cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A356639 Number of integer sequences b with b(1) = 1, b(m) > 0 and b(m+1) - b(m) > 0, of length n which transform under the map S into a nonnegative integer sequence. The transform c = S(b) is defined by c(m) = Product_{k=1..m} b(k) / Product_{k=2..m} (b(k) - b(k-1)).

Original entry on oeis.org

1, 1, 3, 17, 155, 2677, 73327, 3578339, 329652351
Offset: 1

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Author

Thomas Scheuerle, Aug 19 2022

Keywords

Comments

This sequence can be calculated by a recursive algorithm:
Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').
The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).
There exist some transformation pairs of infinite sequences in the database:
A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;
A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;
A079352 <--> A------; A082458 <--> A------; A008233 <--> A264635;
A138278 <--> A------; A006501 <--> A264557; A336496 <--> A------;
A019464 <--> A------; A062112 <--> A------; A171647 <--> A359039;
A279312 <--> A------; A031923 <--> A------.
These transformation pairs are conjectured:
A137326 <--> A------; A066332 <--> A300902; A208147 <--> A308546;
A057895 <--> A------; A349080 <--> A------; A019442 <--> A------;
A349079 <--> A------.
("A------" means not yet in the database.)
Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.
If we allowed signed rational numbers, further interesting transformation pairs could be observed. For example, 1/n will transform into factorials with alternating sign. 2^(-n) transforms into ones with alternating sign and 1/A000045(n) into A000045 with alternating sign.

Examples

			a(4) = 17. The 17 transformation pairs of length 4 are:
  {1, 2, 3, 4}  = S({1, 2, 6, 24}).
  {1, 2, 3, 5}  = S({1, 2, 6, 15}).
  {1, 2, 3, 6}  = S({1, 2, 6, 12}).
  {1, 2, 3, 9}  = S({1, 2, 6, 9}).
  {1, 2, 3, 12} = S({1, 2, 6, 8}).
  {1, 2, 3, 21} = S({1, 2, 6, 7}).
  {1, 2, 4, 5}  = S({1, 2, 4, 20}).
  {1, 2, 4, 6}  = S({1, 2, 4, 12}).
  {1, 2, 4, 8}  = S({1, 2, 4, 8}).
  {1, 2, 4, 12} = S({1, 2, 4, 6}).
  {1, 2, 4, 20} = S({1, 2, 4, 5}).
  {1, 2, 6, 7}  = S({1, 2, 3, 21}).
  {1, 2, 6, 8}  = S({1, 2, 3, 12}).
  {1, 2, 6, 9}  = S({1, 2, 3, 9}).
  {1, 2, 6, 12} = S({1, 2, 3, 6}).
  {1, 2, 6, 15} = S({1, 2, 3, 5}).
  {1, 2, 6, 24} = S({1, 2, 3, 4}).
b(1) = 1 by definition, b(2) = 1+1 as 1 has only 1 as divisor.
a(3) = A000005(b(2)*b(2)) = 3.
The divisors of b(2) are 1,2,4. So b(3) can be b(2)+1, b(2)+2 and b(2)+4.
a(4) = A000005((b(2)+1)*(b(2)+4)) + A000005((b(2)+2)*(b(2)+2)) + A000005((b(2)+4)*(b(2)+1)) = 17.
		

Crossrefs

A329178 Sum of the products of pairs of Padovan numbers which are two apart, starting from A000931(5).

Original entry on oeis.org

1, 3, 5, 11, 19, 34, 62, 107, 191, 335, 587, 1035, 1812, 3184, 5589, 9803, 17213, 30199, 52999, 93014, 163214, 286439, 502655, 882095, 1547991, 2716503, 4767160, 8365776, 14680889, 25763219, 45211237, 79340227, 139232411, 244335770, 428779502, 752455475
Offset: 0

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Author

David Nacin, Nov 08 2019

Keywords

Examples

			For n=3, a(3) = 1*1 + 1*2 + 1*2 + 2*3 = 11.
		

Crossrefs

Programs

  • Python
    p = lambda x:[1, 1, 1][x] if x<3 else p(x-2)+p(x-3)
    a = lambda x:sum(p(i)*p(i+2) for i in range(x+1))

Formula

a(n) = Sum_{i=5..n+5} A000931(i)*A000931(i+2).
a(n) = A329227(n+7) - 1.
Conjectures from Colin Barker, Nov 09 2019: (Start)
G.f.: (1 + x - x^2 + x^3 - x^4) / ((1 - x)*(1 - 2*x + x^2 - x^3)*(1 + x - x^3)).
a(n) = 2*a(n-1) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + a(n-7) for n>6.
(End)
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