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%I #16 Oct 23 2022 11:55:22
%S 1,1,3,19,67,251,1137,4803,20035,87013,377753,1634469,7134385,
%T 31261114,137121113,603206144,2660097603,11749336328,51981371895,
%U 230336544210,1021976441817,4539784391763,20188837618799,89871081815631,400427435522737,1785639575031501
%N a(n) = [x^n] (1 + x + x^2 + x^3)^(4*n)/(1 + x + x^2)^(3*n).
%C This sequence is the third in an infinite family of sequences defined as follows. Let k be a positive integer. Define the rational function G_k(x) = (1 + x + ... + x^k)^(k+1)/(1 + x + ... + x^(k-1))^k, so that G_1(x) = (1 + x)^2, and define the sequence u_k by u_k(n) = [x^n] G_k(x)^n. See A000984, the sequence of central binomial coefficients, for the case k = 1 and A351858 for the case k = 2. The present sequence is the case k = 3.
%C Given a power series G(x) with integer coefficients it is known that the sequence (g(n))n>=1 defined by g(n) := [x^n] G(x)^n satisfies the Gauss congruences g(n*p^r) == g(n*p^(r-1) (mod p^r) for all primes p and positive integers n and r.
%C Thus a(n) satisfies the Gauss congruences. Calculation suggests that, in fact, the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 5 and positive integers n and r. These supercongruences are known to hold for the central binomial coefficients A000984(n) = [x^n] ((1 + x)^2)^n (Meštrović, equation 39).
%C More generally, if r is a positive integer and s an integer then the sequence defined by a(r,s;n) = [x^(r*n)] G_3(x)^(s*n) may satisfy the same supercongruences.
%D R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
%H Paolo Xausa, <a href="/A351859/b351859.txt">Table of n, a(n) for n = 0..425</a>
%H R. Meštrović, <a href="https://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012)</a>, arXiv:1111.3057 [math.NT], 2011.
%F a(n) = Sum_{i = 0..n} Sum_{j = 0..n} Sum_{k = 0..j} (-1)^j* C(4n,n-2*i-j-k) *C(4n,i)*C(3n+j-1,j)*C(j,k).
%F The o.g.f. A(x) = 1 + x + 3*x^2 + 19*x^3 + 67*x^4 + ... is the diagonal of the bivariate rational function 1/(1 - t*(1 + x + x^2 + x^3)^4/(1 + x + x^2)^3) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.
%F Let F(x) = (1/x)*Series_Reversion( x*(1 + x + x^2)^3/(1 + x + x^2 + x^3)^4 ) = 1 + x + 2*x^2 + 8*x^3 + 25*x^4 + 81*x^5 + 305*x^6 + .... Then A(x) = 1 + x*F'(x)/F(x).
%e Examples of supercongruences:
%e a(5) - a(1) = 251 - 1 = 2*(5^3) == 0 (mod 5^3)
%e a(2*7)- a(2) = 137121113 - 3 = 2*5*(7^4)*5711 == 0 (mod 7^4)
%e a(5^2) - a(5) = 1785639575031501 - 251 = 2*(5^6)*1373*3989*10433 == 0 (mod 5^6)
%p seq(add(add(add((-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)* binomial(3*n+j-1,j)*binomial(j,k), k = 0..j), j = 0..n), i = 0..n), n = 0..25);
%t nterms=25;Table[Sum[Sum[Sum[(-1)^j*Binomial[4n,n-2i-j-k]Binomial[4n,i]Binomial[3n+j-1,j]Binomial[j,k],{k,0,j}],{j,0,n}],{i,0,n}],{n,0,nterms-1}] (* _Paolo Xausa_, Apr 11 2022 *)
%o (PARI)
%o a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,j,(-1)^j*binomial(4*n,n-2*i-j-k)*binomial(4*n,i)*binomial(3*n+j-1,j)*binomial(j,k))));
%o vector(25,n,a(n-1)) \\ _Paolo Xausa_, May 04 2022
%Y Cf. A000984, A103885, A333090, A333564, A333565, A333579, A333715, A351856, A351857, A351858.
%K nonn,easy
%O 0,3
%A _Peter Bala_, Mar 01 2022
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