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A201555

a(n) = C(2*n^2,n^2) = A000984(n^2), where A000984 is the central binomial coefficients.

1, 2, 70, 48620, 601080390, 126410606437752, 442512540276836779204, 25477612258980856902730428600, 23951146041928082866135587776380551750, 365907784099042279561985786395502921046971688680, 90548514656103281165404177077484163874504589675413336841320

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

Central coefficients of triangle A228832.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..40

Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Richard Obláth, Congruences with binomial coefficients, Proceedings of the Indian Academy of Science, Section A, Vol. 1 No. 6, 383-386.

FORMULA

L.g.f.: ignoring initial term, equals the logarithmic derivative of A201556.

a(n) = (2*n^2)! / (n^2)!^2.

a(n) = Sum_{k=0..n^2} binomial(n^2,k)^2.

For primes p >= 5: a(p) == 2 (mod p^3), Oblath, Corollary II; a(p) == binomial(2*p,p) (mod p^6) - see Mestrovic, Section 5, equation 31. - Peter Bala, Dec 28 2014

A007814(a(n)) = A159918(n). - Antti Karttunen, Apr 27 2017, based on Vladimir Shevelev's Jul 20 2009 formula in A000984.

EXAMPLE

L.g.f.: L(x) = 2*x + 70*x^2/2 + 48620*x^3/3 + 601080390*x^4/4 + ...

where exponentiation equals the g.f. of A201556:

exp(L(x)) = 1 + 2*x + 37*x^2 + 16278*x^3 + 150303194*x^4 + ... + A201556(n)*x^n + ...

MATHEMATICA

Table[Binomial[2n^2, n^2], {n, 0, 10}] (* Harvey P. Dale, Dec 10 2011 *)

PROG

(PARI) a(n) = binomial(2*n^2, n^2)

(Python)

from math import comb

def A201555(n): return comb((m:=n**2)<<1, m) # Chai Wah Wu, Jul 08 2022