A348410 Number of nonnegative integer solutions to n = Sum_{i=1..n} (a_i + b_i), with b_i even.

1, 1, 5, 19, 85, 376, 1715, 7890, 36693, 171820, 809380, 3830619, 18201235, 86770516, 414836210, 1988138644, 9548771157, 45948159420, 221470766204, 1069091485500, 5167705849460, 25009724705460, 121171296320475, 587662804774890, 2852708925078675, 13859743127937876

Offset: 0

Author

César Eliud Lozada, Oct 17 2021

Keywords

• nonn
• easy

Comments

Suppose n objects are to be distributed into 2n baskets, half of these white and half black. White baskets may contain 0 or any number of objects, while black baskets may contain 0 or an even number of objects. a(n) is the number of distinct possible distributions.

Examples

			Some examples (semicolon separates white basket from black baskets):
For n=1: {{1 ; 0}} - Total possible ways: 1.
For n=2: {{0, 0 ; 0, 2}, {0, 0 ; 2, 0}, {0, 2 ; 0, 0}, {1, 1 ; 0, 0}, {2, 0 ; 0, 0}} - Total possible ways: 5.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1441

Crossrefs

Cf. A033715, A033716, A034297, A063020, A088218, A234839, A294860, A348474, A351856, A351857, A352373.

Programs

Maple

b:= proc(n, t) option remember; `if`(t=0, 1-signum(n),
      add(b(n-j, t-1)*(1+iquo(j, 2)), j=0..n))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..25);  # Alois P. Heinz, Oct 17 2021

Mathematica

(* giveList=True produces the list of solutions *)
(* giveList=False gives the number of solutions *)
counter[objects_, giveList_: False] :=
  Module[{n = objects, nb, eq1, eqa, eqb, eqs, var, sol, var2, list},
   nb = n;
   eq1 = {Total[Map[a[#] + 2*b[#] &, Range[nb]]] - n == 0};
   eqa = {And @@ Map[0 <= a[#] <= n &, Range[nb]]};
   eqb = {And @@ Map[0 <= b[#] <= n &, Range[nb]]};
   eqs = {And @@ Join[eq1, eqa, eqb]};
   var = Flatten[Map[{a[#], b[#]} &, Range[nb]]];
   var = Join[Map[a[#] &, Range[nb]], Map[b[#] &, Range[nb]]];
   sol = Solve[eqs, var, Integers];
   var2 = Join[Map[a[#] &, Range[nb]], Map[2*b[#] &, Range[nb]]];
   list = Sort[Map[var2 /. # &, sol]];
   list = Map[StringReplace[ToString[#], {"," -> " ;"}, n] &, list];
   list = Map[StringReplace[#, {";" -> ","}, n - 1] &, list];
   Return[
    If[giveList, Print["Total: ", Length[list]]; list, Length[sol]]];
   ];
(* second program: *)
b[n_, t_] := b[n, t] = If[t == 0, 1 - Sign[n], Sum[b[n - j, t - 1]*(1 + Quotient[j, 2]), {j, 0, n}]];
a[n_] := b[n, n];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Aug 16 2023, after Alois P. Heinz *)

Formula

Conjecture: D-finite with recurrence +7168*n*(2*n-1)*(n-1)*a(n) -64*(n-1)*(1759*n^2-5294*n+5112)*a(n-1) +12*(7561*n^3-75690*n^2+165271*n-101070)*a(n-2) +5*(110593*n^3-743946*n^2+1659971*n-1232778)*a(n-3) +2680*(4*n-15)*(2*n-7)*(4*n-13)*a(n-4)=0. - R. J. Mathar, Oct 19 2021

From Vaclav Kotesovec, Nov 01 2021: (Start)

Recurrence (of order 2): 16*(n-1)*n*(2*n - 1)*(51*n^2 - 162*n + 127)*a(n) = (n-1)*(5457*n^4 - 22791*n^3 + 32144*n^2 - 17536*n + 3072)*a(n-1) + 8*(2*n - 3)*(4*n - 7)*(4*n - 5)*(51*n^2 - 60*n + 16)*a(n-2).

a(n) ~ sqrt(3 + 5/sqrt(17)) * (107 + 51*sqrt(17))^n / (sqrt(Pi*n) * 2^(6*n+2)). (End)

From Peter Bala, Feb 21 2022: (Start)

a(n) = [x^n] ( (1 - x)*(1 - x^2) )^(-n). Cf. A234839.

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

exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 9*x^3 + 32*x^4 + 119*x^5 + ... is the g.f. of A063020.

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) is the diagonal of the bivariate rational function 1/(1 - t/((1-x)*(1-x^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)*(1-x^2) ). Then A(x) = 1 + x*d/dx (log(F(x))). (End)

a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n+k-1, k)*binomial(2*n-k-1, n-k). Cf. A352373. - Peter Bala, Jun 05 2024

Extensions

More terms from Alois P. Heinz, Oct 17 2021