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

0,3

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.

Links

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

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)

Example

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.

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 *)

Crossrefs

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

Sequence in context: A149794 A149795 A149796 * A005191 A324595 A275027

Adjacent sequences: A348407 A348408 A348409 * A348411 A348412 A348413

Keyword

nonn,easy

Author

César Eliud Lozada, Oct 17 2021

Extensions

More terms from Alois P. Heinz, Oct 17 2021

Status

approved