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.

A376317 a(n) = P(n+1, n+1) where P(n, m) = P(n, m-1) + P(n-1, m + f(m-n)) for n < m with P(n, m) = P(n-1, m) for 0 <= m <= n and P(0, m) = 1 for m >= 0 and where f(n) = [(n mod 4) > 0].

Original entry on oeis.org

1, 2, 6, 23, 100, 466, 2270, 11413, 58776, 308473, 1643922, 8872220, 48393710, 266357916, 1477471248, 8251090171, 46353709956, 261783417427, 1485374891782, 8463626764923, 48408809202918, 277834282516061, 1599585546691518, 9235769089804804, 53466364700486982
Offset: 0

Views

Author

Mikhail Kurkov, Sep 22 2024

Keywords

Comments

Conjecture: cases f(n) = n mod 2 and f(n) = [(n mod 3) > 0] both gives A006318.

Crossrefs

Programs

  • PARI
    upto(n) = my(v1); v1 = vector(2*(n+1), i, 1); v2 = vector(n+1, i, 0); v2[1] = 1; for(i=1, n, for(j=i+1, 2*(n+1)-i, v1[j] = v1[j+(((j-i)%4)>0)] + v1[j-1]); v2[i+1] = v1[i+1]); v2

Formula

Conjecture: a(n+2) = a(n) + A215973(n+2) - A215973(n+1) (noticed by advanced OEIS search).
Recurrence: (n+1)*a(n) = 4*(2*n-1)*a(n-1) - (11*n-25)*a(n-2) - 2*(2*n+5)*a(n-3) + 12*(n-2)*a(n-4) - 2*(2*n-7)*a(n-5). - Vaclav Kotesovec, Sep 23 2024