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.

A347205 a(2n+1) = a(n) for n >= 0, a(2n) = a(n) + a(n - 2^A007814(n)) for n > 0 with a(0) = 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 6, 3, 4, 1, 5, 4, 7, 3, 9, 5, 7, 2, 10, 6, 9, 3, 10, 4, 5, 1, 6, 5, 9, 4, 12, 7, 10, 3, 14, 9, 14, 5, 16, 7, 9, 2, 15, 10, 16, 6, 19, 9, 12, 3, 20, 10, 14, 4, 15, 5, 6, 1, 7, 6, 11, 5, 15, 9, 13, 4, 18, 12, 19, 7, 22, 10, 13
Offset: 0

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Author

Mikhail Kurkov, Aug 23 2021

Keywords

Comments

Scatter plot might be called "Cypress forest on a windy day". - Antti Karttunen, Nov 30 2021

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = If[OddQ[n], a[(n - 1)/2], a[n/2] + a[n/2 - 2^IntegerExponent[n/2, 2]]]; Array[a, 100, 0] (* Amiram Eldar, Sep 06 2021 *)
  • PARI
    a(n) = if (n==0, 1, if (n%2, a(n\2), a(n/2) + a(n/2 - 2^valuation(n/2, 2)))); \\ Michel Marcus, Sep 09 2021

Formula

a(2n+1) = a(n) for n >= 0.
a(2n) = a(n) + a(n - 2^A007814(n)) = a(2*A059894(n)) for n > 0 with a(0) = 1.
Sum_{k=0..2^n - 1} a(k) = A000108(n+1) for n >= 0.
a((4^n - 1)/3) = A000108(n) for n >= 0.
a(2^m*(2^n - 1)) = binomial(n + m, n) for n >= 0, m >= 0.
Generalization:
b(2n+1, p, q) = b(n, p, q) for n >= 0.
b(2n, p, q) = p*b(n, p, q) + q*b(n - 2^A007814(n), p, q) = for n > 0 with b(0, p, q) = 1.
Conjectured formulas: (Start)
Sum_{k=0..2^n - 1} b(k, 2, 1) = A006318(n) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 2, 2) = A115197(n) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 3, 1) = A108524(n+1) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 3, 3) = A116867(n) for n >= 0.
b((4^n - 1)/3, p, q) is generalized Catalan number C(p, q; n). (End)
Conjecture: a(n) = T(n, wt(n)+1), a(2n) = Sum_{k=1..wt(n)+1} T(n, k) where T(2n+1, k) = T(n, k) for 1 <= k <= wt(n)+1, T(2n+1, wt(n)+2) = T(n, wt(n)+1), T(2n, k) = Sum_{i=1..k} T(n, i) for 1 <= k <= wt(n)+1 with T(0, 1) = 1. - Mikhail Kurkov, Dec 13 2024