A229111 Expansion of the g.f. of A053723 in powers of the g.f. of A121591.
1, -5, 35, -275, 2275, -19255, 163925, -1385725, 11483875, -91781375, 688658785, -4581861025, 22550427925, 8852899375, -2431720493125, 47471706909725, -699843878180125, 9141002535744625, -111232778205154375, 1288777160650004375, -14372445132730778975
Offset: 1
Keywords
Examples
G.f. = x - 5*x^2 + 35*x^3 - 275*x^4 + 2275*x^5 - 19255*x^6 + 163925*x^7 + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..958
- Shaun Cooper, Apéry-like sequences defined by four-term recurrence relations, arXiv:2302.00757 [math.NT], 2023. See Table 2 p. 7.
- Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, 2016, 2:5.
- Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See eta p. 3.
- L. O'Brien, Modular forms and two new integer sequences at level 7, Massey University, 2016.
- H. Verrill, Some Congruences related to modular forms, Max Planck Institute, 1999.
Crossrefs
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692, A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Programs
-
Mathematica
a[n_] := a[n] = Switch[n, 1, 1, 2, -5, _, (1/(n-1)^3) ((1-2(n-1)) (11(n-2) (n-1)+5) a[n-1] - 125 (n-2)^3 a[n-2])]; a /@ Range[21] (* Jean-François Alcover, Jan 13 2020 *)
-
PARI
{a(n) = my(m = n-1); if( n<1, 0, if( n<3, [1, -5][n], -( (5*(m - 1))^3*a(n-2) + (2*m - 1)*(11*(m^2 - m) +5)*a(n-1) )/ m^3))};
-
PARI
{a(n) = sum(k=0, n-1, (-1)^k*binomial(n-1, k)^3*binomial(5*k-(n-1), 3*(n-1)))} \\ Seiichi Manyama, Sep 02 2020
Formula
n^3 * a(n+1) = -(2*n - 1)*(11*n*(n - 1) + 5) * a(n) - 125 * (n - 1)^3 * a(n-1).
a(n*p^k) == (p^3 + Kronecker(p, 5)) * a(n*p^(k-1)) - Kronecker(p, 5) * p^3*a(n*p^(-2)) (mod p^k). [Verrill, 1999]
a(n) = Sum_{k=0..n-1} (-1)^k * binomial(n-1,k)^3 * binomial(5*k-(n-1),3*(n-1)). - Seiichi Manyama, Sep 02 2020
Comments