Category Archives: Integer Sequences

Friday is Φiday

by in Arithmetic, Fibonacci Sequence, Integer Sequences, φ, Mathematics, Phi and tagged , , , , , | Leave a comment

The 11th, 12th and 23rd day of a month can be called a φ-day (pronounced fy-day). Why so? Because those numbers are consecutive entries in the famous Fibonacci sequence, which offers better and better approximations to a mathematical constant called φ = (√5 + 1) / 2 = 1.6180339887498948…:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, …

Each number after the second is the sum of the preceding two (so 11, 12, 23… could be the start of a similar sequence). When you divide fib(i) by fib(i-1), you get these approximations to φ:

2 = 2/1
1.5 = 3/2
1.6 = 5/3
1.6 = 8/5
1.625 = 13/8
1.6153846… = 21/13
1.619047619… = 34/21
1.6176470588235294117647… = 55/34
1.618… = 89/55
1.617977528… = 144/89
1.61805… = 233/144
1.618025751… = 377/233
1.618037135… = 610/377
1.618032786… = 987/610
1.618034447… = 1597/987
1.618033813… = 2584/1597
1.618034055… = 4181/2584
1.618033963… = 6765/4181
1.618033998… = 10946/6765
1.618033985… = 17711/10946

Today is the 23rd and not just a φ-day but a Friday (or φriday). So here’s one of the interesting results I’ve recently found while playing with the Fibonacci sequence. As any recreational mathematician kno, you can also find the Fibonacci sequence — and φ — with this little algorithm:

f = 0
LOOP
f = 1 / (f + 1)
print(f)
goto LOOP

The algorithm returns these values:

1, 1/2, 2/3, 3/5, 5/8, 8/13, 13/21, 21/34, 34/55, 55/89, 89/144, 144/233, 233/377, 377/610, 610/987, 987/1597, 1597/2584, 2584/4181, 4181/6765, 6765/10946, 10946/17711, …

I was playing with that algorithm and got an unexpected result with a simple adaptation of it:

f = 0
LOOP
f = 1 / (3 – f)
print(f)
goto LOOP

The values of f generated by this adapted algorithm are:

1/3, 3/8, 8/21, 21/55, 55/144, 144/377, 377/987, 987/2584, 2584/6765, 6765/17711, 17711/46368, 46368/121393, 121393/317811, 317811/832040, 832040/2178309, 2178309/5702887, 5702887/14930352, 14930352/39088169, 39088169/102334155, 102334155/267914296, …

The numerator and denominator in each fraction are next-but-one Fibonacci numbers, beautifully generated at each step:

3 – 0 = 3 → 1/3
3 – 1/3 = (3*3)/3 – 1/3 = 9/3 – 1/3 = (9-1) / 3 = 8 / 3 → 1/(8/3) = 3/8
3 – 3/8 = (3*8)/3 – 3/8 = 24/8 – 3/8 = (24-3) / 8 = 21/8 → 1/(21/8) = 8/21
3 – 8/21 = (3*21)/21 – 8/21 = 63/21 – 8/21 = (63-8)/21 = 55/21 → 1/(55/21) = 21/55
3 – 21/55 = (3*55)/55 – 21/55 = 165/55 – 21/55 = (165-21)/55 = 144/55 → 1/(144/55) = 55/144
3 – 55/144 = (3*144)/144 – 55/144 = (432-55)/144 = 377/144 → 1/(377/144) = 144/377
3 – 144/377 = (3*377)/377 – 144/377 = (1131-144)/377 = 987/377 → 1/(987/377) = 377/987
[…]

If you reverse numerator and denominator, the limit of the fraction is φ^2 = 2.6180339887498948… = φ+1:

3 = 3/1
2.6 = 8/3
2.625 = 21/8
2.6190476190476190476190476… = 55/21
2.6181818181818181818181818… = 144/55
2.6180555555555555555555555… = 377/144
2.6180371352785145888594164… = 987/377
2.6180344478216818642350557… = 2584/987
2.6180340557275541795665634… = 6765/2584
2.6180339985218033998521803… = 17711/6765
2.6180339901755970865563773… = 46368/17711
2.6180339889579020013802622… = 121393/46368
2.6180339887802426828565073… = 317811/121393
2.6180339887543225376088304… = 832040/317811
2.6180339887505408393827219… = 2178309/832040
2.6180339887499890970472967… = 5702887/2178309
2.6180339887499085989254214… = 14930352/5702887
2.6180339887498968544077192… = 39088169/14930352
2.6180339887498951409056791… = 102334155/39088169
2.6180339887498948909091006… = 267914296/102334155

The Call of CFulhu

by in Arithmetic, √2, Continued Fractions, Fibonacci Sequence, H.P. Lovecraft, Integer Sequences, φ, Mathematics, Phi, Square root of 2 and tagged , , , , , | Leave a comment

“The most merciful thing in the world, I think, is the inability of the human mind to correlate all its contents.” So said HPL in “The Call of Cthulhu” (1926). But I’d still like to correlate the contents of mine a bit better. For example, I knew that φ, the golden ratio, is the most irrational of all numbers, in that it is the slowest to be approximated with rational fractions. And I also knew that continued fractions, or CFs, were a way of representing both rationals and irrationals as a string of numbers, like this:

contfrac(10/7) = [1; 2, 3]
10/7 = 1 + 1/(2 + 1/3)
10/7 = 1.428571428571…

contfrac(3/5) = [0; 1, 1, 2]
4/5 = 0 + 1/(1 + 1/(1 + 1/2))
4/5 = 0.8

contfrac(11/8) = [1; 2, 1, 2]
11/8 = 1 + 1/(2 + 1/(1 + 1/2))
11/8 = 1.375

contfrac(4/7) = [0; 1, 1, 3]
4/7 = 0 + 1/(1 + 1/(1 + 1/3))
4/7 = 0.57142857142…

contfrac(17/19) = [0; 1, 8, 2]
17/19 = 0 + 1/(1 + 1/(8 + 1/2))
17/19 = 0.8947368421052…

contfrac(8/25) = [0; 3, 8]
8/25 = 0 + 1/(3 + 1/8)
8/25 = 0.32

contfrac(√2) = [1; 2, 2, 2, 2, 2, 2, 2…] = [1; 2]

√2 = 1 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/(2 + 1/2 + …))))))

√2 = 1.41421356237309504…

contfrac(φ) = [1; 1, 1, 1, 1, 1, 1, 1, 1…]

φ = 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/1 + …)))))))

φ = 1.6180339887498948…

But I didn’t correlate those two contents of my mind: the maximal irrationality of φ and the way continued fractions work.

That’s why I was surprised when I was looking at the continued fractions of 2..(n-1) / n for 3,4,5,6,7… That is, I was looking at the continued fractions of 2/3, 3/4, 2/5, 3/5, 4/5, 5/6, 2/7, 3/7… (skipping fractions like 2/4, 2/6, 3/6 etc, because they’re reducible: 2/4 = ½, 2/6 = 1/3, 3/6 = ½ etc). I wondered which fractions set successive records for the length of their continued fractions as one worked through ½, 2/3, 3/4, 2/5, 3/5, 4/5, 5/6, 2/7, 3/7… And because I hadn’t correlated the contents of my mind, I was surprised at the result. I shouldn’t have been, of course:

contfrac(1/2) = [0; 2] (cfl=1)
1/2 = 0 + 1/2
1/2 = 0.5

contfrac(2/3) = [0; 1, 2] (cfl=2)
2/3 = 0 + 1/(1 + 1/2)
2/3 = 0.666666666…

contfrac(3/5) = [0; 1, 1, 2] (cfl=3)
3/5 = 0 + 1/(1 + 1/(1 + 1/2))
3/5 = 0.6

contfrac(5/8) = [0; 1, 1, 1, 2] (cfl=4)
5/8 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/2)))
5/8 = 0.625

contfrac(8/13) = [0; 1, 1, 1, 1, 2] (cfl=5)
8/13 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))
8/13 = 0.615384615…

contfrac(13/21) = [0; 1, 1, 1, 1, 1, 2] (cfl=6)
13/21 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2)))))
13/21 = 0.619047619…

contfrac(21/34) = [0; 1, 1, 1, 1, 1, 1, 2] (cfl=7)
21/34 = 0 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))))
21/34 = 0.617647059…

contfrac(34/55) = [0; 1, 1, 1, 1, 1, 1, 1, 2] (cfl=8)
contfrac(55/89) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=9)
contfrac(89/144) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=10)
contfrac(144/233) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=11)
contfrac(233/377) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=12)
contfrac(377/610) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=13)
contfrac(610/987) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=14)
contfrac(987/1597) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=15)
contfrac(1597/2584) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=16)
contfrac(2584/4181) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=17)
contfrac(4181/6765) = [0; 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=18)
[…]

Which n1/n2 set records for the length of their continued fractions (with n2 > n1)? It’s the successive Fibonacci fractions, fib(i)/fib(i+1), of course. I didn’t anticipate that answer because I didn’t understand φ and continued fractions properly. And I still don’t, because I’ve been surprised again today looking at palindromic CFs like these:

contfrac(2/5) = [0; 2, 2] (cfl=2)
2/5 = 0 + 1/(2 + 1/2)
2/5 = 0.4

contfrac(3/8) = [0; 2, 1, 2] (cfl=3)
3/8 = 0 + 1/(2 + 1/(1 + 1/2))
3/8 = 0.375

contfrac(3/10) = [0; 3, 3] (cfl=2)
3/10 = 0 + 1/(3 + 1/3)
3/10 = 0.3

contfrac(5/12) = [0; 2, 2, 2] (cfl=3)
5/12 = 0 + 1/(2 + 1/(2 + 1/2))
5/12 = 0.416666666…

contfrac(5/13) = [0; 2, 1, 1, 2] (cfl=4)
5/13 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/2)))
5/13 = 0.384615384…

contfrac(4/15) = [0; 3, 1, 3] (cfl=3)
4/15 = 0 + 1/(3 + 1/(1 + 1/3))
4/15 = 0.266666666…

contfrac(7/16) = [0; 2, 3, 2] (cfl=3)
7/16 = 0 + 1/(2 + 1/(3 + 1/2))
7/16 = 0.4375

contfrac(4/17) = [0; 4, 4] (cfl=2)
4/17 = 0 + 1/(4 + 1/4)
4/17 = 0.235294117…

Again, I wondered which of these fractions set successive records for the length of their palindromic continued fractions. Here’s the answer:

contfrac(1/2) = [0; 2] (cfl=1)
1/2 = 0 + 1/2
1/2 = 0.5

contfrac(2/5) = [0; 2, 2] (cfl=2)
2/5 = 0 + 1/(2 + 1/2)
2/5 = 0.4

contfrac(3/8) = [0; 2, 1, 2] (cfl=3)
3/8 = 0 + 1/(2 + 1/(1 + 1/2))
3/8 = 0.375

contfrac(5/13) = [0; 2, 1, 1, 2] (cfl=4)
5/13 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/2)))
5/13 = 0.384615384…

contfrac(8/21) = [0; 2, 1, 1, 1, 2] (cfl=5)
8/21 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(1 + 1/2))))
8/21 = 0.380952380…

contfrac(13/34) = [0; 2, 1, 1, 1, 1, 2] (cfl=6)
13/34 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(
1 + 1/(1 + 1/2)))))
13/34 = 0.382352941..

contfrac(21/55) = [0; 2, 1, 1, 1, 1, 1, 2] (cfl=7)
21/55 = 0 + 1/(2 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/(1 + 1/2))))))
21/55 = 0.381818181…

contfrac(34/89) = [0; 2, 1, 1, 1, 1, 1, 1, 2] (cfl=8)
contfrac(55/144) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=9)
contfrac(89/233) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=10)
contfrac(144/377) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=11)
contfrac(233/610) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=12)
contfrac(377/987) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=13)
contfrac(610/1597) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=14)
contfrac(987/2584) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=15)
contfrac(1597/4181) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=16)
contfrac(2584/6765) = [0; 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2] (cfl=17)
[…]

Now it’s the successive Fibonacci skip-one fractions, fib(i)/fib(i+2), that set records for the length of their palindromic continued fractions. But I think you’d have to be very good at maths not to be surprised by that result.

After that, I continued to be compelled by the Call of CFulhu and started to look at the CFs of Fibonacci skip-n fractions in general. That’s contfrac(fib(i)/fib(i+n)) for n = 1,2,3,… And I’ve found more interesting patterns, as I’ll describe in a follow-up post.

hail(Satan)!

by in 666, Arithmetic, Integer Sequences, Mathematics, Number of the Beast and tagged , , , , , , | Leave a comment

It’s a very simple function that raises a very difficult question. An unanswered question, in fact. Take any whole number. If it’s odd, multiply it by 3 and add 1. If it’s even, divide it by 2. Repeat until you reach 1. That’s the hailstone function, because the numbers rise and fall like hailstones being formed in a cloud. Here are some examples:

5 → 16 → 8 → 4 → 2 → 1 (steps=5)

3 → 10 → 5 → 16 → 8 → 4 → 2 → 1 (st=7)

7 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 (st=16)

Graph for hail(7) = 16 (mx=52)

25 → 76 → 38 → 19 → 58 → 29 → 88 → 44 → 22 → 11 → 34 → 17 → 52 → 26 → 13 → 40 →
20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 (st=23)

Graph for hail(25) = 23 (mx=88)

But is this function truly a hailstone function? That is, does every number fall finally to earth and reach 1? So far, for every number tested, the answer has been yes. But do all numbers reach 1? The Collatz conjecture says they do. But no-one can prove it. Or disprove it. All it would take is one number failing to fall to earth. Mathematicians don’t think there is one, but numbers can take a surprising length of time to get to the ground. Here’s 27:

27 → 82 → 41 → 124 → 62 → 31 → 94 → 47 → 142 → 71 → 214 → 107 → 322 → 161 → 484 → 242 → 121 → 364 → 182 → 91 → 274 → 137 → 412 → 206 → 103 → 310 → 155 → 466 → 233 → 700 → 350 → 175 → 526 → 263 → 790 → 395 → 1186 → 593 → 1780 → 890 → 445 → 1336 → 668 → 334 → 167 → 502 → 251 → 754 → 377 → 1132 → 566 → 283 → 850 → 425 → 1276 → 638 → 319 → 958 → 479 → 1438 → 719 → 2158 → 1079 → 3238 → 1619 → 4858 → 2429 → 7288 → 3644 → 1822 → 911 → 2734 → 1367 → 4102 → 2051 → 6154 → 3077 → 9232 → 4616 → 2308 → 1154 → 577 → 1732 → 866 → 433 → 1300 → 650 → 325 → 976 → 488 → 244 → 122 → 61 → 184 → 92 → 46 → 23 → 70 → 35 → 106 → 53 → 160 → 80 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1 (st=111)

Graph for hail(27) = 111 (mx=9232)

27 takes 111 steps to reach 1. And the 111 made me think of another question. If the function hail(n) returns the number of steps required for n to reach 1, then hail(27) = 111. But what about hail(n) = 666? That is, what is the first number that requires 666 steps to reach 1? I say “first number”, because one very big number is guaranteed to take 666 steps:

666 = hail(306,180,206,916,083,902,309,240,650,087,602,475,282,639,486,413,866,622,
577,088,471,913,520,022,894,784,390,350,900,738,050,555,138,105,234,536,857,820,245,
071,373,614,031,482,942,161,565,170,086,143,298,589,738,273,508,330,367,307,539,078,
392,896,587,187,265,470,464)

Put another way, 666 = hail(2^666), because for any power of 2, hail(2^p) = p. But is there a smaller number, which I’ll call satan, for which hail(satan) = 666? Here’s a tantalizing taster of the task:

hail(27) = 111 (mx=9232)
hail(30262) = 222 (mx=2484916)
hail(164521) = 333 (mx=21933016)
hail(886953) = 444 (mx=52483285312)
hail(5143151) = 555 (mx=125218704148)
hail(satan) = 666 (mx=?)

But what is satan? Before I answer, here are some more graphs for interesting hail(n):

hail(231) = 127 (mx=9232)

hail(327) = 143 (mx=9232)

hail(703) = 170 (mx=250504)

hail(871) = 178 (mx=190996)

hail(2223) = 182 (mx=250504)

hail(3711) = 237 (mx=481624)

hail(35655) = 323 (mx=41163712)

hail(142587) = 374 (mx=593279152)

Now I’ll answer the question. If satan = 26597116, then hail(satan) = 666:

hail(26597116) = 666 (mx=15208728208)

Therefore:

hail(satan)! =
1,010,632,056,840,781,493,390,822,708,129,876,451,757,582,398,324,145,411,
340,420,807,357,413,802,103,697,022,989,202,806,801,491,012,040,989,802,
203,557,527,039,339,704,057,130,729,302,834,542,423,840,165,856,428,740,
661,530,297,972,410,682,828,699,397,176,884,342,513,509,493,787,480,774,
903,493,389,255,262,878,341,761,883,261,899,426,484,944,657,161,693,131,
380,311,117,619,573,051,526,423,320,389,641,805,410,816,067,607,893,067,
483,259,816,815,364,609,828,668,662,748,110,385,603,657,973,284,604,842,
078,094,141,556,427,708,745,345,100,598,829,488,472,505,949,071,967,727,
270,911,965,060,885,209,294,340,665,506,480,226,426,083,357,901,503,097,
781,140,832,497,013,738,079,112,777,615,719,116,203,317,542,199,999,489,
227,144,752,667,085,796,752,482,688,850,461,263,732,284,539,176,142,365,
823,973,696,764,537,603,278,769,322,286,708,855,475,069,835,681,643,710,
846,140,569,769,330,065,775,414,413,083,501,043,659,572,299,454,446,517,
242,824,002,140,555,140,464,296,291,001,901,438,414,675,730,552,964,914,
569,269,734,038,500,764,140,551,143,642,836,128,613,304,734,147,348,086,
095,123,859,660,926,788,460,671,181,469,216,252,213,374,650,499,557,831,
741,950,594,827,147,225,699,896,414,088,694,251,261,045,196,672,567,495,
532,228,826,719,381,606,116,974,003,112,642,111,561,332,573,503,212,960,
729,711,781,993,903,877,416,394,381,718,464,765,527,575,014,252,129,040,
283,236,963,922,624,344,456,975,024,058,167,368,431,809,068,544,577,258,
472,983,979,437,818,072,648,213,608,650,098,749,369,761,056,961,203,791,
265,363,665,664,696,802,245,199,962,040,041,544,438,210,327,210,476,982,
203,348,458,596,093,079,296,569,561,267,409,473,914,124,132,102,055,811,
493,736,199,668,788,534,872,321,705,360,511,305,248,710,796,441,479,213,
354,542,583,576,076,596,250,213,454,667,968,837,996,023,273,163,069,094,
700,429,467,106,663,925,419,581,193,136,339,860,545,658,673,623,955,231,
932,399,404,809,404,108,767,232,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,
000,000,000,000,000,000,000,000,000 = 666!

Here’s a question I haven’t answered: if satanic numbers are those n satisfying hail(n) = 666, how many satanic numbers are there? We’ve already seen two of them: 666 = hail(2^666) = hail(26597116). But how many more are there? Not infinitely many, because for n > 2^666, hail(n) > 666. In fact, after satan = 26597116, the next three satanic numbers arrive very quickly:

hail(satan+0) = 666 = hail(26597116)
hail(satan+1) = 666 = hail(26597117)
hail(satan+2) = 666 = hail(26597118)
hail(satan+3) = 666 = hail(26597119)

hail(satan-1) = 180 = hail(26597115)
hail(satan+4) = 180 = hail(26597120)

So there are four consecutive satanic numbers. But it isn’t unusual for a run of consecutive numbers to have the same hail(). Here’s a graph of the values of hail(n) for n = 1,2,3… (running left-to-right, down-up, with 1,2,3… in the lower lefthand corner). When n is divisible by 10, hail(n) is represented in red; when n is odd and divisible by 5, hail(n) is green. Note how many runs of identical hail(n) there are:

Graph for hail(n)

Here are successive records for runs of identical hail(n):

hail(12..13) = 9 (run=2)
hail(28..30) = 18 (run=3)
hail(98..102) = 25 (r=5)
hail(386..391) = 120 (r=6)
hail(943..949) = 36 (r=7)
hail(1494..1501) = 47 (r=8)
hail(1680..1688) = 42 (r=9)
hail(2987..3000) = 48 (r=14)
hail(7083..7099) = 57 (r=17)
hail(57346..57370) = 78 (r=25)
hail(252548..252574) = 181 (r=27)
hail(331778..331806) = 91 (r=29)
hail(524289..524318) = 102 (r=30)
hail(596310..596349) = 97 (r=40)

Finally, here’s Poland’s finest putting the function of 26597116 to music:

“Hail Satan!” by Dopelord

Elsewhere Other-Accessible…

• Dopelord at Bandcamp

Mods and Clockers

by in Fractals, Geometry, Integer Sequences, Mathematics, Triangles and tagged , , , , , | Leave a comment

To understand clock-arithmetic, simply picture a clock-face with one hand and a big fat 0 in place of the 12. Now you can do some clock-arithmetic. For example, set the hour-hand to 5, then move on 4 hours. You’ve done this sum:

5 + 4 → 9

Now try 9 + 7. The hour-hand is already on 9, so move forward 7 hours:

9 + 7 → 4

Now try 3 + 8 + 1:

3 + 8 + 1 → 0

And 3 * 4:

4 * 3 = 4 + 4 + 4 → 0

That’s clock-arithmetic. But you’re not confined to 12-hour clocks. Here’s a 7-hour clock, where the 7 is replaced with a 0:

3 + 1 → 4
4 + 5 → 2
2 + 4 + 1 → 0
3 * 3 = 3 + 3 + 3 → 2

Another name for clock-arithmetic is modular arithmetic, because the clocks model the process of dividing a number by 12 or 7 and finding the remainder or residue — 12 or 7 is known as the modulus (and modulo is Latin for “by the modulus”).

5 + 4 = 9 → 9 / 12 = 0*12 + 9

(5 + 4) modulo 12 = 9

3 + 8 + 1 = 12 → 12 / 12 = 1*12 + 0

(3 + 8 + 1) modulo 12 = 0

19 / 12 = 1*12 + 7

19 mod 12 = 7

3 + 1 = 4 → 4 / 7 = 0*7 + 4

(3 + 1) mod 7 = 4

2 + 4 + 1 = 7 → 7 / 7 = 1*7 + 0

(2 + 4 + 1) mod 7 = 0

19 / 7 = 2*7 + 5

19 mod 7 = 5

Modular arithmetic can do wonderful things. One small but beautiful example is the way it can uncover hidden fractals in Pascal’s triangle:

Pascal’s Triangle (via Desmos)

How to create Pascal’s triangle (via Wikipedia)

If you color all numbers n mod 2 = 1 (i.e., odd numbers) in the triangle, they create the famous Sierpiński triangle:

The Sierpiński triangle in Pascal’s triangle (via Fractal Foundation)

↓

Pascal’s triangle, n mod 2 = 1 (click for larger)

The Sierpiński triangle appears like this for all n mod 4 = 2 in Pascal’s triangle:

Pascal’s triangle, n mod 4 = 2 (click for larger)

And so on:

Pascal’s triangle, n mod 8 = 4

Pascal’s triangle, n mod 16 = 8

Pascal’s triangle, n mod 32 = 16

Pascal’s triangle, n mod 64 = 32

Pascal’s triangle, n mod 128 = 64

Pascal’s triangle, n mod 256 = 128

Pascal’s triangle, n mod 2,4,8… = 1,2,4… (animated via EzGif)

Post-Performative Post-Scriptum

There’s no need to calculate Pascal’s triangle in full to find the fractals above. The 10th row of Pascal’s triangle is this:

1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1

The 20th row is this:

1, 20, 190, 1140, 4845, 15504, 38760, 77520, 125970, 167960, 184756, 167960, 125970, 77520, 38760, 15504, 4845, 1140, 190, 20, 1

And the 29th is this:

1, 29, 406, 3654, 23751, 118755, 475020, 1560780, 4292145, 10015005, 20030010, 34597290, 51895935, 67863915, 77558760, 77558760, 67863915, 51895935, 34597290, 20030010, 10015005, 4292145, 1560780, 475020, 118755, 23751, 3654, 406, 29, 1

But you don’t need to consider those ever-growing numbers in the triangle when you’re finding fractals with modular arithmetic. When the modulus is 2, you just work with 0 and 1, that is, you add the previous numbers in the triangle and find the sum modulo 2. When the modulus is 4, you just work with 0, 1, 2 and 3, adding the numbers and finding the sum modulo 4. When it’s 8, you just work with 0, 1, 2, 3, 4, 5, 6 and 7, finding the sum modulo 8. And so on.

Fabulous Furry Fibonacci Fractal

by in Arithmetic, Fibonacci Sequence, Fractals, Integer Sequences, Mathematics and tagged , , , , , , , , | Leave a comment

At least, I think it’s a fractal. I came across it when I was counting the ways in which the integers can be the sum of distinct Fibonacci numbers. Here for reference is the Fibonacci sequence, the beautiful and endlessly fertile sequence that’s seeded with “1, 1” and continued by summing the two previous numbers:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040…

I noticed some interesting patterns in the distinct-fib-num-sum count for the integers:

1 = 1 (count=1)
2 = 2 (count=1)
3 = 1+2 = 3 (count=2)
4 = 1+3 (c=1)
5 = 2+3 = 5 (c=2)
6 = 1+2+3 = 1+5 (c=2)
7 = 2+5 (c=1)
8 = 1+2+5 = 3+5 = 8 (c=3)
9 = 1+3+5 = 1+8 (c=2)
10 = 2+3+5 = 2+8 (c=2)
11 = 1+2+3+5 = 1+2+8 = 3+8 (c=3)
12 = 1+3+8 (c=1)
13 = 2+3+8 = 5+8 = 13 (c=3)
14 = 1+2+3+8 = 1+5+8 = 1+13 (c=3)
15 = 2+5+8 = 2+13 (c=2)
16 = 1+2+5+8 = 3+5+8 = 1+2+13 = 3+13 (c=4)
17 = 1+3+5+8 = 1+3+13 (c=2)
18 = 2+3+5+8 = 2+3+13 = 5+13 (c=3)
19 = 1+2+3+5+8 = 1+2+3+13 = 1+5+13 (c=3)
20 = 2+5+13 (c=1)
21 = 1+2+5+13 = 3+5+13 = 8+13 = 21 (c=4)
22 = 1+3+5+13 = 1+8+13 = 1+21 (c=3)
23 = 2+3+5+13 = 2+8+13 = 2+21 (c=3)
24 = 1+2+3+5+13 = 1+2+8+13 = 3+8+13 = 1+2+21 = 3+21 (c=5)
25 = 1+3+8+13 = 1+3+21 (c=2)
26 = 2+3+8+13 = 5+8+13 = 2+3+21 = 5+21 (c=4)
27 = 1+2+3+8+13 = 1+5+8+13 = 1+2+3+21 = 1+5+21 (c=4)
28 = 2+5+8+13 = 2+5+21 (c=2)
29 = 1+2+5+8+13 = 3+5+8+13 = 1+2+5+21 = 3+5+21 = 8+21 (c=5)
30 = 1+3+5+8+13 = 1+3+5+21 = 1+8+21 (c=3)
31 = 2+3+5+8+13 = 2+3+5+21 = 2+8+21 (c=3)
32 = 1+2+3+5+8+13 = 1+2+3+5+21 = 1+2+8+21 = 3+8+21 (c=4)
33 = 1+3+8+21 (c=1)
34 = 2+3+8+21 = 5+8+21 = 13+21 = 34 (c=4)
35 = 1+2+3+8+21 = 1+5+8+21 = 1+13+21 = 1+34 (c=4)
36 = 2+5+8+21 = 2+13+21 = 2+34 (c=3)
37 = 1+2+5+8+21 = 3+5+8+21 = 1+2+13+21 = 3+13+21 = 1+2+34 = 3+34
(c=6)
38 = 1+3+5+8+21 = 1+3+13+21 = 1+3+34 (c=3)
39 = 2+3+5+8+21 = 2+3+13+21 = 5+13+21 = 2+3+34 = 5+34 (c=5)
40 = 1+2+3+5+8+21 = 1+2+3+13+21 = 1+5+13+21 = 1+2+3+34 = 1+5+34
(c=5)
41 = 2+5+13+21 = 2+5+34 (c=2)
42 = 1+2+5+13+21 = 3+5+13+21 = 8+13+21 = 1+2+5+34 = 3+5+34 = 8+3
4 (c=6)
43 = 1+3+5+13+21 = 1+8+13+21 = 1+3+5+34 = 1+8+34 (c=4)
44 = 2+3+5+13+21 = 2+8+13+21 = 2+3+5+34 = 2+8+34 (c=4)
45 = 1+2+3+5+13+21 = 1+2+8+13+21 = 3+8+13+21 = 1+2+3+5+34 = 1+2+
8+34 = 3+8+34 (c=6)
46 = 1+3+8+13+21 = 1+3+8+34 (c=2)
47 = 2+3+8+13+21 = 5+8+13+21 = 2+3+8+34 = 5+8+34 = 13+34 (c=5)
48 = 1+2+3+8+13+21 = 1+5+8+13+21 = 1+2+3+8+34 = 1+5+8+34 = 1+13+
34 (c=5)
49 = 2+5+8+13+21 = 2+5+8+34 = 2+13+34 (c=3)
50 = 1+2+5+8+13+21 = 3+5+8+13+21 = 1+2+5+8+34 = 3+5+8+34 = 1+2+1
3+34 = 3+13+34 (c=6)
51 = 1+3+5+8+13+21 = 1+3+5+8+34 = 1+3+13+34 (c=3)
52 = 2+3+5+8+13+21 = 2+3+5+8+34 = 2+3+13+34 = 5+13+34 (c=4)
53 = 1+2+3+5+8+13+21 = 1+2+3+5+8+34 = 1+2+3+13+34 = 1+5+13+34 (c=4)
54 = 2+5+13+34 (c=1)
55 = 1+2+5+13+34 = 3+5+13+34 = 8+13+34 = 21+34 = 55 (c=5)
56 = 1+3+5+13+34 = 1+8+13+34 = 1+21+34 = 1+55 (c=4)

The patterns are easier to see when the counts are set out like this:

1, 1, 2, 1, 2, 2, 1, 3, 2, 2, 3, 1, 3, 3, 2, 4, 2, 3, 3, 1, 4, 3, 3, 5, 2, 4, 4, 2, 5, 3, 3, 4, 1, 4, 4, 3, 6, 3, 5, 5, 2, 6, 4, 4, 6, 2, 5, 5, 3, 6, 3, 4, 4, 1, 5, 4, 4, 7, 3, 6, 6, 3, 8, 5, 5, 7, 2, 6, 6, 4, 8, 4, 6, 6, 2, 7, 5, 5, 8, 3, 6, 6, 3, 7, 4, 4, 5, 1, 5, 5, 4, 8, 4, 7, 7, 3, 9, 6, 6, 9, 3, 8, 8, 5, 10, 5, 7, 7, 2, 8, 6, 6, 10, 4, 8, 8, 4, 10, 6, 6, 8, 2, 7, 7, 5, 10, 5, 8, 8, 3, 9, 6, 6, 9, 3, 7, 7, 4, 8, 4, 5, 5, 1, 6, 5, 5, 9, 4, 8, 8… 1… — See A000119, Number of representations of n as a sum of distinct Fibonacci numbers, at the Online Encyclopedia of Integer Sequences (OEIS)

The numbers between each pair of 1s are symmetrical:

1, 2, 1,
1, 2, 2, 1,
1, 3, 2, 2, 3, 1,
1, 3, 3, 2, 4, 2, 3, 3, 1
1, 4, 3, 3, 5, 2, 4, 4, 2, 5, 3, 3, 4, 1

And when fibsumcount(n) = 1, then n = fib(i)-1, i.e. n is one less than a Fibonacci number:

1 = 1 (c=1)
2 = 2 (c=1)
4 = 1+3 (c=1)
7 = 2+5 (c=1)
12 = 1+3+8 (c=1)
20 = 2+5+13 (c=1)
33 = 1+3+8+21 (c=1)
54 = 2+5+13+34 (c=1)
88 = 1+3+8+21+55 (c=1)
143 = 2+5+13+34+89 (c=1)
232 = 1+3+8+21+55+144 (c=1)
376 = 2+5+13+34+89+233 (c=1)
609 = 1+3+8+21+55+144+377 (c=1)
986 = 2+5+13+34+89+233+610 (c=1)
1596 = 1+3+8+21+55+144+377+987 (c=1)
2583 = 2+5+13+34+89+233+610+1597 (c=1)
4180 = 1+3+8+21+55+144+377+987+2584 (c=1)
6764 = 2+5+13+34+89+233+610+1597+4181 (c=1)
10945 = 1+3+8+21+55+144+377+987+2584+6765 (c=1)
17710 = 2+5+13+34+89+233+610+1597+4181+10946 (c=1)
[…]

I also noticed a pattern relating to the maximum count reached in the numbers between the 1s. Suppose the function max(fib(i)-1..fib(i+1)-1) returns the highest count of ways to represent the numbers from fib(i)-1 to fib(i+1)-1. Notice how max() increases:

max(2..4) = 2
max(4..7) = 2
max(7..12) = 3
max(12..20) = 4
max(20..33) = 5
max(33..54) = 6
max(54..88) = 8
max(88..143) = 10
max(143..232) = 13
max(232..376) = 16
max(376..609) = 21
max(609..986) = 26
max(986..1596) = 34
max(1596..2583) = 42
max(2583..4180) = 55
max(4180..6764) = 68
[…]

The pattern is described like this at the Online Encyclopedia of Integer Sequences:

a(n) = 1 if and only if n+1 is a Fibonacci number. The length of such a quasi-period (from Fib(i)-1 to Fib(i+1)-1, inclusive) is a Fibonacci number + 1. The maximum value of a(n) within each subsequent quasi-period increases by a Fibonacci number. For example, from n = 143 to n = 232, the maximum is 13. From 232 to 376, the maximum is 16, an increase of 3. From 376 to 609, 21, an increase of 5. From 609 to 986, 26, increasing by 5 again. Each two subsequent maxima seem to increase by the same increment, the next Fibonacci number. – Kerry Mitchell, Nov 14 2009

The maxima of the quasi-periods are in A096748. – Max Barrentine, Sep 13 2015 — See commentary for A000119 at OEIS

Here is A096748:

1, 2, 2, 2, 3, 4, 5, 6, 8, 10, 13, 16, 21, 26, 34, 42, 55, 68, 89, 110, 144, 178, 233, 288, 377, 466, 610, 754, 987, 1220, 1597, 1974, 2584, 3194, 4181, 5168, 6765, 8362, 10946, 13530, 17711, 21892, 28657, 35422, 46368, 57314, 75025, 92736, 121393, 150050 — A096748, Expansion of (1+x)^2/(1-x^2-x^4), at OEIS

These maxima are the succesive highest points in a graph of A000119, Number of representations of n as a sum of distinct Fibonacci numbers:

Graph of count of ways in which 1,2,3… can be sum of distinct Fibonacci numbers

The graph looks like a furry caterpillar or similar and the symmetry of counts between the 1s is more obvious there:

fibsumcounts for 33..54

fibsumcounts for 54..88

fibsumcounts for 88..143

fibsumcounts for 143..232

fibsumcounts for 232..376

fibsumcounts for 376..609

And the fractal nature of the counts is more obvious when the graph is rotated by 90° and then mirrored:

Rotated and mirrored graph of count of ways in which 1,2,3… can be sum of distinct Fibonacci numbers

Abounding in Abundants

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

This is the famous Ulam spiral, invented by the Jewish mathematician Stanisław Ulam (pronounced OO-lam) to represent prime numbers on a square grid:

The Ulam spiral of prime numbers

The red square represents 1, with 2 as the white block immediately to its right and 3 immediately above 2. Then 5 is the white block one space to the left of 3 and 7 the white block one space below 5. Then 11 is the white block right beside 2 and 13 the white block one space above 11. And so on. The primes aren’t regularly spaced on the spiral but patterns are nevertheless appearing. Here’s the Ulam spiral at higher resolutions:

The Ulam spiral x2

The Ulam spiral x4

The primes are neither regular nor random in their distribution on the spiral. They stand tantalizingly betwixt and between. So the numbers represented on this Ulam-like spiral, which looks like an aerial view of a city designed by architects who occasionally get drunk:

Ulam-like spiral of abundant numbers

The distribution of abundant numbers is much more regular than the primes, but is far from wholly predictable. And what are abundant numbers? They’re numbers n such that sum(divisors(n)-n) > n. In other words, when you add the divisors of n less than n, the sum is greater than n. The first abundant number is 12:

12 is divisible by 1, 2, 3, 4, 6 → 1 + 2 + 3 + 4 + 6 = 16 > 12

The abundant numbers go like this:

12, 18, 20, 24, 30, 36, 40, 42, 48, 54, 56, 60, 66, 70, 72, 78, 80, 84, 88, 90, 96, 100, 102, 104, 108, 112, 114, 120, 126, 132, 138, 140, 144, 150, 156, 160, 162, 168, 174, 176, 180, 186, 192, 196, 198, 200, 204, 208, 210, 216, 220, 222, 224, 228, 234, 240, 246, 252, 258, 260, 264, 270… — A005101 at the Online Encyclopedia of Integer Sequences

Are all abundant numbers even? No, but the first odd abundant number takes a long time to arrive: it’s 45045. The abundance of 45045 was first discovered by the French mathematician Charles de Bovelles or Carolus Bovillus (c. 1475-1566), according to David Wells in his wonderful Penguin Dictionary of Curious and Interesting Numbers (1986):

45045 = 3^2 * 5 * 7 * 11 * 13 → 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 21 + 33 + 35 + 39 + 45 + 55 + 63 + 65 + 77 + 91 + 99 + 105 + 117 + 143 + 165 + 195 + 231 + 273 + 315 + 385 + 429 + 455 + 495 + 585 + 693 + 715 + 819 + 1001 + 1155 + 1287 + 1365 + 2145 + 3003 + 3465 + 4095 + 5005 + 6435 + 9009 + 15015 = 59787 > 45045

Here’s the spiral of abundant numbers at higher resolutions:

Abundant numbers x2

Abundant numbers x4

Negating the spiral of the abundant numbers — almost — is the spiral of the deficient numbers, where sum(divisors(n)-n) < n. Like most odd numbers, 15 is deficient:

15 = 3 * 5 → 1 + 3 + 5 = 9 < 15

Here’s the spiral of deficient numbers at various resolutions:

Deficient numbers on Ulam-like spiral

Deficient numbers x2

Deficient numbers x4

The spiral of deficient numbers doesn’t quite negate (reverse the colors of) the spiral of abundant numbers because of the very rare perfect numbers, where sum(divisors(n)-n) = n. That is, their factor-sums are exactly equal to themselves:

• 6 = 2 * 3 → 1 + 2 + 3 = 6
• 28 = 2^2 * 7 → 1 + 2 + 4 + 7 + 14 = 28
• 496 = 2^4 * 31 → 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 = 496

Now let’s try numbers n such than sum(divisors(n)) mod 2 = 1 (“n mod 2″ gives the remainder when n is divided by 2, i.e. n mod 2 is either 0 or 1). For example:

• 4 = 2^2 → 1 + 2 + 4 = 7 → 7 mod 2 = 1
• 18 = 2 * 3^2 → 1 + 2 + 3 + 6 + 9 + 18 = 39 → 39 mod 2 = 1
• 72 = 2^3 * 3^2 → 1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18 + 24 + 36 + 72 = 195 → 195 mod 2 = 1

Here are spirals for these numbers:

Ulam-like spiral for n such than sum(divisors(n)) mod 2 = 1

sum(divisors(n)) mod 2 = 1 x2

sum(divisors(n)) mod 2 = 1 x4

sum(divisors(n)) mod 2 = 1 x8

sum(divisors(n)) mod 2 = 1 x16

Pyramids for Pi

by in Arithmetic, Integer Sequences, π, Mathematics, Squares and tagged , , , , | Leave a comment

These are the odd numbers:


1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59...

If you add the odd numbers, 1+3+5+7…, you get the square numbers:


1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900...

And if you add the square numbers, 1+4+9+16…, you get what are called the square pyramidal numbers:


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...

There’s not a circle in sight, so you wouldn’t expect to find π amid the pyramids. But it’s there all the same. You can get π from this formula using the square pyramidal numbers:

π from a formula using square pyramidal numbers (Wikipedia)

Here are the approximations getting nearer and near to π:


3.1415926535897932384... = π
3.1666666666666666666... = sqpyra2pi(i=1) / 6 + 3
1 = sqpyra(1)

3.1415926535897932384... = π
3.1452380952380952380... = sqpyra2pi(i=3) / 6 + 3
14 = sqpyra(3)

3.1415926535897932384... = π
3.1412548236077647842... = sqpyra2pi(i=8) / 6 + 3
204 = sqpyra(8)

3.1415926535897932384... = π
3.1415189855952756236... = sqpyra2pi(i=14) / 6 + 3
1,015 = sqpyra(14)

3.1415926535897932384... = π
3.1415990074057163751... = sqpyra2pi(i=33) / 6 + 3
12,529 = sqpyra(33)

3.1415926535897932384... = π
3.1415920110950124679... = sqpyra2pi(i=72) / 6 + 3
127,020 = sqpyra(72)

3.1415926535897932384... = π
3.1415926017980070553... = sqpyra2pi(i=168) / 6 + 3
1,594,684 = sqpyra(168)

3.1415926535897932384... = π
3.1415926599504002195... = sqpyra2pi(i=339) / 6 + 3
13,043,590 = sqpyra(339)

3.1415926535897932384... = π
3.1415926530042565359... = sqpyra2pi(i=752) / 6 + 3
142,035,880 = sqpyra(752)

3.1415926535897932384... = π
3.1415926535000384883... = sqpyra2pi(i=1406) / 6 + 3
927,465,791 = sqpyra(1406)

3.1415926535897932384... = π
3.1415926535800054618... = sqpyra2pi(i=2944) / 6 + 3
8,509,683,520 = sqpyra(2944)

3.1415926535897932384... = π
3.1415926535890006043... = sqpyra2pi(i=6806) / 6 + 3
105,111,513,491 = sqpyra(6806)

3.1415926535897932384... = π
3.1415926535897000092... = sqpyra2pi(i=13892) / 6 + 3
893,758,038,910 = sqpyra(13892)

3.1415926535897932384... = π
3.1415926535897999990... = sqpyra2pi(i=33315) / 6 + 3
12,325,874,793,790 = sqpyra(33315)

3.1415926535897932384... = π
3.1415926535897939999... = sqpyra2pi(i=68985) / 6 + 3
109,433,980,000,485 = sqpyra(68985)

3.1415926535897932384... = π
3.1415926535897932999... = sqpyra2pi(i=159563) / 6 + 3
1,354,189,390,757,594 = sqpyra(159563)

3.1415926535897932384... = π
3.1415926535897932300... = sqpyra2pi(i=309132) / 6 + 3
9,847,199,658,130,890 = sqpyra(309132)

3.1415926535897932384... = π
3.1415926535897932389... = sqpyra2pi(i=774865) / 6 + 3
155,080,688,289,901,465 = sqpyra(774865)

3.1415926535897932384... = π
3.1415926535897932384... = sqpyra2pi(i=1586190) / 6 + 3
1,330,285,259,163,175,415 = sqpyra(1586190)

Summer Sets (and Truncated Triangulars)

by in Arithmetic, Integer Sequences, Mathematics, Triangular Numbers and tagged , , , | Leave a comment

Here is the sequence of triangular numbers, created by summing consecutive integers from 1 (i.e., 1+2+3+4+5…):


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, 1485, 1540, 1596, 1653, 1711, 1770, 1830, 1891, 1953, 2016, 2080, 2145, 2211, 2278, 2346, 2415, 2485, 2556, 2628, 2701, 2775, 2850, 2926, 3003, 3081, 3160, 3240, 3321, 3403, 3486, 3570, 3655, 3741, 3828, 3916, 4005, 4095, 4186, 4278, 4371, 4465, 4560, 4656, 4753, 4851, 4950, 5050, 5151, 5253, 5356, 5460, 5565, 5671, 5778, 5886, 5995...

And here is a sequence of truncated triangulars, created by summing consecutive integers from 15 (i.e., 15+16+17+18+19…):


15, 31, 48, 66, 85, 105, 126, 148, 171, 195, 220, 246, 273, 301, 330, 360, 391, 423, 456, 490, 525, 561, 598, 636, 675, 715, 756, 798, 841, 885, 930, 976, 1023, 1071, 1120, 1170, 1221, 1273, 1326, 1380, 1435, 1491, 1548, 1606, 1665, 1725, 1786, 1848, 1911, 1975, 2040, 2106, 2173, 2241, 2310, 2380, 2451, 2523, 2596, 2670, 2745, 2821, 2898, 2976, 3055, 3135, 3216, 3298, 3381, 3465, 3550, 3636, 3723, 3811, 3900, 3990, 4081, 4173, 4266, 4360, 4455, 4551, 4648, 4746, 4845, 4945, 5046, 5148, 5251, 5355, 5460, 5566, 5673, 5781...

It’s obvious that the sequences are different at each successive step: 1 ≠ 15, 3 ≠ 31, 6 ≠ 48, 10 ≠ 66, 21 ≠ 85, and so on. But seven numbers occur in both sequences: 15, 66, 105, 171, 561, 1326 and 5460. And that’s it — 7 is the 14-th entry in A309507 at the Encyclopedia of Integer Sequences:


0, 1, 1, 1, 3, 3, 1, 2, 5, 3, 3, 3, 3, 7, 3, 1, 5, 5, 3, 7, 7, 3, 3, 5, 5, 7, 7, 3, 7, 7, 1, 3, 7, 7, 11, 5, 3, 7, 7, 3, 7, 7, 3, 11, 11, 3, 3, 5, 8, 11, 7, 3, 7, 15, 7, 7, 7, 3, 7, 7, 3, 11, 5, 3, 15, 7, 3, 7, 15, 7, 5, 5, 3, 11, 11, 7, 15, 7, 3, 9, 9, 3, 7 — A309507

I decided to take create graphs of shared numbers in compared sequences like this. In the 135×135 grid below, the brightness of the squares corresponds to the count of shared numbers in the sequence-pair sum(x..x+n) and sum(y..y+n), where x and y are the coordinates of each individual square. I think the grid looks like a city of skyscrapers bisected by a highway:

Count of shared numbers in sequence-pairs sum(x..x+n) and sum(y..y+n)

Note that the bright white diagonal in the grid corresponds to the sequence-pairs where x = y. Because the sequences are identical in each pair, the count of shared numbers is infinite. The grid is symmetrically reflected along the diagonal because, for example, the sequence-pair for x=12, y=43, where sum(12..12+n) is compared with sum(43..43+n), corresponds to the sequence pair for x=43, y=12, where sum(43..43+n) is compared with sum(12..12+n). The scale of brightness runs from 0 (black) to 255 (full white) and increases by 32 for each shared number in the sequence. Obviously, then, the brightness can’t increase indefinitely and some maximally bright squares will represent sequence-pairs that have different counts of shared pairs.

Now try altering the size of the step in brightness. You get grids in which the width of the central strip increases (smaller step) or decreases (bigger step). Here are grids for steps for 1, 2, 4, 8, 16, 32 and 64 (I’ve removed the bright x=y diagonal for the first few grids, because it’s too prominent against duller shades):

Brightness-step = 1

Brightness-step = 2

Brightness-step = 4

Brightness-step = 8

Brightness-step = 16

Brightness-step = 32

Brightness-step = 63

Brightness-step = 1, 2, 4, 8, 16, 32, 63 (animated)

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.”