A084081 Sum of lists created by n substitutions k -> Range[k+1,0,-2] starting with {0}, counting down from k+1 to 0 step -2.

0, 1, 2, 5, 10, 24, 50, 121, 260, 637, 1400, 3468, 7752, 19380, 43890, 110561, 253000, 641355, 1480050, 3771885, 8765250, 22439040, 52451256, 134796060, 316663760, 816540124, 1926501200, 4982228488, 11798983280, 30593078076, 72690164850

Offset

0,3

Comments

Lengths of lists is A047749.

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Formula

Equals A093951(n) - A047749(n).

From G. C. Greubel, Oct 17 2022: (Start)

a(2*n+1) = (3*n-1)*binomial[3*n+1, n]/((n+1)*(3*n+1)).

a(2*n) = 10*binomial(3*n+1, n-1)/(2*n+3). (End)

Example

Lists {0}, {1}, {2, 0}, {3, 1, 1}, {4, 2, 0, 2, 0, 2, 0} sum to 0, 1, 2, 5, 10.

Mathematica

Plus@@@Flatten/@NestList[ # /. k_Integer :> Range[k+1, 0, -2]&, {0}, 8]
A084081[n_]:= If[EvenQ[n], 10*Binomial[(3*n+2)/2, (n-2)/2]/(n+3), 2*(3*n + 1)*Binomial[(3*n+5)/2, (n+1)/2]/((n+3)*(3*n+5))];
Table[A084081[n], {n, 40}] (* G. C. Greubel, Oct 17 2022 *)

Prog

(Magma)
F:=Floor; B:=Binomial;
function A084081(n)
  if (n mod 2) eq 0 then return 10*B(F((3*n+2)/2), F((n-2)/2))/(n+3);
  else return 2*(3*n+1)*B(F((3*n+5)/2), F((n+1)/2))/((n+3)*(3*n+5));
  end if; return A084081;
end function;
[A084081(n): n in [0..40]]; // G. C. Greubel, Oct 17 2022
(SageMath)
def A084081(n):
    if (n%2==0): return 10*binomial(int((3*n+2)/2), int((n-2)/2))/(n+3)
    else: return 2*(3*n+1)*binomial(int((3*n+5)/2), int((n+1)/2))/((n+3)*(3*n+5))
[A084081(n) for n in range(40)] # G. C. Greubel, Oct 17 2022

Crossrefs

Cf. A047749, A093951.

Sequence in context: A291247 A316697 A032170 * A106376 A359068 A151514

Adjacent sequences: A084078 A084079 A084080 * A084082 A084083 A084084

Keyword

nonn

Author

Wouter Meeussen, May 11 2003

Status

approved