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A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_(2*n).

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%I A351857 #20 Mar 22 2025 08:39:14
%S A351857 2,14,92,654,4752,35204,264112,2000526,15264866,117161264,903533380,
%T A351857 6995547780,54343476072,423360920528,3306313730592,25876855432846,
%U A351857 202909132368942,1593755466338030,12537009118650016,98753463725849904,778825917274945408,6149069826564738780
%N A351857 Number of nonnegative integer solutions to n = x_1 + x_2 + ... + x_(2*n) + 2*y_1 + 2*y_2 + ... + 2*y_(2*n).
%C A351857 This is a companion sequence to A348410.
%C A351857 Suppose n identical objects are distributed in 4*n labeled baskets, 2*n colored white and 2*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 A351857 R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
%F A351857 a(n) = [x^n] ( 1/((1 - x)*(1 - x^2)) )^(2*n).
%F A351857 a(n) = Sum_{k = 0..floor(n/2)} C(3*n-2*k-1,n-2*k)*C(2*n+k-1,k).
%F A351857 a(n) = Sum_{k = 0..n} (-1)^k*C(5*n-k-1,n-k)*C(2*n+k-1,k).
%F A351857 1024*n*(n-1)*(2*n-1)*(2*n-3)*(4*n-1)*(4*n-3)*P(n-1)*a(n) = 8*(n-1)*(2*n-3)*Q(n)*a(n-1) + 7*(7*n-8)*(7*n-9)*(7*n-10)*(7*n-11)*(7*n-12)*(7*n-13)*P(n)*a(n-2), with a(1) = 2, a(2) = 14, P(n) = 1744*n^4-3815*n^3+ 2920*n^2-912*n+96 and Q(n) = 46599680*n^8-381534880*n^7+1306363456*n^6- 2428492279*n^5+2661904813*n^4 -1747232452*n^3+664205312*n^2- 132046848*n+10321920.
%F A351857 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 A351857 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 A351857 The o.g.f. A(x) = 2*x + 14*x^2 + 92*x^3 + ... is the diagonal of the bivariate rational function x*t/(1 - t/((1 - x)*(1 - x^2))^2 ) and hence is an algebraic function over Q(x) by Stanley 1999, Theorem 6.33, p. 197.
%F A351857 Let F(x) = (1/x)*Series_Reversion( x*(1 - x)^2*(1 - x^2)^2 ) = A365855(x). Then A(x) = x*(d/dx log(F(x))). [corrected by _Jason Yuen_, Mar 22 2025]
%e A351857 n = 2: 14 distributions of 2 identical objects in 4 white and 4 black baskets
%e A351857              White             Black
%e A351857    1)   (0) (0) (0) (0)   [2] [0] [0] [0]
%e A351857    2)   (0) (0) (0) (0)   [0] [2] [0] [0]
%e A351857    3)   (0) (0) (0) (0)   [0] [0] [2] [0]
%e A351857    4)   (0) (0) (0) (0)   [0] [0] [0] [2]
%e A351857    5)   (2) (0) (0) (0)   [0] [0] [0] [0]
%e A351857    6)   (0) (2) (0) (0)   [0] [0] [0] [0]
%e A351857    7)   (0) (0) (2) (0)   [0] [0] [0] [0]
%e A351857    8)   (0) (0) (0) (2)   [0] [0] [0] [0]
%e A351857    9)   (1) (1) (0) (0)   [0] [0] [0] [0]
%e A351857   10)   (1) (0) (1) (0)   [0] [0] [0] [0]
%e A351857   11)   (1) (0) (0) (1)   [0] [0] [0] [0]
%e A351857   12)   (0) (1) (1) (0)   [0] [0] [0] [0]
%e A351857   13)   (0) (1) (0) (1)   [0] [0] [0] [0]
%e A351857   14)   (0) (0) (1) (1)   [0] [0] [0] [0]
%p A351857 seq(add( binomial(3*n-2*k-1,n-2*k)*binomial(2*n+k-1,k), k = 0..floor(n/2) ), n = 0..20);
%o A351857 (PARI) a(n) = sum(k = 0, n\2, binomial(3*n-2*k-1, n-2*k)*binomial(2*n+k-1, k)); \\ _Michel Marcus_, Feb 27 2022
%Y A351857 Cf. A234839, A348410, A351856, A365855.
%K A351857 nonn,easy
%O A351857 1,1
%A A351857 _Peter Bala_, Feb 22 2022