Tag Archives: Online Encyclopedia of Integer Sequences

Moniliform Maths

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2 = 1/2 + 2/4 + 3/8 + 4/16 + 5/32…

sum(np / 2n)

2 = prime = sum(n / 2n)
6 = 2·3 = sum(n2 / 2n)
26 = 2·13 = sum(n3 / 2n)
150 = 2·3·52 = sum(n4 / 2n)
1082 = 2·541 = sum(n5 / 2n)
9366 = 2·3·7·223 = sum(n6 / 2n)
94586 = 2·47293 = sum(n7 / 2n)
1091670 = 2·3·5·36389 = sum(n8 / 2n)
14174522 = 2·7087261 = sum(n9 / 2n)
204495126 = 2·3·11·41·75571 = sum(n10 / 2n)

• A000629 Number of necklaces of partitions of n+1 labeled beads.

1, 2, 6, 26, 150, 1082, 9366, 94586, 1091670, 14174522, 204495126, 3245265146, 56183135190, 1053716696762, 21282685940886, 460566381955706, 10631309363962710, 260741534058271802, 6771069326513690646, 185603174638656822266, 5355375592488768406230

• moniliform ← French moniliforme (1800 or earlier) ← classical Latin monīle necklace

sum(np / 3n)

3/4 = prime / (22) = sum(n / 3n)
3/2 = prime / prime = sum(n2 / 3n)
33/8 = (3·11) / (23) = sum(n3 / 3n)
15 = 3·5 = sum(n4 / 3n)
273/4 = (3·7·13) / (22) = sum(n5 / 3n)
1491/4 = (3·7·71) / (22) = sum(n6 / 3n)
38001/16 = (3·53·239) / (24) = sum(n7 / 3n)
17295 = 3·5·1153 = sum(n8 / 3n)
566733/4 = (3·188911) / (22) = sum(n9 / 3n)
2579313/2 = (3·11·47·1663) / prime = sum(n10 / 3n)

sum(np / 4n)

4/9 = (22) / (32) = sum(n / 4n)
20/27 = (22·5) / (33) = sum(n2 / 4n)
44/27 = (22·11) / (33) = sum(n3 / 4n)
380/81 = (22·5·19) / (34) = sum(n4 / 4n)
4108/243 = (22·13·79) / (35) = sum(n5 / 4n)
17780/243 = (22·5·7·127) / (35) = sum(n6 / 4n)
269348/729 = (22·172·233) / (36) = sum(n7 / 4n)
4663060/2187 = (22·5·107·2179) / (37) = sum(n8 / 4n)
10091044/729 = (22·2522761) / (36) = sum(n9 / 4n)
218374420/2187 = (22·5·11·23·103·419) / (37) = sum(n10 / 4n)

sum(np / 5n)

5/16 = prime / (24) = sum(n / 5n)
15/32 = (3·5) / (25) = sum(n2 / 5n)
115/128 = (5·23) / (27) = sum(n3 / 5n)
285/128 = (3·5·19) / (27) = sum(n4 / 5n)
3535/512 = (5·7·101) / (29) = sum(n5 / 5n)
26355/1024 = (3·5·7·251) / (210) = sum(n6 / 5n)
458555/4096 = (5·91711) / (212) = sum(n7 / 5n)
1139685/2048 = (3·5·75979) / (211) = sum(n8 / 5n)
25492435/8192 = (5·17·443·677) / (213) = sum(n9 / 5n)
316786305/16384 = (3·5·11·1919917) / (214) = sum(n10 / 5n)

sum(np / 6n)

6/25 = (2·3) / (52) = sum(n / 6n)
42/125 = (2·3·7) / (53) = sum(n2 / 6n)
366/625 = (2·3·61) / (54) = sum(n3 / 6n)
4074/3125 = (2·3·7·97) / (55) = sum(n4 / 6n)
11334/3125 = (2·3·1889) / (55) = sum(n5 / 6n)
189714/15625 = (2·3·7·4517) / (56) = sum(n6 / 6n)
3706518/78125 = (2·3·181·3413) / (57) = sum(n7 / 6n)
82749954/390625 = (2·3·7·1970237) / (58) = sum(n8 / 6n)
2078250726/1953125 = (2·3·31·1061·10531) / (59) = sum(n9 / 6n)
11598884682/1953125 = (2·3·7·11·232·47459) / (59) = sum(n10 / 6n)

sum(np / 7n)

7/36 = prime / (22·32) = sum(n / 7n)
7/27 = prime / (33) = sum(n2 / 7n)
91/216 = (7·13) / (23·33) = sum(n3 / 7n)
70/81 = (2·5·7) / (34) = sum(n4 / 7n)
2149/972 = (7·307) / (22·35) = sum(n5 / 7n)
3311/486 = (7·11·43) / (2·35) = sum(n6 / 7n)
285929/11664 = (7·40847) / (24·36) = sum(n7 / 7n)
220430/2187 = (2·5·7·47·67) / (37) = sum(n8 / 7n)
1359337/2916 = (7·17·11423) / (22·36) = sum(n9 / 7n)
5239157/2187 = (7·11·68041) / (37) = sum(n10 / 7n)

sum(np / 8n)

8/49 = (23) / (72) = sum(n / 8n)
72/343 = (23·32) / (73) = sum(n2 / 8n)
776/2401 = (23·97) / (74) = sum(n3 / 8n)
10440/16807 = (23·32·5·29) / (75) = sum(n4 / 8n)
174728/117649 = (23·21841) / (76) = sum(n5 / 8n)
3525192/823543 = (23·32·11·4451) / (77) = sum(n6 / 8n)
11870648/823543 = (23·41·36191) / (77) = sum(n7 / 8n)
319735800/5764801 = (23·32·52·19·9349) / (78) = sum(n8 / 8n)
9686934584/40353607 = (23·1210866823) / (79) = sum(n9 / 8n)
326084753016/282475249 = (23·32·11·16273·25301) / (710) = sum(n10 / 8n)

sum(np / 9n)

9/64 = (32) / (26) = sum(n / 9n)
45/256 = (32·5) / (28) = sum(n2 / 9n)
531/2048 = (32·59) / (211) = sum(n3 / 9n)
1935/4096 = (32·5·43) / (212) = sum(n4 / 9n)
34983/32768 = (32·132·23) / (215) = sum(n5 / 9n)
381465/131072 = (32·5·72·173) / (217) = sum(n6 / 9n)
9725787/1048576 = (32·67·1272) / (220) = sum(n7 / 9n)
35420535/1048576 = (32·5·787123) / (220) = sum(n8 / 9n)
1160703963/8388608 = (32·47·409·6709) / (223) = sum(n9 / 9n)
21129845715/33554432 = (32·5·11·2213·19289) / (225) = sum(n10 / 9n)

sum(np / 10n)

10/81 = (2·5) / (34) = sum(n / 10n)
110/729 = (2·5·11) / (36) = sum(n2 / 10n)
470/2187 = (2·5·47) / (37) = sum(n3 / 10n)
7370/19683 = (2·5·11·67) / (39) = sum(n4 / 10n)
142870/177147 = (2·5·7·13·157) / (311) = sum(n5 / 10n)
1114190/531441 = (2·5·7·11·1447) / (312) = sum(n6 / 10n)
30495890/4782969 = (2·5·3049589) / (314) = sum(n7 / 10n)
953934190/43046721 = (2·5·11·569·15241) / (316) = sum(n8 / 10n)
3728765410/43046721 = (2·5·372876541) / (316) = sum(n9 / 10n)
145739620510/387420489 = (2·5·11·1324905641) / (318) = sum(n10 / 10n)

Pi’s Guys

by in Arithmetic, Continued Fractions, Irrational numbers, π, Mathematics, Pi and tagged , , , , , , | Leave a comment

3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2, 1, 1, 2, 2, 2, 2, 1, 84, 2, 1, 1, 15, 3, 13, 1, 4, 2, 6, 6, 99, 1, 2, 2, 6, 3, 5, 1, 1, 6, 8, 1, 7, 1, 2, 3, 7, 1, 2, 1, 1, 12, 1, 1, 1, 3, 1, 1, 8, 1, 1, 2, 1, 6, 1, 1, 5, 2, 2, 3, 1, 2, 4, 4, 16, 1, 161, 45, 1, 22, 1, 2, 2, 1, 4, 1, 2, 24, 1, 2, 1, 3, 1, 2, 1, …

The first 5821569425 terms were computed by Eric W. Weisstein on Sep 18 2011.
The first 10672905501 terms were computed by Eric W. Weisstein on Jul 17 2013.
The first 15000000000 terms were computed by Eric W. Weisstein on Jul 27 2013.
The first 30113021586 terms were computed by Syed Fahad on Apr 27 2021.
The first 653520000000 terms were computed by Max Frank, Nov 01 2025.

• A001203 Simple continued fraction expansion of Pi.

Sequence Unfurls…

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

The Fibonacci sequence is beautiful like clockwork. There’s a perfectly clear, rigorously defined mechanism ticking out an entirely predictable result for ever:

0, 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, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986, 102334155, … — A000045 at the Online Encyclopedia of Integer Sequences (OEIS)

And there’s a formula to calculate any term in the sequence without calculating all the terms that precede it:

Binet’s formula for Fn, the n-th Fibonacci number

But I also like sequences that you might call definitely arbitrary. That is, there’s a perfectly clear, rigorously defined mechanism, but the results seem arbitrary — not predictable at all:

6, 15, 5, 22, 6, 3, 30, 9, 7, 2, 45, 15, 6, 5, 1, 36, 14, 6, 5, 3, 1, 62, 22, 16, 6, 5, 3, 2, 69, 21, 15, 4, 9, 5, 2, 1, 84, 30, 15, 9, 6, 7, 2, 2, 1, 56, 22, 13, 7, 3, 5, 2, 0, 0, 0, 142, 45, 22, 15, 12, 6, 9, 5, 3, 1, 2, 53, 17, 8, 4, 5, 1, 6, 3, 1, 1, 1, 0, 124, 36, 27, 14, 18, 6, 6, 5, 2, 3, 1, 1, 0, … A349083 at OEIS

What’s the formula there? That sequence is defined at the OEIS as “The number of three-term Egyptian fractions of rational numbers x/y, 0 < x/y < 1, ordered as below. The sequence is the number of (p,q,r) such that x/y = 1/p + 1/q + 1/r where p, q, and r are integers with p < q < r.” For example: “The sixth rational number is 3/4 [and] 3/4 = 1/2 + 1/5 + 1/20 = 1/2 + 1/6 + 1/12 = 1/3 + 1/4 + 1/5, so a(6)=3.”

Factory Façades

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Practically speaking, I’d never heard of them. Practical numbers, that is. They’re defined like this at the Online Encyclopedia of Integer Sequences:

A005153 Practical numbers: positive integers m such that every k <= sigma(m) is a sum of distinct divisors of m. Also called panarithmic numbers. […] Equivalently, positive integers m such that every number k <= m is a sum of distinct divisors of m. — A005153 at OEIS

In other words, if you take, say, divisors(12) = 1, 2, 3, 4, 6, you can find partial sums of those divisors that equal every number from 1 to 16, where 16 = 1+2+3+4+6. Here are all those sums, with c as the count of divisor-sums equalling a particular k (to simplify things, I’m excluding 12 as a divisor of 12):

1, 2, 3, 4, 6 = divisors(12)

01 = 1 (c=1)
02 = 2 (c=1)
03 = 1 + 2 = 3 (c=2)
04 = 1 + 3 = 4 (c=2)
05 = 2 + 3 = 1 + 4 (c=2)
06 = 1 + 2 + 3 = 2 + 4 = 6 (c=3)
07 = 1 + 2 + 4 = 3 + 4 = 1 + 6 (c=3)
08 = 1 + 3 + 4 = 2 + 6 (c=2)
09 = 2 + 3 + 4 = 1 + 2 + 6 = 3 + 6 (c=3)
10 = 1 + 2 + 3 + 4 = 1 + 3 + 6 = 4 + 6 (c=3)
11 = 2 + 3 + 6 = 1 + 4 + 6 (c=2)
12 = 1 + 2 + 3 + 6 = 2 + 4 + 6 (c=2)
13 = 1 + 2 + 4 + 6 = 3 + 4 + 6 (c=2)
14 = 1 + 3 + 4 + 6 (c=1)
15 = 2 + 3 + 4 + 6 (c=1)
16 = 1 + 2 + 3 + 4 + 6 (c=1)

Learning about practical numbers inspired me to look at the graphs of the count of the divisor-sums for 12. If you include count(0) = 1 (there is one way of choosing divisors of 12 to equal 0, namely, by choosing none of the divisors), the graph looks like this:

counts of divisorsum(12) = k, where 12 = 2^2 * 3 → 1, 2, 3, 4, 6

Here are some more graphs for partialsumcount(n), adjusted for a standardized y-max. They remind me variously of skyscrapers, pyramids, stupas, factories and factory façades, forts bristling with radar antennae, and the Houses of Parliament. All in an art-deco style:

18 = 2 * 3^2 → 1, 2, 3, 6, 9

24 = 2^3 * 3 → 1, 2, 3, 4, 6, 8, 12

30 = 2 * 3 * 5 → 1, 2, 3, 5, 6, 10, 15

36 = 2^2 * 3^2 → 1, 2, 3, 4, 6, 9, 12, 18

48 = 2^4 * 3 → 1, 2, 3, 4, 6, 8, 12, 16, 24

54 = 2 * 3^3 → 1, 2, 3, 6, 9, 18, 27

60 = 2^2 * 3 * 5 → 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30

72 = 2^3 * 3^2 → 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36

88 = 2^3 * 11 → 1, 2, 4, 8, 11, 22, 44, 88

96 = 2^5 * 3 → 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48

100 = 2^2 * 5^2 → 1, 2, 4, 5, 10, 20, 25, 50

108 = 2^2 * 3^3 → 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54

120 = 2^3 * 3 * 5 → 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60

126 = 2 * 3^2 * 7 → 1, 2, 3, 6, 7, 9, 14, 18, 21, 42, 63

162 = 2 * 3^4 → 1, 2, 3, 6, 9, 18, 27, 54, 81

220 = 2^2 * 5 * 11 → 1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110

And what about im-practical numbers, where the partial sums of divisors(m) don’t equal every number 1..sigma(m)? There are interesting fractal patterns to be uncovered there, as you can see from the graph for 190 (because all divsumcount(k) = 1, the graph looks like a bar-code):

190 = 2 * 5 * 19 → 1, 2, 5, 10, 19, 38, 95

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

Piles of Prime Pairs

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

A087641 Start of the first sequence of exactly n consecutive pairs of twin primes

29, 101, 5, 9419, 909287, 325267931, 678771479, 1107819732821, 170669145704411, 3324648277099157, 789795449254776509

Example: a(6)=325267931 is the starting point of the first occurrence of 6 consecutive pairs of twin primes: (325267931 325267933) (325267937 325267939) (325267949 325267951) (325267961 325267963) (325267979 325267981) (325267991 325267993).

• A087641 at the Encyclopedia of Integer Sequences

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

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.”

Nexcelsior

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

In “The Trivial Troot”, I looked at what happens when tri(k), the k-th triangular number, is one digit longer than the previous triangular number, tri(k-1):


6 = tri(3)
10 = tri(4)

91 = tri(13)
105 = tri(14)

990 = tri(44)
1035 = tri(45)
[...]

↓

10 ← 4
105 ← 14
1035 ← 45
10011 ← 141
100128 ← 447
1000405 ← 1414
10001628 ← 4472
100005153 ← 14142
1000006281 ← 44721
10000020331 ← 141421
100000404505 ← 447214
1000001326005 ← 1414214
10000002437316 ← 4472136
100000012392316 ← 14142136
[...]

What’s going on with k? In a sense, it’s calculating the square roots of 2 and 20:

√2 = 1·414213562373095048801688724209698078569671875376948073176679738...
√20 = 4·472135954999579392818347337462552470881236719223051448541794491...

Now let’s say “Excelsior!” and go higher with a related sequence. A006003 is defined at the Online Encyclopedia of Integer Sequences as the “sum of the next n natural numbers”. Here it is:


1 = 1
5 = 2 + 3
15 = 4 + 5 + 6
34 = 7 + 8 + 9 + 10
65 = 11 + 12 + 13 + 14 + 15
111 = 16 + 17 + 18 + 19 + 20 + 21
175 = 22 + 23 + 24 + 25 + 26 + 27 + 28
260 = 29 + 30 + 31 + 32 + 33 + 34 + 35 + 36
369 = 37 + 38 + 39 + 40 + 41 + 42 + 43 + 44 + 45
505 = 46 + 47 + 48 + 49 + 50 + 51 + 52 + 53 + 54 + 55
671 = 56 + 57 + 58 + 59 + 60 + 61 + 62 + 63 + 64 + 65 + 66
870 = 67 + 68 + 69 + 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77 + 78
1105 = 79 + 80 + 81 + 82 + 83 + 84 + 85 + 86 + 87 + 88 + 89 + 90 + 91
1379 = 92 + 93 + 94 + 95 + 96 + 97 + 98 + 99 + 100 + 101 + 102 + 103 + 104 + 105
1695 = 106 + 107 + 108 + 109 + 110 + 111 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 119 + 120
2056 = 121 + 122 + 123 + 124 + 125 + 126 + 127 + 128 + 129 + 130 + 131 + 132 + 133 + 134 + 135 + 136
2465 = 137 + 138 + 139 + 140 + 141 + 142 + 143 + 144 + 145 + 146 + 147 + 148 + 149 + 150 + 151 + 152 + 153
2925 = 154 + 155 + 156 + 157 + 158 + 159 + 160 + 161 + 162 + 163 + 164 + 165 + 166 + 167 + 168 + 169 + 170 + 171
3439 = 172 + 173 + 174 + 175 + 176 + 177 + 178 + 179 + 180 + 181 + 182 + 183 + 184 + 185 + 186 + 187 + 188 + 189 + 190
4010 = 191 + 192 + 193 + 194 + 195 + 196 + 197 + 198 + 199 + 200 + 201 + 202 + 203 + 204 + 205 + 206 + 207 + 208 + 209 + 210
[...]

If you’re familiar with triangular numbers, you’ll see that sumnext(k) is always higher than tri(k), except for sumnext(1) = 1 = tri(k). Now, this is what happens when sumnext(k) is one digit longer than sumnext(k-1):


5 = sumnext(2)
15 = sumnext(3)

65 = sumnext(5)
111 = sumnext(6)

870 ← 12
1105 ← 13

9855 ← 27
10990 ← 28

97585 ← 58
102719 ← 59

976625 ← 125
1000251 ← 126

9951391 ← 271
10061960 ← 272

99588644 ← 584
100101105 ← 585

997809119 ← 1259
1000188630 ← 1260

9995386529 ← 2714
10006439295 ← 2715
[...]

↓

15 ← 3
111 ← 6
1105 ← 13
10990 ← 28
102719 ← 59
1000251 ← 126
10061960 ← 272
100101105 ← 585
1000188630 ← 1260
10006439295 ← 2715
100049490449 ← 5849
1000188006300 ← 12600
10000910550385 ← 27145
100003310078561 ← 58481
1000021311323825 ← 125993
10000026341777165 ← 271442
100000232056567634 ← 584804
1000002262299152685 ← 1259922
10000004237431278525 ← 2714418
100000026858987459346 ← 5848036
1000000119305407615071 ← 12599211
10000000921801015908705 ← 27144177
100000001209342964609615 ← 58480355
1000000000250317736274865 ← 125992105
10000000037633414521952245 ← 271441762
100000000183357362892853070 ← 584803548
1000000000250317673908773025 ← 1259921050
[...]

What’s going on now? In a sense, the digits of k are approximating the cube roots of 20, 200 and 2000:


2.714417616594906571518089469679489204805107769489096957284365443... = cuberoot(20)
5.848035476425732131013574720275845557060997270202060082845147020... = cuberoot(200)
12.59921049894873164767210607278228350570251464701507980081975112... = cuberoot(2000)

cuberoot(20) = 2.714417616594906571518089469679489204805107769489096957284365443...
cuberoot(200) = 5.848035476425732131013574720275845557060997270202060082845147020...
cuberoot(2000) = 12.59921049894873164767210607278228350570251464701507980081975112...

So you could say that this sequence has gone nexcelsior: sumnext(k) > tri(k); cubes are higher than squares; and (20, 200, 2000) is bigger than (2, 20).

Previously Pre-Posted…

• “The Trivial Troot” — explaining the earlier pattern in triangular numbers

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.