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).

2, 14, 92, 654, 4752, 35204, 264112, 2000526, 15264866, 117161264, 903533380, 6995547780, 54343476072, 423360920528, 3306313730592, 25876855432846, 202909132368942, 1593755466338030, 12537009118650016, 98753463725849904, 778825917274945408, 6149069826564738780

Offset: 1

Author

Peter Bala, Feb 22 2022

Keywords

• nonn
• easy

Comments

This is a companion sequence to A348410.

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.

Examples

			n = 2: 14 distributions of 2 identical objects in 4 white and 4 black baskets
             White             Black
   1)   (0) (0) (0) (0)   [2] [0] [0] [0]
   2)   (0) (0) (0) (0)   [0] [2] [0] [0]
   3)   (0) (0) (0) (0)   [0] [0] [2] [0]
   4)   (0) (0) (0) (0)   [0] [0] [0] [2]
   5)   (2) (0) (0) (0)   [0] [0] [0] [0]
   6)   (0) (2) (0) (0)   [0] [0] [0] [0]
   7)   (0) (0) (2) (0)   [0] [0] [0] [0]
   8)   (0) (0) (0) (2)   [0] [0] [0] [0]
   9)   (1) (1) (0) (0)   [0] [0] [0] [0]
  10)   (1) (0) (1) (0)   [0] [0] [0] [0]
  11)   (1) (0) (0) (1)   [0] [0] [0] [0]
  12)   (0) (1) (1) (0)   [0] [0] [0] [0]
  13)   (0) (1) (0) (1)   [0] [0] [0] [0]
  14)   (0) (0) (1) (1)   [0] [0] [0] [0]
		

References

• R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Cf. A234839, A348410, A351856, A365855.

Programs

Maple

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);

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

Formula

a(n) = [x^n] ( 1/((1 - x)*(1 - x^2)) )^(2*n).

a(n) = Sum_{k = 0..floor(n/2)} C(3*n-2*k-1,n-2*k)*C(2*n+k-1,k).

a(n) = Sum_{k = 0..n} (-1)^k*C(5*n-k-1,n-k)*C(2*n+k-1,k).

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.

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.

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.

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.

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]