Category Archives: Base Behavior

In 10 Words: Im-Precise

by in Base Behavior, Binary numbers, π, Mathematics, Pi and tagged , , , | Leave a comment

I was looking at the best rational approximations for π when I was puzzled for a moment or two by the way the precision of digits didn’t always improve:

22/7 →
3.1428571... = 22/7 (precision = 3 digits)
3.1415926... = π
333/106 →
3.141509433... = 333/106 (pr=5)
3.141592653... = π
355/113 →
3.14159292035... = 355/113 (pr=7)
3.14159265358... = π
103993/33102 →
3.14159265301190... (pr=10)
3.14159265358979... = π
104348/33215 →
3.14159265392142... (pr=10)
3.14159265358979... = π
208341/66317 →
3.14159265346743... (pr=10)
3.14159265358979... = π
312689/99532 →
3.14159265361893... (pr=10)
3.14159265358979... = π
833719/265381 →
3.1415926535810777... (pr=12)
3.1415926535897932... = π
1146408/364913 →
3.141592653591403... (pr=11)
3.141592653589793... = π
4272943/1360120 →
3.14159265358938917... (pr=13)
3.14159265358979323... = π
5419351/1725033 →
3.14159265358981538... (pr=13)
3.14159265358979323... = π
80143857/25510582 →
3.1415926535897926593... (pr=15)
3.1415926535897932384... = π
165707065/52746197 →
3.14159265358979340254... (pr=16)
3.14159265358979323846... = π
245850922/78256779 →
3.14159265358979316028... (pr=16)
3.14159265358979323846... = π
411557987/131002976 →
3.141592653589793257826... (pr=17)
3.141592653589793238462... = π
1068966896/340262731 →
3.1415926535897932353925... (pr=18)
3.1415926535897932384626... = π
2549491779/811528438 →
3.1415926535897932390140... (pr=18)
3.1415926535897932384626... = π
6167950454/1963319607 →
3.14159265358979323838637... (pr=19)
3.14159265358979323846264... = π
14885392687/4738167652 →
3.141592653589793238493875... (pr=20)
3.141592653589793238462643... = π

But it was my precision that was wrong, of course. I wasn’t thinking about digits precisely enough. One approximation can be closer to π with fewer precise digits than another (e.g. 3.14201… is closer to π than 3.14101…). The same applies in binary, but there the precision tends to increase much more obviously:

22/7 →
3.1428571... = 22/7 in base 10 (pr=3)
3.1415926... = π in base 10
11.0010010010010... = 22/7 in base 2 (pr=9)
11.0010010000111... = π in base 2
333/106 →
3.141509433... = 333/106 in b10 (pr=5)
3.141592653... = π in b10
11.001001000011100111... = 333/106 in b2 (pr=14)
11.001001000011111101... = π in b2
355/113 →
3.14159292035... (pr=7)
3.14159265358... = π
11.00100100001111110110111100... = 355/113 in b2 (pr=22)
11.00100100001111110110101010... = π in b2
103993/33102 →
3.14159265301190... (pr=10)
3.14159265358979... = π
11.001001000011111101101010100001100... (pr=29)
11.001001000011111101101010100010001... = π
104348/33215 →
3.14159265392142... (pr=10)
3.14159265358979... = π
11.001001000011111101101010100010011111... (pr=32)
11.001001000011111101101010100010001000... = π
208341/66317 →
3.14159265346743... (pr=10)
3.14159265358979... = π
11.001001000011111101101010100001111... (pr=29)
11.001001000011111101101010100010001... = π
312689/99532 →
3.14159265361893... (pr=10)
3.14159265358979... = π
11.001001000011111101101010100010001010010... (pr=35)
11.001001000011111101101010100010001000010... = π
833719/265381 →
3.1415926535810777... (pr=12)
3.1415926535897932... = π
11.0010010000111111011010101000100001111... (pr=33)
11.0010010000111111011010101000100010000... = π
1146408/364913 →
3.141592653591403... (pr=11)
3.141592653589793... = π
11.0010010000111111011010101000100010000111011... (pr=39)
11.0010010000111111011010101000100010000101101... = π
4272943/1360120 →
3.14159265358938917... (pr=13)
3.14159265358979323... = π
11.001001000011111101101010100010001000010100110... (pr=41)
11.001001000011111101101010100010001000010110100... = π
5419351/1725033 →
3.14159265358981538... (pr=13)
3.14159265358979323... = π
11.0010010000111111011010101000100010000101101010010... (pr=45)
11.0010010000111111011010101000100010000101101000110... = π
80143857/25510582 →
3.1415926535897926593... (pr=15)
3.1415926535897932384... = π
11.0010010000111111011010101000100010000101101000101101... (pr=48)
11.0010010000111111011010101000100010000101101000110000... = π
165707065/52746197 →
3.14159265358979340254... (pr=16)
3.14159265358979323846... = π
11.00100100001111110110101010001000100001011010001100010100... (pr=52)
11.00100100001111110110101010001000100001011010001100001000... = π
245850922/78256779 →
3.14159265358979316028... (pr=16)
3.14159265358979323846... = π
11.001001000011111101101010100010001000010110100011000000110... (pr=53)
11.001001000011111101101010100010001000010110100011000010001... = π
411557987/131002976 →
3.141592653589793257826... (pr=17)
3.141592653589793238462... = π
11.00100100001111110110101010001000100001011010001100001010001... (pr=55)
11.00100100001111110110101010001000100001011010001100001000110... = π
1068966896/340262731 →
3.1415926535897932353925... (pr=18)
3.1415926535897932384626... = π
11.00100100001111110110101010001000100001011010001100001000100110... (pr=58)
11.00100100001111110110101010001000100001011010001100001000110100... = π
2549491779/811528438 →
3.1415926535897932390140... (pr=18)
3.1415926535897932384626... = π
11.00100100001111110110101010001000100001011010001100001000110111010... (pr=61)
11.00100100001111110110101010001000100001011010001100001000110100110... = π
6167950454/1963319607 →
3.14159265358979323838637... (pr=19)
3.14159265358979323846264... = π
11.0010010000111111011010101000100010000101101000110000100011010001101... (pr=63)
11.0010010000111111011010101000100010000101101000110000100011010011000... = π
14885392687/4738167652 →
3.141592653589793238493875... (pr=20)
3.141592653589793238462643... = π
11.001001000011111101101010100010001000010110100011000010001101001110100... (pr=65)
11.001001000011111101101010100010001000010110100011000010001101001100010... = π

Post-Performative Post-Scriptum…

The title of this terato-toxic post is a maximal mash-up (wow) of two well-known toxico-teratic tropes:

• “There are 10 kinds of people in the world. Those who understand binary and those who don’t.”
• Sam Goldwyn’s malapropism: “In two words: im-possible!”

Power Flip

by in Arithmetic, Base Behavior, Integer Sequences, Mathematics, Powers, Prime numbers and tagged , , , , , , , , , , , , , , | Leave a comment

12 is an interesting number in a lot of ways. Here’s one way I haven’t seen mentioned before:

12 = 3^1 * 2^2


The digits of 12 represent the powers of the primes in its factorization, if primes are represented from right-to-left, like this: …7, 5, 3, 2. But I couldn’t find any more numbers like that in base 10, so I tried a power flip, from right-left to left-right. If the digits from left-to-right represent the primes in the order 2, 3, 5, 7…, then this number is has prime-power digits too:

81312000 = 2^8 * 3^1 * 5^3 * 7^1 * 11^2 * 13^0 * 17^0 * 19^0


Or, more simply, given that n^0 = 1:

81312000 = 2^8 * 3^1 * 5^3 * 7^1 * 11^2


I haven’t found any more left-to-right prime-power digital numbers in base 10, but there are more in other bases. Base 5 yields at least three (I’ve ignored numbers with just two digits in a particular base):

110 in b2 = 2^1 * 3^1 (n=6)
130 in b6 = 2^1 * 3^3 (n=54)
1010 in b2 = 2^1 * 3^0 * 5^1 (n=10)
101 in b3 = 2^1 * 3^0 * 5^1 (n=10)
202 in b7 = 2^2 * 3^0 * 5^2 (n=100)
3020 in b4 = 2^3 * 3^0 * 5^2 (n=200)
330 in b8 = 2^3 * 3^3 (n=216)
13310 in b14 = 2^1 * 3^3 * 5^3 * 7^1 (n=47250)
3032000 in b5 = 2^3 * 3^0 * 5^3 * 7^2 (n=49000)
21302000 in b5 = 2^2 * 3^1 * 5^3 * 7^0 * 11^2 (n=181500)
7810000 in b9 = 2^7 * 3^8 * 5^1 (n=4199040)
81312000 in b10 = 2^8 * 3^1 * 5^3 * 7^1 * 11^2


Post-Performative Post-Scriptum

When I searched for 81312000 at the Online Encyclopedia of Integer Sequences, I discovered that these are Meertens numbers, defined at A246532 as the “base n Godel encoding of x [namely,] 2^d(1) * 3^d(2) * … * prime(k)^d(k), where d(1)d(2)…d(k) is the base n representation of x.”

Period Panes

by in Arithmetic, Base Behavior, Circles, Geometry, Mathematics, Reciprocals and tagged , , , , , , , , , , , , , , , , , , , , , | Leave a comment

In his Penguin Dictionary of Curious and Interesting Numbers (1986), David Wells says that 142857 is “beloved of all recreational mathematicians”. He then says it’s the decimal period of the reciprocal of the fourth prime: “1/7 = 0·142857142857142…” And the reciprocal has maximum period. There are 6 = 7-1 digits before repetition begins, unlike the earlier prime reciprocals:


1/2 = 0·5
1/3 = 0·333...
1/5 = 0·2
1/7 = 0·142857 142857 142...

In other words, all possible remainders appear when you calculate the decimals of 1/7:


1*10 / 7 = 1 remainder 3 → 0·1
3*10 / 7 = 4 remainder 2 → 0·14
2*10 / 7 = 2 remainder 6 → 0·142
6*10 / 7 = 8 remainder 4 → 0·1428
4*10 / 7 = 5 remainder 5 → 0·14285
5*10 / 7 = 7 remainder 1 → 0·142857
1*10 / 7 = 1 remainder 3 → 0·142857 1
3*10 / 7 = 4 remainder 2 → 0·142857 14
2*10 / 7 = 2 remainder 6 → 0·142857 142...

That happens again with 1/17 and 1/19, but Wells says that “surprisingly, there is no known method of predicting which primes have maximum period.” It’s a simple question that involves some deep mathematics. Looking at prime reciprocals is like peering through a small window into a big room. Some things are easy to see, some are difficult and some are presently impossible.

In his discussion of 142857, Wells mentions one way of peering through a period pane: “The sequence of digits also makes a striking pattern when the digits are arranged around a circle.” Here is the pattern, with ten points around the circle representing the digits 0 to 9:

The digits of 1/7 = 0·142857142…

But I prefer, for further peers through the period-panes, to create the period-panes using remainders rather than digits. That is, the number of points around the circle is determined by the prime itself rather than the base in which the reciprocal is calculated:

The remainders of 1/7 = 1, 3, 2, 6, 4, 5…

Period-panes can look like butterflies or bats or bivalves or spiders or crabs or even angels. Try the remainders of 1/13. This prime reciprocal doesn’t have maximum period: 1/13 = 0·076923 076923 076923… So there are only six remainders, creating this pattern:

remainders(1/13) = 1, 10, 9, 12, 3, 4

The multiple 2/13 has different remainders and creates a different pattern:

remainders(2/13) = 2, 7, 5, 11, 6, 8

But 1/17, 1/19 and 1/23 all have maximum period and yield these period-panes:

remainders(1/17) = 1, 10, 15, 14, 4, 6, 9, 5, 16, 7, 2, 3, 13, 11, 8, 12

remainders(1/19) = 1, 10, 5, 12, 6, 3, 11, 15, 17, 18, 9, 14, 7, 13, 16, 8, 4, 2

remainders(1/23) = 1, 10, 8, 11, 18, 19, 6, 14, 2, 20, 16, 22, 13, 15, 12, 5, 4, 17, 9, 21, 3, 7

It gets mixed again with the prime 73, which doesn’t have maximum period and yields a plethora of period-panes (some patterns repeat with different n * 1/73, so I haven’t included them):

remainders(1/73)

remainders(2/73)

remainders(3/73)

remainders(4/73)

remainders(5/73)

remainders(6/73)

remainders(9/73)

remainders(11/73) (identical to pattern of 5/73)

remainders(12/73)

remainders(18/73)

101 yields a plethora of period-panes, but they’re variations on a simple theme. They look like flapping wings in this animated gif:

remainders of n/101 (animated)

The remainders of 137 yield more complex period-panes:

remainders of n/137 (animated)

And what about different bases? Here are period-panes for the remainders of 1/17 in bases 2 to 16:

remainders(1/17) in base 2

remainders(1/17) in b3

remainders(1/17) in b4

remainders(1/17) in b5

remainders(1/17) in b6

remainders(1/17) in b7

remainders(1/17) in b8

remainders(1/17) in b9

remainders(1/17) in b10

remainders(1/17) in b11

remainders(1/17) in b12

remainders(1/17) in b13

remainders(1/17) in b14

remainders(1/17) in b15

remainders(1/17) in b16

remainders(1/17) in bases 2 to 16 (animated)

But the period-panes so far have given a false impression. They’ve all been symmetrical. That isn’t the case with all the period-panes of n/19:

remainders(1/19) in b2

remainders(1/19) in b3

remainders(1/19) in b4 = 1, 4, 16, 7, 9, 17, 11, 6, 5 (asymmetrical)

remainders(1/19) in b5 = 1, 5, 6, 11, 17, 9, 7, 16, 4 (identical pattern to that of b4)

remainders(1/19) in b6

remainders(1/19) in b7

remainders(1/19) in b8

remainders(1/19) in b9

remainders(1/19) in b10 (identical pattern to that of b2)

remainders(1/19) in b11

remainders(1/19) in b12

remainders(1/19) in b13

remainders(1/19) in b14

remainders(1/19) in b15

remainders(1/19) in b16

remainders(1/19) in b17

remainders(1/19) in b18

remainders(1/19) in bases 2 to 18 (animated)

Here are a few more period-panes in different bases:

remainders(1/11) in b2

remainders(1/11) in b7

remainders(1/13) in b6

remainders(1/43) in b6

remainders in b2 for reciprocals of 29, 37, 53, 59, 61, 67, 83, 101, 107, 131, 139, 149 (animated)

And finally, to performativize the pun of “period pane”, here are some period-panes for 1/29, whose maximum period will be 28 (NASA says that the “Moon takes about one month to orbit Earth … 27.3 days to complete a revolution, but 29.5 days to change from New Moon to New Moon”):

remainders(1/29) in b4

remainders(1/29) in b5

remainders(1/29) in b8

remainders(1/29) in b9

remainders(1/29) in b11

remainders(1/29) in b13

remainders(1/29) in b14

remainders(1/29) in various bases (animated)

Pyramidic Palindromes

by in Base Behavior, Integer Sequences, π, Mathematics, Palindromic numbers, Pi, Powers, Squares and tagged , , , , , , , , , , , | Leave a comment

As I’ve said before on Overlord of the Über-Feral: squares are boring. As I’ve shown before on Overlord of the Über-Feral: squares are not so boring after all.

Take A000330 at the Online Encyclopedia of Integer Sequences:

1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370… — A000330 at OEIS

The sequence shows the square pyramidal numbers, formed by summing the squares of integers:

• 1 = 1^2
• 5 = 1^2 + 2^2 = 1 + 4
• 14 = 1^2 + 2^2 + 3^2 = 1 + 4 + 9
• 30 = 1^2 + 2^2 + 3^2 + 4^2 = 1 + 4 + 9 + 16

[…]

You can see the pyramidality of the square pyramidals when you pile up oranges or cannonballs:

Square pyramid of 91 cannonballs at Rye Castle, East Sussex (Wikipedia)

I looked for palindromes in the square pyramidals. These are the only ones I could find:

1 (k=1)
5 (k=2)
55 (k=5)
1992991 (k=181)

The only ones in base 10, that is. When I looked in base 9 = 3^2, I got a burst of pyramidic palindromes like this:

1 (k=1)
5 (k=2)
33 (k=4) = 30 in base 10 (k=4)
111 (k=6) = 91 in b10 (k=6)
122221 (k=66) = 73810 in b10 (k=60)
123333321 (k=666) = 54406261 in b10 (k=546)
123444444321 (k=6,666) = 39710600020 in b10 (k=4920)
123455555554321 (k=66,666) = 28952950120831 in b10 (k=44286)
123456666666654321 (k=666,666) = 21107018371978630 in b10 (k=398580)
123456777777777654321 (k=6,666,666) = 15387042129569911801 in b10 (k=3587226)
123456788888888887654321 (k=66,666,666) = 11217155797104231969640 in b10 (k=32285040)

The palindromic pattern from 6[…]6 ends with 66,666,666, because 8 is the highest digit in base 9. When you look at the 666,666,666th square pyramidal in base 9, you’ll find it’s not a perfect palindrome:

123456801111111111087654321 (k=666,666,666) = 8177306744945450299267171 in b10 (k=290565366)

But the pattern of pyramidic palindromes is good while it lasts. I can’t find any other base yielding a pattern like that. And base 9 yields another burst of pyramidic palindromes in a related sequence, A000537 at the OEIS:

1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, 4356, 6084, 8281, 11025, 14400, 18496, 23409, 29241, 36100, 44100, 53361, 64009, 76176, 90000, 105625, 123201, 142884, 164836, 189225, 216225, 246016, 278784, 314721, 354025, 396900, 443556, 494209, 549081… — A000537 at OEIS

The sequence is what you might call the cubic pyramidal numbers, that is, the sum of the cubes of integers:

• 1 = 1^2
• 9 = 1^2 + 2^3 = 1 + 8
• 36 = 1^3 + 2^3 + 3^3 = 1 + 8 + 27
• 100 = 1^3 + 2^3 + 3^3 + 4^3 = 1 + 8 + 27 + 64

[…]

I looked for palindromes there in base 9:

1 (k=1) = 1 (k=1)
121 (k=4) = 100 in base 10 (k=4)
12321 (k=14) = 8281 (k=13)
1234321 (k=44) = 672400 (k=40)
123454321 (k=144) = 54479161 (k=121)
12345654321 (k=444) = 4412944900 (k=364)
1234567654321 (k=1444) = 357449732641 (k=1093)
123456787654321 (k=4444) = 28953439105600 (k=3280)
102012022050220210201 (k=137227) = 12460125198224404009 (k=84022)

But while palindromes are fun, they’re not usually mathematically significant. However, this result using the square pyramidals is certainly significant:

Previously Pre-Posted…

More posts about how squares aren’t so boring after all:

• Curvous Energy
• Back to Drac #1
• Back to Drac #2
• Square’s Flair

Nuts for Numbers

by in Arithmetic, Base Behavior, Integer Sequences, Mathematics and tagged , , , , , , , , , , , , , , , , , , , | 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)

Bi-Bell Basics

by in Arithmetic, Base Behavior, Binary numbers, Blancmange Curves, Fibonacci Sequence, Fractals, Geometry, Mathematics, Powers and tagged , , , , , , , , , , , , , , , , , | Leave a comment

Here’s what you might call a Sisyphean sequence. It struggles upward, then slips back, over and over again:

1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 1, 2, 2, 3, 2, 3, 3, 4, 2...


The struggle goes on for ever. Every time it reaches a new maximum, it will fall back to 1 at the next step. And in fact 1, 2, 3 and all other integers occur infinitely often in the sequence, because it represents the digit-sums of binary numbers:

1 ← 1
1 = 1+0 ← 10 in binary = 2 in base ten
2 = 1+1 ← 11 = 3
1 = 1+0+0 ← 100 = 4
2 = 1+0+1 ← 101 = 5
2 = 1+1+0 ← 110 = 6
3 = 1+1+1 ← 111 = 7
1 = 1+0+0+0 ← 1000 = 8
2 = 1+0+0+1 ← 1001 = 9
2 = 1+0+1+0 ← 1010 = 10
3 = 1+0+1+1 ← 1011 = 11
2 = 1+1+0+0 ← 1100 = 12
3 = 1+1+0+1 ← 1101 = 13
3 = 1+1+1+0 ← 1110 = 14
4 = 1+1+1+1 ← 1111 = 15
1 = 1+0+0+0+0 ← 10000 = 16
2 = 1+0+0+0+1 ← 10001 = 17
2 = 1+0+0+1+0 ← 10010 = 18
3 = 1+0+0+1+1 ← 10011 = 19
2 = 1+0+1+0+0 ← 10100 = 20


Now here’s a related sequence in which all integers do not occur infinitely often:

1, 2, 3, 3, 4, 5, 6, 4, 5, 6, 7, 7, 8, 9, 10, 5, 6, 7, 8, 8, 9, 10, 11, 9, 10, 11, 12, 12, 13, 14, 15, 6, 7, 8, 9, 9, 10, 11, 12, 10, 11, 12, 13, 13, 14, 15, 16, 11, 12, 13, 14, 14, 15, 16, 17, 15, 16, 17, 18, 18, 19, 20, 21, 7, 8, 9, 10, 10, 11, 12, 13, 11, 12, 13, 14, 14, 15, 16, 17, 12, 13, 14, 15, 15, 16, 17, 18, 16, 17, 18, 19, 19, 20, 21, 22, 13, 14, 15, 16, 16, 17, 18, 19, 17, 18, 19, 20, 20, 21, 22, 23, 18, 19, 20, 21, 21, 22, 23, 24, 22, 23, 24, 25, 25, 26, 27, 28, 8, 9, 10, 11, 11, 12, 13, 14, 12, 13, 14, 15, 15...


The sequence represents the sum of the values of occupied columns in the binary numbers, reading from right to left:

10 in binary = 2 in base ten
21 (column values from right to left)
2*1 + 1*0 = 2

11 = 3
21
2*1 + 1*1 = 3

100 = 4
321 (column values from right to left)
3*1 + 2*0 + 1*0 = 3

101 = 5
321
3*1 + 2*0 + 1*1 = 4

110 = 6
321
3*1 + 2*1 + 1*0 = 5

111 = 7
321
3*1 + 2*1 + 1*1 = 6

1000 = 8
4321
4*1 + 3*0 + 2*0 + 1*0 = 4

1001 = 9
4321
4*1 + 3*0 + 2*0 + 1*1 = 5

1010 = 10
4321
4*1 + 3*0 + 2*1 + 1*0 = 6

1011 = 11
4321
4*1 + 3*0 + 2*1 + 1*1 = 7

1100 = 12
4321
4*1 + 3*1 + 2*0 + 1*0 = 7

1101 = 13
4321
4*1 + 3*1 + 2*0 + 1*1 = 8

1110 = 14
4321
4*1 + 3*1 + 2*1 + 1*0 = 9

1111 = 15
4321
4*1 + 3*1 + 2*1 + 1*1 = 10

10000 = 16
54321
5*1 + 4*0 + 3*0 + 2*0 + 1*0 = 5


In that sequence, although no number occurs infinitely often, some numbers occur more often than others. If you represent the count of sums up to a certain digit-length as a graph, you get a famous shape:

Bell curve formed by the count of column-sums in base 2

Bi-bell curves for 1 to 16 binary digits (animated)

In “Pi in the Bi”, I looked at that way of forming the bell curve and called it the bi-bell curve. Now I want to go further. Suppose that you assign varying values to the columns and try other bases. For example, what happens if you assign the values 2^p + 1 to the columns, reading from right to left, then use base 3 to generate the sums? These are the values of 2^p + 1:

2, 3, 5, 9, 17, 33, 65, 129, 257, 513, 1025...


And here’s an example of how you generate a column-sum in base 3:

2102 in base 3 = 65 in base ten
9532 (column values from right to left)
2*9 + 1*5 + 0*3 + 2*2 = 27


The graphs for these column-sums using base 3 look like this as the digit-length rises. They’re no longer bell-curves (and please note that widths and heights have been normalized so that all graphs fit the same space):

Graph for the count of column-sums in base 3 using 2^p + 1 (digit-length <= 7)

(width and height are normalized)

Graph for base 3 and 2^p + 1 (dl<=8)

Graph for base 3 and 2^p + 1 (dl<=9)

Graph for base 3 and 2^p + 1 (dl<=10)

Graph for base 3 and 2^p + 1 (dl<=11)

Graph for base 3 and 2^p + 1 (dl<=12)

Graph for base 3 and 2^p + 1 (animated)

Now try base 3 and column-values of 2^p + 2 = 3, 4, 6, 10, 18, 34, 66, 130, 258, 514, 1026…

Graph for base 3 and 2^p + 2 (dl<=7)

Graph for base 3 and 2^p + 2 (dl<=8)

Graph for base 3 and 2^p + 2 (dl<=9)

Graph for base 3 and 2^p + 2 (dl<=10)

Graph for base 3 and 2^p + 2 (animated)

Now try base 5 and 2^p + 1 for the columns. The original bell curve has become like a fractal called the blancmange curve:

Graph for base 5 and 2^p + 1 (dl<=7)

Graph for base 5 and 2^p + 1 (dl<=8)

Graph for base 5 and 2^p + 1 (dl<=9)

Graph for base 5 and 2^p + 1 (dl<=10)

Graph for base 5 and 2^p + 1 (animated)

And finally, return to base 2 and try the Fibonacci numbers for the columns:

Graph for base 2 and Fibonacci numbers = 1,1,2,3,5… (dl<=7)

Graph for base 2 and Fibonacci numbers (dl<=9)

Graph for base 2 and Fibonacci numbers (dl<=11)

Graph for base 2 and Fibonacci numbers (dl<=13)

Graph for base 2 and Fibonacci numbers (dl<=15)

Graph for base 2 and Fibonacci numbers (animated)

Previously Pre-Posted…

• Pi in the Bi — bell curves generated by binary digits

A Walk on the Wide Side

by in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Irrational numbers, φ, Mathematics, Phi and tagged , , , , , , , , , , , , , , , | 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…

by in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, Mathematics, Phi and tagged , , , , , , , , , , , | 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?

Agogic Arithmetic

by in Base Behavior, Integer Sequences, Mathematics, Triangular Numbers and tagged , , , , , | 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.