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A107863
Column 1 of triangle A107862; a(n) = binomial(n*(n+1)/2 + n, n).
4
1, 2, 10, 84, 1001, 15504, 296010, 6724520, 177232627, 5317936260, 179013799328, 6681687099710, 273897571557780, 12233149001721760, 591315394579074378, 30756373941461374800, 1712879663609111933495, 101696990867999141755140
OFFSET
0,2
FORMULA
a(n) = [x^(n*(n+1)/2)] 1/(1 - x)^(n+1). - Ilya Gutkovskiy, Oct 10 2017
From Peter Bala, Feb 23 2020: (Start)
Put b(n) = a(n-1). We have the congruences:
b(p) == 1 (mod p^3) for prime p >= 5 (uses Mestrovic, equation 35);
b(2*p) == 2*p (mod p^4) for prime p >= 5 (uses Mestrovic, equation 44 and the von Staudt-Clausen theorem).
Conjectural congruences:
b(3*p) == (81*p*2 - 1)/8 (mod p^3) for prime p >= 3;
3*b(4*p) == -4*p (mod p^3) for all prime p. Cf. A135860 and A135861. (End)
MATHEMATICA
Table[Binomial[n*(n+3)/2, n], {n, 0, 40}] (* G. C. Greubel, Feb 19 2022 *)
PROG
(PARI) a(n)=binomial(n*(n+1)/2+n, n)
(Sage) [binomial(n*(n+3)/2, n) for n in (0..40)] # G. C. Greubel, Feb 19 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jun 04 2005
STATUS
approved