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A351856 Number of nonnegative integer solutions to 2*n = x_1 + x_2 + ... + x_n + 2*y_1 + 2*y_2 + ... + 2*y_n.

%I A351856 #15 Mar 03 2022 11:43:51
%S A351856 2,14,119,1086,10252,98735,963832,9502014,94386908,943206264,
%T A351856 9471346755,95491466655,966026045376,9800968460024,99685873633744,
%U A351856 1016118049037630,10377363759903252,106161722891946356,1087696666197827374,11159365823946907336,114631982782490824420
%N A351856 Number of nonnegative integer solutions to 2*n = x_1 + x_2 + ... + x_n + 2*y_1 + 2*y_2 + ... + 2*y_n.
%C A351856 This is a companion sequence to A348410.
%C A351856 Suppose 2*n identical objects are distributed in 2*n labeled baskets, n colored white and n colored black. White baskets can contain any number of objects (or be empty), while black baskets must contain an even number of objects (or be empty). a(n) is the number of distinct possible distributions.
%D A351856 R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
%F A351856 a(n) = [x^(2*n)] ( 1/((1 - x)*(1 - x^2)) )^n.
%F A351856 a(n) = Sum_{k = 0..n} C(3*n-2*k-1,2*n-2*k)*C(n+k-1,k).
%F A351856 a(n) = Sum_{k = 0..2*n} (-1)^k*C(4*n-k-1,2*n-k)*C(n+k-1,k).
%F A351856 32*n*(n-1)*(2*n-1)*(2*n-3)*(41*n^2-126*n+93)*a(n) = 2*(n-1)*(2*n-3)*(16851*n^4-68637*n^3+93680*n^2-49024*n+7680)*a(n-1) - 5*(5*n-9)*(5*n-8)*(5*n-7)*(5*n-6)*(41*n^2-44*n+8)*a(n-2) with a(1) = 2 and a(2) = 14.
%F A351856 The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k.
%F A351856 Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k.
%F A351856 The o.g.f. A(x) = 2*x + 14*x^2 + 119*x^3 + ... is the diagonal of the bivariate rational function x*t*(x - 1)*((x - 1)^2 - t)/((x - 1)^3 - t*(2*x + t - 2)) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
%F A351856 Let F(x) = (1/x)*Series_Reversion( x*sqrt((1-x)*(1-x^2)) ) and put G(x) = x*(d/dx)(log(F(x))). Then A(x^2) = (G(x) + G(-x))/2.
%e A351856 n = 2: 14 distributions of 4 identical objects in 2 white and 2 black baskets
%e A351856         White     Black
%e A351856    1)  (0) (0)   [4] [0]
%e A351856    2)  (0) (0)   [0] [4]
%e A351856    3)  (0) (0)   [2] [2]
%e A351856    4)  (2) (0)   [2] [0]
%e A351856    5)  (0) (2)   [2] [0]
%e A351856    6)  (1) (1)   [2] [0]
%e A351856    7)  (2) (0)   [0] [2]
%e A351856    8)  (0) (2)   [0] [2]
%e A351856    9)  (1) (1)   [0] [2]
%e A351856   10)  (4) (0)   [0] [0]
%e A351856   11)  (0) (4)   [0] [0]
%e A351856   12)  (3) (1)   [0] [0]
%e A351856   13)  (1) (3)   [0] [0]
%e A351856   14)  (2) (2)   [0] [0]
%p A351856 seq( add(binomial(3*n-2*k-1,2*n-2*k)*binomial(n+k-1,k), k = 0..n), n = 1..20);
%Y A351856 Cf. A234839, A348410, A351857.
%K A351856 nonn,easy
%O A351856 1,1
%A A351856 _Peter Bala_, Feb 22 2022