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
Offset: 0
Keywords
Examples
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 + ...
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..40
- R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
- R. Oblath, Congruences with binomial coefficients, Proceedings of the Indian Academy of Science, Section A, Vol. 1 No. 6, 383-386
Programs
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Mathematica
Table[Binomial[2n^2,n^2],{n,0,10}] (* Harvey P. Dale, Dec 10 2011 *)
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PARI
a(n) = binomial(2*n^2,n^2)
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Python
from math import comb def A201555(n): return comb((m:=n**2)<<1,m) # Chai Wah Wu, Jul 08 2022
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.
Comments