 55 = 1.618... 144 / 89 = 1.617977528089887640449438202 233 / 144 = 1.61805... 377 / 233 = 1.618025751072961373390557940 610 / 377 = 1.618037135278514588859416446 987 / 610 = 1.618032786885245901639344262 1597 / 987 = 1.618034447821681864235055724 2584 / 1597 = 1.618033813400125234815278648 4181 / 2584 = 1.618034055727554179566563468 6765 / 4181 = 1.618033963166706529538387946 [...]

The ratios get closer and closer to Φ = 1.618033988749894848204586834… = (√5 + 1) / 2. In other words, fib(k) ≈ fib(k-1) * Φ = fib(k-1) * 1.618… in base 10. This means that the digit-length of fib(k) ≈ integer(k * log(&Phi)) + 1. In base b, the average value of a digit in a Fibonacci number is (b^2-b) / 2b. Therefore in base 10, the average value of a digit is (10^2-10) / 20 = 90 / 20 = 4.5. The average value of digsum(fib(k)) ≈ 4.5 * log(&Phi) * k = 4.5 * 0.20898764… * k = 0.940444… * k. It isn’t surprising that as fib(k) gets larger, digsum(fib(k)) tends to get smaller than k.

In base 10, anyway. But what about base 11? In base 11, log(Φ) = 0.20068091818623… and the average value of a base-11 digit in fib(k) is 5 = 110 / 22 = (11^2 – 11) / 22. Therefore the average value of digsum(fib(k)) in base 11 is 5 * log(&Phi) * k = 5 * 0.20068091818623… * k = 1.00340459… * k. The average value of digsum(fib(k)) is much closer to k and it’s not surprising that for so many fib(k) in base 11, digsum(fib(k)) = k. In base 11, log(Φ) ≈ 1/5 and because the average digval is 5, digsum(fib(k)) ≈ 5 * 1/5 * k = 1 * k = k. As we’ve seen, that isn’t true in base 10. Nor is it true in base 12, where log(Φ) = 0.1936538843826… and average digval is 5.5 = (12^2 – 12) / 24 = 132 / 24. Therefore the average value in base 12 of digsum(fib(k)) = 1.0650963641… * k. The function digsum(fib(k)) = k rapidly dries up in base 12, just as it does in base 10:

digsum(fib(1),b=12) = 1 = digsum(1) (k=1) digsum(fib(5),b=12) = 5 = digsum(5) (k=5) digsum(fib(11) = 11 = digsum(175) (k=13) digsum(fib(12) = 12 = digsum(275) (k=14) digsum(fib(75) = 75 = digsum(976446538A0863811) (k=89) (c=5) digsum(fib(80) = 80 = digsum(1B3643B50939808B400) (k=96) digsum(fib(A3) = A3 = digsum(35147A566682BB9529034402) (k=123) digsum(fib(165) = 165 (k=221) digsum(fib(283) = 283 (k=387) digsum(fib(2AB) = 2AB (k=419) (c=10) digsum(fib(39A) = 39A (k=550) digsum(fib(460) = 460 (k=648) digsum(fib(525) = 525 (k=749) digsum(fib(602) = 602 (k=866) digsum(fib(624) = 624 (k=892) (c=15) digsum(fib(781) = 781 (k=1105) digsum(fib(1219) = 1219 (k=2037)

Previously Pre-Posted…

• Mötley Vüe — more on digsum(fib(k)) = k

Two be Continued…

Saturday, 17th Jun, 2023 by Krilling for Company in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Mathematics, Phi and tagged base 10, base 11, digit-sums, Fibonacci index, Fibonacci numbers, Integer Sequences, recreational math, recreational mathematics, recreational maths, undecimal, undenary, unodecimal | Leave a comment

Here’s a useless fact that nobody interested in mathematics would ever forget: digsum(fib(2222)) = 2222. That is, if you add the digits of the 2222nd Fibonacci number, you get 2222:

fib(2222) = 104,966,721,620,282,584,734,867,037,988,863,914,269,721,309,244,628,258,918,225,835,217,264,239,539,186,480,867,849,267,122,885,365,019,934,494,625,410,255,045,832,359,715,759,649,385,824,745,506,982,513,773,397,742,803,445,080,995,617,047,976,796,168,678,756,479,470,761,439,513,575,962,955,568,645,505,845,492,393,360,201,582,183,610,207,447,528,637,825,187,188,815,786,270,477,935,419,631,184,553,635,981,047,057,037,341,800,837,414,913,595,584,426,355,208,257,232,868,908,837,817,478,483,039,310,790,967,631,454,123,105,472,742,221,897,397,857,677,674,619,381,961,429,837,434,434,636,098,678,708,225,493,682,469,561
2222 = 1 + 0 + 4 + 9 + 6 + 6 + 7 + 2 + 1 + 6 + 2 + 0 + 2 + 8 + 2 + 5 + 8 + 4 + 7 + 3 + 4 + 8 + 6 + 7 + 0 + 3 + 7 + 9 + 8 + 8 + 8 + 6 + 3 + 9 + 1 + 4 + 2 + 6 + 9 + 7 + 2 + 1 + 3 + 0 + 9 + 2 + 4 + 4 + 6 + 2 + 8 + 2 + 5 + 8 + 9 + 1 + 8 + 2 + 2 + 5 + 8 + 3 + 5 + 2 + 1 + 7 + 2 + 6 + 4 + 2 + 3 + 9 + 5 + 3 + 9 + 1 + 8 + 6 + 4 + 8 + 0 + 8 + 6 + 7 + 8 + 4 + 9 + 2 + 6 + 7 + 1 + 2 + 2 + 8 + 8 + 5 + 3 + 6 + 5 + 0 + 1 + 9 + 9 + 3 + 4 + 4 + 9 + 4 + 6 + 2 + 5 + 4 + 1 + 0 + 2 + 5 + 5 + 0 + 4 + 5 + 8 + 3 + 2 + 3 + 5 + 9 + 7 + 1 + 5 + 7 + 5 + 9 + 6 + 4 + 9 + 3 + 8 + 5 + 8 + 2 + 4 + 7 + 4 + 5 + 5 + 0 + 6 + 9 + 8 + 2 + 5 + 1 + 3 + 7 + 7 + 3 + 3 + 9 + 7 + 7 + 4 + 2 + 8 + 0 + 3 + 4 + 4 + 5 + 0 + 8 + 0 + 9 + 9 + 5 + 6 + 1 + 7 + 0 + 4 + 7 + 9 + 7 + 6 + 7 + 9 + 6 + 1 + 6 + 8 + 6 + 7 + 8 + 7 + 5 + 6 + 4 + 7 + 9 + 4 + 7 + 0 + 7 + 6 + 1 + 4 + 3 + 9 + 5 + 1 + 3 + 5 + 7 + 5 + 9 + 6 + 2 + 9 + 5 + 5 + 5 + 6 + 8 + 6 + 4 + 5 + 5 + 0 + 5 + 8 + 4 + 5 + 4 + 9 + 2 + 3 + 9 + 3 + 3 + 6 + 0 + 2 + 0 + 1 + 5 + 8 + 2 + 1 + 8 + 3 + 6 + 1 + 0 + 2 + 0 + 7 + 4 + 4 + 7 + 5 + 2 + 8 + 6 + 3 + 7 + 8 + 2 + 5 + 1 + 8 + 7 + 1 + 8 + 8 + 8 + 1 + 5 + 7 + 8 + 6 + 2 + 7 + 0 + 4 + 7 + 7 + 9 + 3 + 5 + 4 + 1 + 9 + 6 + 3 + 1 + 1 + 8 + 4 + 5 + 5 + 3 + 6 + 3 + 5 + 9 + 8 + 1 + 0 + 4 + 7 + 0 + 5 + 7 + 0 + 3 + 7 + 3 + 4 + 1 + 8 + 0 + 0 + 8 + 3 + 7 + 4 + 1 + 4 + 9 + 1 + 3 + 5 + 9 + 5 + 5 + 8 + 4 + 4 + 2 + 6 + 3 + 5 + 5 + 2 + 0 + 8 + 2 + 5 + 7 + 2 + 3 + 2 + 8 + 6 + 8 + 9 + 0 + 8 + 8 + 3 + 7 + 8 + 1 + 7 + 4 + 7 + 8 + 4 + 8 + 3 + 0 + 3 + 9 + 3 + 1 + 0 + 7 + 9 + 0 + 9 + 6 + 7 + 6 + 3 + 1 + 4 + 5 + 4 + 1 + 2 + 3 + 1 + 0 + 5 + 4 + 7 + 2 + 7 + 4 + 2 + 2 + 2 + 1 + 8 + 9 + 7 + 3 + 9 + 7 + 8 + 5 + 7 + 6 + 7 + 7 + 6 + 7 + 4 + 6 + 1 + 9 + 3 + 8 + 1 + 9 + 6 + 1 + 4 + 2 + 9 + 8 + 3 + 7 + 4 + 3 + 4 + 4 + 3 + 4 + 6 + 3 + 6 + 0 + 9 + 8 + 6 + 7 + 8 + 7 + 0 + 8 + 2 + 2 + 5 + 4 + 9 + 3 + 6 + 8 + 2 + 4 + 6 + 9 + 5 + 6 + 1

Numbers like this, where k = digsum(fib(k)), are rare. And 2222 is almost certainly the last of them. These are the relevant listings at the Online Encyclopedia of Integer Sequences:

0, 1, 5, 10, 31, 35, 62, 72, 175, 180, 216, 251, 252, 360, 494, 504, 540, 946, 1188, 2222 — A020995, Numbers k such that the sum of the digits of Fibonacci(k) is k.
0, 1, 5, 55, 1346269, 9227465, 4052739537881, 498454011879264, 1672445759041379840132227567949787325, 18547707689471986212190138521399707760, 619220451666590135228675387863297874269396512... — A067515, Fibonacci numbers with index = digit sum.

At least, they’re rare in base 10. What about other bases? Well, they’re rare in all other bases except one: base 11. When I looked there, I quickly found more than 450 numbers where digsum(fib(k),b=11) = k. So here’s an interesting little problem: Why is base 11 so productive? Or maybe I should say: Φ is base 11 so productive?

Primal Stream

Sunday, 29th Jan, 2023 by Krilling for Company in Integer Sequences, Mathematics, Prime numbers and tagged consecutive integers, highly composite odd numbers, primes, recreational math, recreational mathematics, recreational maths, sums of consecutive integers | Leave a comment

It’s obvious when you think about: an even number can never be the sum of two consecutive integers. Conversely, an odd number (except 1) is always the sum of two consecutive integers: 3 = 1 + 2; 5 = 2 + 3; 7 = 3 + 4; 9 = 4 + 5; and so on. The sum of three consecutive integers can be either odd or even: 6 = 1 + 2 + 3; 9 = 2 + 3 + 4. The sum of four consecutive integers must always be even: 1 + 2 + 3 + 4 = 10; 2 + 3 + 4 + 5 = 14. And so on.

But notice that 9 is the sum of consecutive integers in two different ways: 9 = 4 + 5 = 2 + 3 + 4. Having spotted that, I decided to look for numbers that were the sums of consecutive integers in the most different ways. These are the first few:

3 = 1 + 2 (number of sums = 1) 9 = 2 + 3 + 4 = 4 + 5 (s = 2) 15 = 1 + 2 + 3 + 4 + 5 = 4 + 5 + 6 = 8 + 7 = (s = 3) 45 (s = 5) 105 (s = 7) 225 (s = 8) 315 (s = 11) 945 (s = 15) 1575 (s = 17) 2835 (s = 19) 3465 (s = 23) 10395 (s = 31)

It was interesting that the number of different consecutive-integer sums for n was most often a prime number. Next I looked for the sequence at the Online Encyclopedia of Integer Sequences and discovered something that I hadn’t suspected:

A053624 Highly composite odd numbers: where d(n) increases to a record.

1, 3, 9, 15, 45, 105, 225, 315, 945, 1575, 2835, 3465, 10395, 17325, 31185, 45045, 121275, 135135, 225225, 405405, 675675, 1576575, 2027025, 2297295, 3828825, 6891885, 11486475, 26801775, 34459425, 43648605, 72747675, 130945815 — A053624 at OEIS

The notes add that the sequence is “Also least number k such that the number of partitions of k into consecutive integers is a record. For example, 45 = 22+23 = 14+15+16 = 7+8+9+10+11 = 5+6+7+8+9+10 = 1+2+3+4+5+6+7+8+9, six such partitions, but all smaller terms have fewer such partitions (15 has four).” When you don’t count the number n itself as a partition of n, you get 3 partitions for 15, i.e. consecutive integers sum to 15 in 3 different ways, so s = 3. I looked at more values for s and found that the stream of primes continued to flow:

3 → s = 1 9 = 3^2 → s = 2 (prime) 15 = 3 * 5 → s = 3 (prime) 45 = 3^2 * 5 → s = 5 (prime) 105 = 3 * 5 * 7 → s = 7 (prime) 225 = 3^2 * 5^2 → s = 8 = 2^3 315 = 3^2 * 5 * 7 → s = 11 (prime) 945 = 3^3 * 5 * 7 → s = 15 = 3 * 5 1575 = 3^2 * 5^2 * 7 → s = 17 (prime) 2835 = 3^4 * 5 * 7 → s = 19 (prime) 3465 = 3^2 * 5 * 7 * 11 → s = 23 (prime) 10395 = 3^3 * 5 * 7 * 11 → s = 31 (prime) 17325 = 3^2 * 5^2 * 7 * 11 → s = 35 = 5 * 7 31185 = 3^4 * 5 * 7 * 11 → s = 39 = 3 * 13 45045 = 3^2 * 5 * 7 * 11 * 13 → s = 47 (prime) 121275 = 3^2 * 5^2 * 7^2 * 11 → s = 53 (prime) 135135 = 3^3 * 5 * 7 * 11 * 13 → s = 63 = 3^2 * 7 225225 = 3^2 * 5^2 * 7 * 11 * 13 → s = 71 (prime) 405405 = 3^4 * 5 * 7 * 11 * 13 → s = 79 (prime) 675675 = 3^3 * 5^2 * 7 * 11 * 13 → s = 95 = 5 * 19 1576575 = 3^2 * 5^2 * 7^2 * 11 * 13 → s = 107 (prime) 2027025 = 3^4 * 5^2 * 7 * 11 * 13 → s = 119 = 7 * 17 2297295 = 3^3 * 5 * 7 * 11 * 13 * 17 → s = 127 (prime) 3828825 = 3^2 * 5^2 * 7 * 11 * 13 * 17 → s = 143 = 11 * 13 6891885 = 3^4 * 5 * 7 * 11 * 13 * 17 → s = 159 = 3 * 53 11486475 = 3^3 * 5^2 * 7 * 11 * 13 * 17 → s = 191 (prime) 26801775 = 3^2 * 5^2 * 7^2 * 11 * 13 * 17 → s = 215 = 5 * 43 34459425 = 3^4 * 5^2 * 7 * 11 * 13 * 17 → s = 239 (prime) 43648605 = 3^3 * 5 * 7 * 11 * 13 * 17 * 19 → s = 255 = 3 * 5 * 17 72747675 = 3^2 * 5^2 * 7 * 11 * 13 * 17 * 19 → s = 287 = 7 * 41 130945815 = 3^4 * 5 * 7 * 11 * 13 * 17 * 19 → s = 319 = 11 * 29

I can’t spot any way of predicting when n will yield a primal s, but I like the way that a simple question took an unexpected turn. When a number sets a record for the number of different ways it can be the sum of consecutive integers, that number will also be a highly composite odd number.

You Sixy Beast

Friday, 6th Jan, 2023 by Krilling for Company in 666, Integer Sequences, Mathematics, Number of the Beast, Palindromic numbers, Repdigits, Triangular Numbers and tagged 222111, 36, 6, 666, Arithmetic, number 6, Number of the Beast, number six, palindromes, palindromic numbers, repdigits, six hundred and sixty six, sixes | 2 Comments

666 is the Number of the Beast. But it’s much more than that. After all, it’s a number, so it has mathematical properties (everything has mathematical properties, but it’s a sine-qua-non of numbers). For example, 666 is a palindromic number, reading the same forwards and backwards. And it’s a repdigit, consisting of a single repeated digit. Now try answering this question: how many pebbles are there in this triangle?

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Counting the pebbles one by one would take a long time, but there’s a short-cut. Each line of the triangle after the first is one pebble longer than the previous line. There are 36 lines and therefore 36 pebbles in the final line. So the full number of pebbles = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36. And there’s an easy formula for that sum: (36^2 + 36) / 2 = (1296 + 36) / 2 = 1332 / 2 = 666.

So 666 is the 36th triangular number:

1 = 1 1+2 = 3 1+2+3 = 6 1+2+3+4 = 10 1+2+3+4+5 = 15 1+2+3+4+5+6 = 21 1+2+3+4+5+6+7 = 28 1+2+3+4+5+6+7+8 = 36 1+2+3+4+5+6+7+8+9 = 45 1+2+3+4+5+6+7+8+9+10 = 55 [...] 1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30+31+32+33+34+35+36 = 666

But what’s tri(666), the 666th triangular number? By the formula above, it equals (666^2 + 666) / 2 = (443556 + 666) / 2 = 444222 / 2 = 222111. But recall something else from above: tri(6) = 1+2+3+4+5+6 = 21. Is it a coincidence that tri(6) = 21 and tri(666) = 222111? No, it isn’t:

tri(6) = 21 = (6^2 + 6) / 2 = (36 + 6) / 2 = 42 / 2 tri(66) = 2211 = (66^2 + 66) / 2 = (4356 + 66) / 2 = 4422 / 2 tri(666) = 222111 = (666^2 + 666) / 2 = (443556 + 666) / 2 = 444222 / 2 tri(6666) = 22221111 tri(66666) = 2222211111 tri(666666) = 222222111111 tri(6666666) = 22222221111111 tri(66666666) = 2222222211111111 tri(666666666) = 222222222111111111 tri(6666666666) = 22222222221111111111 tri(66666666666) = 2222222222211111111111 tri(666666666666) = 222222222222111111111111 tri(6666666666666) = 22222222222221111111111111 tri(66666666666666) = 2222222222222211111111111111 tri(666666666666666) = 222222222222222111111111111111

So we’ve looked at tri(36) = 666 and tri(666) = 222111. Let’s go a step further: tri(222111) = 24666759216. So 666 appears again. And the sixiness carries on here:

tri(36) = 666 tri(3366) = 5666661 tri(333666) = 55666666611 tri(33336666) = 555666666666111 tri(3333366666) = 5555666666666661111 tri(333333666666) = 55555666666666666611111 tri(33333336666666) = 555555666666666666666111111 tri(3333333366666666) = 5555555666666666666666661111111 tri(333333333666666666) = 55555555666666666666666666611111111 tri(33333333336666666666) = 555555555666666666666666666666111111111 tri(3333333333366666666666) = 5555555555666666666666666666666661111111111 tri(333333333333666666666666) = 55555555555666666666666666666666666611111111111 tri(33333333333336666666666666) = 555555555555666666666666666666666666666111111111111 tri(3333333333333366666666666666) = 5555555555555666666666666666666666666666661111111111111 tri(333333333333333666666666666666) = 55555555555555666666666666666666666666666666611111111111111

Agogic Arithmetic

Thursday, 29th Dec, 2022 by Krilling for Company in Base Behavior, Integer Sequences, Mathematics, Triangular Numbers and tagged harmonic numbers, harmonic series, leading digits, number sequences, OEIS, Online Encyclopedia of Integer Sequences | Leave a comment

This is one of my favorite integer sequences:
• 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431, ... — A000217 at OEIS

And it’s easy to work out the rule that generates the sequence. It’s the sequence of triangular numbers, of course, which you get by summing the integers:
1 1 + 2 = 3 3 + 3 = 6 6 + 4 = 10 10 + 5 = 15 15 + 6 = 21 21 + 7 = 28 28 + 8 = 36 36 + 9 = 45 [...]

I like this sequence too, but it isn’t a sequence of integers and it’s much harder to work out the rule that generates it:
• 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, 7381/2520, 83711/27720, 86021/27720, 1145993/360360, 1171733/360360...

But you could say that it’s the inverse of the triangular numbers, because you generate it like this:
1 1 + 1/2 = 3/2 3/2 + 1/3 = 11/6 11/6 + 1/4 = 25/12 25/12 + 1/5 = 137/60 137/60 + 1/6 = 49/20 49/20 + 1/7 = 363/140 363/140 + 1/8 = 761/280 761/280 + 1/9 = 7129/2520 [...]
It’s the harmonic series, which is defined at Wikipedia as “the infinite series formed by summing all positive unit fractions”. I can’t understand its subtleties or make any important discoveries about it, but I thought I could ask (and begin to answer) a question that perhaps no-one else in history had ever asked: When are the leading digits of the k-th harmonic number, hs(k), equal to the digits of k in base 10?
hs(1) = 1 hs(43) = 4.349... hs(714) = 7.1487... hs(715) = 7.1501... hs(9763) = 9.76362... hs(122968) = 12.296899... hs(122969) = 12.296907... hs(1478366) = 14.7836639... hs(17239955) = 17.23995590... hs(196746419) = 19.6746419... hs(2209316467) = 22.0931646788...

Do those numbers have any true mathematical significance? I doubt it. But they were fun to find, even though I wasn’t the first person in history to ask about them:
• 1, 43, 714, 715, 9763, 122968, 122969, 1478366, 17239955, 196746419, 2209316467, 24499118645, 268950072605 — A337904 at OEIS, Numbers k such that the decimal expansion of the k-th harmonic number starts with the digits of k, in the same order.

Root Pursuit

Wednesday, 20th Jul, 2022 by Krilling for Company in √2, Base Behavior, Integer Sequences, Irrational numbers, Mathematics, Powers, Square root of 2, Square Roots and tagged Arithmetic, √10, √30, √6, base 10, base 30, base 6, math, mathematics, maths, powers, powers of 2, powers of 3, powers of 5, recreational math, recreational mathematics, recreational maths, roots, Square Roots | Leave a comment

Roots are hard, powers are easy. For example, the square root of 2, or √2, is the mysterious and never-ending number that is equal to 2 when multiplied by itself:
• √2 = 1·414213562373095048801688724209698078569671875376948073...
It’s hard to calculate √2. But the powers of 2, or 2^p, are the straightforward numbers that you get by multiplying 2 repeatedly by itself. It’s easy to calculate 2^p:
• 2 = 2^1 • 4 = 2^2 • 8 = 2^3 • 16 = 2^4 • 32 = 2^5 • 64 = 2^6 • 128 = 2^7 • 256 = 2^8 • 512 = 2^9 • 1024 = 2^10 • 2048 = 2^11 • 4096 = 2^12 • 8192 = 2^13 • 16384 = 2^14 • 32768 = 2^15 • 65536 = 2^16 • 131072 = 2^17 • 262144 = 2^18 • 524288 = 2^19 • 1048576 = 2^20 [...]
But there is a way to find √2 by finding 2^p, as I discovered after I asked a simple question about 2^p and 3^p. What are the longest runs of matching digits at the beginning of each power?
• 131072 = 2^17 • 129140163 = 3^17 • 1255420347077336152767157884641... = 2^193 • 1214512980685298442335534165687... = 3^193 • 2175541218577478036232553294038... = 2^619 • 2177993962169082260270654106078... = 3^619 • 7524389324549354450012295667238... = 2^2016 • 7524012611682575322123383229826... = 3^2016
There’s no obvious pattern. Then I asked the same question about 2^p and 5^p. And an interesting pattern appeared:
• 32 = 2^5 • 3125 = 5^5 • 316912650057057350374175801344 = 2^98 • 3155443620884047221646914261131... = 5^98 • 3162535207926728411757739792483... = 2^1068 • 3162020133383977882730040274356... = 5^1068 • 3162266908803418110961625404267... = 2^127185 • 3162288411569894029343799063611... = 5^127185
The digits 31622 rang a bell. Isn’t that the start of √10? Yes, it is:
• √10 = 3·1622776601683793319988935444327185337195551393252168268575...
I wrote a fast machine-code program to find even longer runs of matching initial digits. Sure enough, the pattern continued:
• 316227... = 2^2728361 • 316227... = 5^2728361 • 3162277... = 2^15917834 • 3162277... = 5^15917834 • 31622776... = 2^73482154 • 31622776... = 5^73482154 • 3162277660... = 2^961700165 • 3162277660... = 5^961700165
But why are powers of 2 and 5 generating the digits of √10? If you’re good at math, that’s a trivial question about a trivial discovery. Here’s the answer: We use base ten and 10 = 2 * 5, 10^2 = 100 = 2^2 * 5^2 = 4 * 25, 10^3 = 1000 = 2^3 * 5^3 = 8 * 125, and so on. When the initial digits of 2^p and 5^p match, those matching digits must come from the digits of √10. Otherwise the product of 2^p * 5^p would be too large or too small. Here are the records for matching initial digits multiplied by themselves:
• 32 = 2^5 • 3125 = 5^5 • 3^2 = 9

• 316912650057057350374175801344 = 2^98 • 3155443620884047221646914261131... = 5^98 • 31^2 = 961 • 3162535207926728411757739792483... = 2^1068 • 3162020133383977882730040274356... = 5^1068 • 3162^2 = 9998244 • 3162266908803418110961625404267... = 2^127185 • 3162288411569894029343799063611... = 5^127185 • 31622^2 = 999950884 • 316227... = 2^2728361 • 316227... = 5^2728361 • 316227^2 = 99999515529 • 3162277... = 2^15917834 • 3162277... = 5^15917834 • 3162277^2 = 9999995824729 • 31622776... = 2^73482154 • 31622776... = 5^73482154 • 31622776^2 = 999999961946176

• 3162277660... = 2^961700165 • 3162277660... = 5^961700165 • 3162277660^2 = 9999999998935075600
The square of each matching run falls short of 10^p. And so when the digits of 2^p and 5^p stop matching, one power must fall below √10, as it were, and one must rise above:
• 3 162266908803418110961625404267... = 2^127185 • 3·162277660168379331998893544432... = √10 • 3 162288411569894029343799063611... = 5^127185
In this way, 2^p * 5^p = 10^p. And that’s why matching initial digits of 2^p and 5^p generate the digits of √10. The same thing, mutatis mutandis, happens in base 6 with 2^p and 3^p, because 6 = 2 * 3:
• 2.24103122055214532500432040411... = √6 (in base 6)

• 24 = 2^4 • 213 = 3^4 • 225522024 = 2^34 in base 6 = 2^22 in base 10 • 22225525003213 = 3^34 (3^22) • 2241525132535231233233555114533... = 2^1303 (2^327) • 2240133444421105112410441102423... = 3^1303 (3^327) • 2241055222343212030022044325420... = 2^153251 (2^15007) • 2241003215453455515322105001310... = 3^153251 (3^15007) • 2241032233315203525544525150530... = 2^233204 (2^20164) • 2241030204225410320250422435321... = 3^233204 (3^20164) • 2241031334114245140003252435303... = 2^2110415 (2^102539) • 2241031103430053425141014505442... = 3^2110415 (3^102539)
And in base 30, where 30 = 2 * 3 * 5, you can find the digits of √30 in three different ways, because 30 = 2 * 15 = 3 * 10 = 5 * 6:
• 5·E9F2LE6BBPBF0F52B7385PE6E5CLN... = √30 (in base 30)

• 55AA4 = 2^M in base 30 = 2^22 in base 10 • 5NO6CQN69C3Q0E1Q7F = F^M = 15^22 • 5E63NMOAO4JPQD6996F3HPLIMLIRL6F... = 2^K6 (2^606) • 5ECQDMIOCIAIR0DGJ4O4H8EN10AQ2GR... = F^K6 (15^606) • 5E9DTE7BO41HIQDDO0NB1MFNEE4QJRF... = 2^B14 (2^9934) • 5E9G5SL7KBNKFLKSG89J9J9NT17KHHO... = F^B14 (15^9934) [...] • 5R4C9 = 3^E in base 30 = 3^14 in base 10 • 52CE6A3L3A = A^E = 10^14 • 5E6SOQE5II5A8IRCH9HFBGO7835KL8A = 3^3N (3^113) • 5EC1BLQHNJLTGD00SLBEDQ73AH465E3... = A^3N (10^113) • 5E9FI455MQI4KOJM0HSBP3GG6OL9T8P... = 3^EJH (3^13187) • 5E9EH8N8D9TR1AH48MT7OR3MHAGFNFQ... = A^EJH (10^13187) [...] • 5OCNCNRAP = 5^I in base 30 = 5^18 in base 10 • 54NO22GI76 = 6^I (6^18) • 5EG4RAMD1IGGHQ8QS2QR0S0EH09DK16... = 5^1M7 (5^1567) • 5E2PG4Q2G63DOBIJ54E4O035Q9TEJGH... = 6^1M7 (6^1567) • 5E96DB9T6TBIM1FCCK8A8J7IDRCTM71... = 5^F9G (5^13786) • 5E9NM222PN9Q9TEFTJ94261NRBB8FCH... = 6^F9G (6^13786) [...]
So that’s √10, √6 and √30. But I said at the beginning that you can find √2 by finding 2^p. How do you do that? By offsetting the powers, as it were. With 2^p and 5^p, you can find the digits of √10. With 2^(p+1) and 5^p, you can find the digits of √2 and √20, because 2^(p+1) * 5^p = 2 * 2^p * 5^p = 2 * 10^p:
• √2 = 1·414213562373095048801688724209698078569671875376948073... • √20 = 4·472135954999579392818347337462552470881236719223051448...

• 16 = 2^4 • 125 = 5^3 • 140737488355328 = 2^47 • 142108547152020037174224853515625 = 5^46 • 1413... = 2^243 • 1414... = 5^242 • 14141... = 2^6651 • 14142... = 5^6650 • 141421... = 2^35389 • 141420... = 5^35388 • 4472136... = 2^162574 • 4472135... = 5^162573 • 141421359... = 2^3216082 • 141421352... = 5^3216081 • 447213595... = 2^172530387 • 447213595... = 5^172530386 [...]

God Give Me Benf’

Wednesday, 29th Jun, 2022 by Krilling for Company in Base Behavior, Integer Sequences, Mathematics, Powers and tagged Arithmetic, Benford's law, first-digit law, integer spiral, math, mathematics, maths, Newcomb-Benford law, power-spiral, powers, powers of 2, recreational math, recreational mathematics, recreational maths, Ulam Spirals | Leave a comment

In “Wake the Snake”, I looked at the digits of powers of 2 and mentioned a fascinating mathematical phenomenon known as Benford’s law, which governs — in a not-yet-fully-explained way — the leading digits of a wide variety of natural and human statistics, from the lengths of rivers to the votes cast in elections. Benford’s law also governs a lot of mathematical data. It states, for example, that the first digit, d, of a power of 2 in base b (except b = 2, 4, 8, 16…) will occur with the frequency log_b(1 + 1/d). In base 10, therefore, Benford’s law states that the digits 1..9 will occur with the following frequencies at the beginning of 2^p:
1: 30.102999% 2: 17.609125% 3: 12.493873% 4: 09.691001% 5: 07.918124% 6: 06.694678% 7: 05.799194% 8: 05.115252% 9: 04.575749%
Here’s a graph of the actual relative frequencies of 1..9 as the leading digit of 2^p (open images in a new window if they appear distorted):

And here’s a graph for the predicted frequencies of 1..9 as the leading digit of 2^p, as calculated by the log(1+1/d) of Benford’s law:

The two graphs agree very well. But Benford’s law applies to more than one leading digit. Here are actual and predicted graphs for the first two leading digits of 2^p, 10..99:

And actual and predicted graphs for the first three leading digits of 2^p, 100..999:

But you can represent the leading digit of 2^p in another way: using an adaptation of the famous Ulam spiral. Suppose powers of 2 are represented as a spiral of squares that begins like this, with 2^0 in the center, 2^1 to the right of center, 2^2 above 2^1, and so on:
←←←⮲ 432↑ 501↑ 6789

If the digits of 2^p start with 1, fill the square in question; if the digits of 2^p don’t start with 1, leave the square empty. When you do this, you get this interesting pattern (the purple square at the very center represents 2^0):

Ulam-like power-spiral for 2^p where 1 is the leading digit

Here’s a higher-resolution power-spiral for 1 as the leading digit:

Power-spiral for 2^p, leading-digit = 1 (higher resolution)

And here, at higher resolution still, are power-spirals for all the possible leading digits of 2^p, 1..9 (some spirals look very similar, so you have to compare those ones carefully):

Power-spiral for 2^p, leading-digit = 1 (very high resolution)

Power-spiral for 2^p, leading-digit = 2

Power-spiral for 2^p, ld = 3

Power-spiral for 2^p, ld = 4

Power-spiral for 2^p, ld = 5

Power-spiral for 2^p, ld = 6

Power-spiral for 2^p, ld = 7

Power-spiral for 2^p, ld = 8

Power-spiral for 2^p, ld = 9

Power-spiral for 2^p, ld = 1..9 (animated)

Now try the power-spiral of 2^p, ld = 1, in some other bases:

Power-spiral for 2^p, leading-digit = 1, base = 9

Power-spiral for 2^p, ld = 1, b = 15

You can also try power-spirals for other n^p. Here’s 3^p:

Power-spiral for 3^p, ld = 1, b = 10

Power-spiral for 3^p, ld = 2, b = 10

Power-spiral for 3^p, ld = 1, b = 4

Power-spiral for 3^p, ld = 1, b = 7

Power-spiral for 3^p, ld = 1, b = 18

Elsewhere Other-Accessible…

• Wake the Snake — an earlier look at the digits of 2^p

Overlord In Terms of Core Issues Around Maximal Engagement with Key Notions of the Über-Feral

მეფე (2) • მსოფლიოს (3) • მათემატიკა (5) •

Category Archives: Integer Sequences