Tag Archives: Online Encyclopedia of Integer Sequences

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

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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.

Magiciprocal

by in Base Behavior, Integer Sequences, Magic squares, Mathematics, Prime numbers, Reciprocals and tagged , , , , , , , , | Leave a comment


A021023 Decimal expansion of 1/19.

0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8, 9, 4, 7, 3, 6, 8, 4, 2, 1, 0, 5, 2, 6, 3, 1, 5, 7, 8 [...] The magic square that uses the decimals of 1/19 is fully magic. — A021023 at the Online Encyclopedia of Integer Sequences

Mötley Vüe

by in Base Behavior, Digit-sums, Fibonacci Sequence, Integer Sequences, φ, Mathematics and tagged , , , , , , , , , , , , , , , , , , , , | Leave a comment

Here’s the Fibonacci sequence, where each term (after the first two) is created by adding the two previous numbers:


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

In “Fib and Let Tri”, I described how my eye was caught by 55, which is a palindrome, reading the same backwards and forwards. “Were there any other Fibonacci palindromes?” I wondered. So I looked to see. Now my eye has been caught by 55 again, but for another reason. It should be easy to spot another interesting aspect to 55 when the Fibonacci numbers are set out like this:


fib(1) = 1
fib(2) = 1
fib(3) = 2
fib(4) = 3
fib(5) = 5
fib(6) = 8
fib(7) = 13
fib(8) = 21
fib(9) = 34
fib(10) = 55
fib(11) = 89
fib(12) = 144
fib(13) = 233
fib(14) = 377
fib(15) = 610
fib(16) = 987
fib(17) = 1597
fib(18) = 2584
fib(19) = 4181
fib(20) = 6765
[...]

55 is fib(10), the 10th Fibonacci number, and 5+5 = 10. That is, digsum(fib(10)) = 10. What other Fibonacci numbers work like that? I soon found some and confirmed my answer at the Online Encyclopedia of Integer Sequences:


1, 5, 10, 31, 35, 62, 72, 175, 180, 216, 251, 252, 360, 494, 504, 540, 946, 1188, 2222 — A020995 at OEIS

And that seems to be the lot, according to the OEIS. In base 10, at least, but why stop at base 10? When I looked at base 11, the numbers of digsum(fib(k)) = k didn’t stop coming, because I couldn’t take the Fibonacci numbers very high on my computer. But the OEIS gives a much longer list, starting like this:


1, 5, 13, 41, 53, 55, 60, 61, 90, 97, 169, 185, 193, 215, 265, 269, 353, 355, 385, 397, 437, 481, 493, 617, 629, 630, 653, 713, 750, 769, 780, 889, 905, 960, 1013, 1025, 1045, 1205, 1320, 1405, 1435, 1501, 1620, 1650, 1657, 1705, 1735, 1769, 1793, 1913, 1981, 2125, 2153, 2280, 2297, 2389, 2413, 2460, 2465, 2509, 2533, 2549, 2609, 2610, 2633, 2730, 2749, 2845, 2893, 2915, 3041, 3055, 3155, 3209, 3360, 3475, 3485, 3521, 3641, 3721, 3749, 3757, 3761, 3840, 3865, 3929, 3941, 4075, 4273, 4301, 4650, 4937, 5195, 5209, 5435, 5489, 5490, 5700, 5917, 6169, 6253, 6335, 6361, 6373, 6401, 6581, 6593, 6701, 6750, 6941, 7021, 7349, 7577, 7595, 7693, 7740, 7805, 7873, 8009, 8017, 8215, 8341, 8495, 8737, 8861, 8970, 8995, 9120, 9133, 9181, 9269, 9277, 9535, 9541, 9737, 9935, 9953, 10297, 10609, 10789, 10855, 11317, 11809, 12029, 12175... — A025490 at OEIS

The list ends with 1636597 = A18666[b11] and the OEIS says that 1636597 almost certainly completes the list. According to David C. Terr’s paper “On the Sums of Fibonacci Numbers” (pdf), published in the Fibonacci Quarterly in 1996, the estimated digit-sum for the k-th Fibonacci number in base b is given by the formula (b-1)/2 * k * log(b,φ), where log(b,φ) is the logarithm in base b of the golden ratio, 1·61803398874… Terr then notes that the simplified formula (b-1)/2 * log(b,φ) gives the estimated average ratio digsum(fib(k)) / k in base b. Here are the estimates for bases 2 to 20:


b02 = 0.3471209568153086...
b03 = 0.4380178794859424...
b04 = 0.5206814352229629...
b05 = 0.5979874356654401...
b06 = 0.6714235829697111...
b07 = 0.7418818776805580...
b08 = 0.8099488992357201...
b09 = 0.8760357589718848...
b10 = 0.9404443811249043...
b11 = 1.0034045909311624...
b12 = 1.0650963641043091...
b13 = 1.1256639207937723...
b14 = 1.1852250528196852...
b15 = 1.2438775226715552...
b16 = 1.3017035880574074...
b17 = 1.3587732842474014...
b18 = 1.4151468584732730...
b19 = 1.4708766105122322...
b20 = 1.5260083080264088...

In base 2, you can expect digsum(fib(k)) to be much smaller than k; in base 20, you can expect digsum(fib(k)) to be much larger. But as you can see, the estimate for base 11, 1.0034045909311624…, is very nearly 1. That’s why base 11 produces so many results for digsum(fib(k)) = k, because only a slight deviation from the estimate might create a perfect ratio of 1 for digsum(fib(k)) / k, i.e. digsum(fib(k)) = k. But in the end the results run out in base 11 too, because as k gets higher and fib(k) gets bigger, the estimate becomes more and more accurate and digsum(fib(k)) > k. With lower k, digsum(fib(k)) can easily fall below k or match k. That happens in other bases, but because their estimates are further from 1, results for digsum(fib(k)) = k run out much more quickly.

To see this base behavior represented visually, I’ve created Ulam-like spirals for k using three colors: blue for digsum(fib(k)) < k, yellow for digsum(fib(k)) > k, and red for digsum(fib(k)) = k (with the green square at the center representing fib(1) = 1). As you can see below, the spiral for base 11 immediately stands out. It’s motley, not dominated by blue or yellow like the other spirals:

Spiral for digsum(fib(k)) in base 9
(blue for digsum(fib(k)) < k, yellow for digsum(fib(k)) > k, red for digsum(fib(k)) = k, green for fib(1))

Spiral for digsum(fib(k)) in base 10

Spiral for digsum(fib(k)) in base 11 — a motley view of blue, yellow and red

Spiral for digsum(fib(k)) in base 12

Spiral for digsum(fib(k)) in base 13

Finally, here are spirals at higher and higher resolution for digsum(fib(k)) = k in base 11:

digsum(fib(k)) = k in base 11 (low resolution)
(green square is fib(1))

digsum(fib(k)) = k in base 11 (x2 resolution)

digsum(fib(k)) = k in base 11 (x4)

digsum(fib(k)) = k in base 11 (x8)

digsum(fib(k)) = k in base 11 (x16)

digsum(fib(k)) = k in base 11 (x32)

digsum(fib(k)) = k in base 11 (x64)

digsum(fib(k)) = k in base 11 (x128)

digsum(fib(k)) = k in base 11 (animated)

Count Amounts

by in Base Behavior, Integer Sequences, Mathematics and tagged , , , , , , , , , , , , | Leave a comment

One of my favourite integer sequences is what I call the digit-line. You create it by taking this very familiar integer sequence:

• 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20…

And turning it into this one:

• 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 2, 0… (A033307 in the Online Encyclopedia of Integer Sequences)

You simply chop all numbers into single digits. What could be simpler? Well, creating the digit-line couldn’t be simpler, but it is in fact a very complex object. There are hidden depths in its patterns, as even a brief look will uncover. For example, you can try counting the digits as they appear one-by-one in the line and seeing whether the digit-counts compare. Do the 1s of the digit-line always outnumber the 0s, as you might expect? Yes, they do (unless you start the digit-line 0, 1, 2, 3…). But do the 2s always outnumber the 0s? No: at position 2, there’s a 2, and at position 11 there’s a 0. So that’s one 2 and one 0. Does it happen again? Yes, it happens again at the 222nd digit of the digit-line, as below:

1, 2count=1, 3, 4, 5, 6, 7, 8, 9, 1, 0count=1, 1, 1, 1, 22, 1, 3, 1, 4, 1, 5, 1, 6, 1, 7, 1, 8, 1, 9, 23, 02, 24, 1, 25, 26, 27, 3, 28, 4, 29, 5, 210, 6, 211, 7, 212, 8, 213, 9, 3, 03, 3, 1, 3, 214, 3, 3, 3, 4, 3, 5, 3, 6, 3, 7, 3, 8, 3, 9, 4, 04, 4, 1, 4, 215, 4, 3, 4, 4, 4, 5, 4, 6, 4, 7, 4, 8, 4, 9, 5, 05, 5, 1, 5, 216, 5, 3, 5,4, 5, 5, 5, 6, 5, 7, 5, 8, 5, 9, 6, 06, 6, 1, 6, 217, 6, 3, 6, 4, 6, 5, 6, 6, 6, 7, 6, 8, 6, 9, 7, 07, 7, 1, 7, 218, 7, 3, 7, 4, 7, 5, 7, 6, 7, 7, 7, 8, 7, 9, 8, 08, 8, 1, 8, 219, 8, 3, 8, 4, 8, 5, 8, 6, 8, 7, 8, 8, 8, 9, 9, 09, 9, 1, 9, 220, 9, 3, 9, 4, 9, 5, 9, 6, 9, 7, 9, 8, 9, 9, 1, 010, 011, 1, 012, 1, 1, 013, 221, 1, 014, 3, 1, 015, 4, 1, 016, 5, 1, 017, 6, 1, 018, 7, 1, 019, 8, 1, 020, 9, 1, 1, 021…

So count(2) = count(0) = 1 at digit 11 of the digit-line in the 0 of what was originally 10. And count(2) = count(0) = 21 @ digit 222 in the 0 of what was originally 110. Is a pattern starting to emerge? Yes, it is. Here are the first few points at which the count(2) = count(0) in the digit-line of base 10:

1 @ 11 in 10
21 @ 222 in 110
321 @ 3333 in 1110
4321 @ 44444 in 11110
54321 @ 555555 in 111110
654321 @ 6666666 in 1111110
7654321 @ 77777777 in 11111110
87654321 @ 888888888 in 111111110
987654321 @ 9999999999 in 1111111110
10987654321 @ 111111111110 in 11111111110
120987654321 @ 1222222222221 in 111111111110
[...]

The count(2) = count(0) = 321 at position 3333 in the digit-line, and 4321 at position 44444, and 54321 at position 555555, and so on. I don’t understand why these patterns occur, but you can predict the count-and-position of 2s and 0s easily until position 9999999999, after which things become more complicated. Related patterns for 2 and 0 occur in all other bases except binary (which doesn’t have a 2 digit). Here’s base 6:

1 @ 11 in 10 (1 @ 7 in 6)
21 @ 222 in 110 (13 @ 86 in 42)
321 @ 3333 in 1110 (121 @ 777 in 258)
4321 @ 44444 in 11110 (985 @ 6220 in 1554)
54321 @ 555555 in 111110 (7465 @ 46655 in 9330)
1054321 @ 11111110 in 1111110 (54121 @ 335922 in 55986)
12054321 @ 122222221 in 11111110 (380713 @ 2351461 in 335922)
132054321 @ 1333333332 in 111111110 (2620201 @ 16124312 in 2015538)
1432054321 @ 14444444443 in 1111111110 (17736745 @ 108839115 in 12093234)
15432054321 @ 155555555554 in 11111111110 (118513705 @ 725594110 in 72559410)
205432054321 @ 2111111111105 in 111111111110 (783641641 @ 4788921137 in 435356466)
2205432054321 @ 22222222222220 in 1111111111110 (5137206313 @ 31345665636 in 2612138802)

And what about comparing other pairs of digits? In fact, the count of all digits except 0 matches infinitely often. To write the numbers 1..9 takes one of each digit (except 0). To write the numbers 1 to 99 takes twenty of each digit (except 0). Here’s the proof:

11, 21, 31, 41, 51, 61, 71, 81, 91, 12, 01, 13, 14, 15, 22, 16, 32, 17, 42, 18, 52, 19, 62, 110, 72, 111, 82, 112, 92, 23, 02, 24, 113, 25, 26, 27, 33, 28, 43, 29, 53, 210, 63, 211, 73, 212, 83, 213, 93, 34, 03, 35, 114, 36, 214, 37, 38, 39, 44, 310, 54, 311, 64, 312, 74, 313, 84, 314, 94, 45, 04, 46, 115, 47, 215, 48, 315, 49, 410, 411, 55, 412, 65, 413, 75, 414, 85, 415, 95, 56, 05, 57, 116, 58, 216, 59, 316, 510, 416, 511, 512, 513, 66, 514, 76, 515, 86, 516, 96, 67, 06, 68, 117, 69, 217, 610, 317, 6
11, 417, 612, 517, 613, 614, 615, 77, 616, 87, 617, 97, 78, 07, 79, 118, 710, 218, 711, 318, 712, 418, 713, 518, 714, 618, 715, 716, 717, 88, 718, 98, 89, 08, 810, 119, 811, 219, 812, 319, 813, 419, 814, 519, 815, 619, 816, 719, 817, 818, 819, 99, 910, 09, 911, 120, 912, 220, 913, 320, 914, 420, 915, 520, 916, 620, 917, 720, 918, 820, 919, 920…

And what about writing 1..999, 1..9999, and so on? If you think about it, for every pair of non-zero digits, d1 and d2, all numbers containing one digit can be matched with a number containing the other. 100 → 200, 111 → 222, 314 → 324, 561189571 → 562289572, and so on. So in counting 1..999, 1..9999, 1..99999, you use the same number of non-zero digits. And once again a pattern emerges:

count(0) = 0; count(1) = 1; count(2) = 1; count(3) = 1; count(4) = 1; count(5) = 1; count(6) = 1; count(7) = 1; count(8) = 1; count(9) = 1 (writing 1..9)
count(0) = 9; count(1) = 20; count(2) = 20; count(3) = 20; count(4) = 20; count(5) = 20; count(6) = 20; count(7) = 20; count(8) = 20; count(9) = 20 (writing 1..99)
0: 189; 1: 300; 2: 300; 3: 300; 4: 300; 5: 300; 6: 300; 7: 300; 8: 300; 9: 300 (writing 1..999)
0: 2889; 1: 4000; 2: 4000; 3: 4000; 4: 4000; 5: 4000; 6: 4000; 7: 4000; 8: 4000; 9: 4000 (writing 1..9999)
0: 38889; 1: 50000; 2: 50000; 3: 50000; 4: 50000; 5: 50000; 6: 50000; 7: 50000; 8: 50000; 9: 50000 (writing 1..99999)
0: 488889; 1: 600000; 2: 600000; 3: 600000; 4: 600000; 5: 600000; 6: 600000; 7: 600000; 8: 600000; 9: 600000 (writing 1..999999)
0: 5888889; 1: 7000000; 2: 7000000; 3: 7000000; 4: 7000000; 5: 7000000; 6: 7000000; 7: 7000000; 8: 7000000; 9: 7000000 (writing 1..9999999)
[...]

And here’s base 6 again:

0: 0; 1: 1; 2: 1; 3: 1; 4: 1; 5: 1 (writing 1..5)
0: 5; 1: 20; 2: 20; 3: 20; 4: 20; 5: 20 (writing 1..55 in base 6)
0: 145; 1: 300; 2: 300; 3: 300; 4: 300; 5: 300 (writing 1..555)
0: 2445; 1: 4000; 2: 4000; 3: 4000; 4: 4000; 5: 4000 (writing 1..5555)
0: 34445; 1: 50000; 2: 50000; 3: 50000; 4: 50000; 5: 50000 (writing 1..55555)
0: 444445; 1: 1000000; 2: 1000000; 3: 1000000; 4: 1000000; 5: 1000000 (writing 1..555555)
0: 5444445; 1: 11000000; 2: 11000000; 3: 11000000; 4: 11000000; 5: 11000000 (writing 1..5555555)
0: 104444445; 1: 120000000; 2: 120000000; 3: 120000000; 4: 120000000; 5: 120000000 (writing 1..55555555)
0: 1144444445; 1: 1300000000; 2: 1300000000; 3: 1300000000; 4: 1300000000; 5: 1300000000 (writing 1..555555555)