A073617 Consider Pascal's triangle A007318; a(n) = product of terms at +45 degrees slope with the horizontal.

1, 1, 1, 2, 3, 12, 30, 240, 1050, 16800, 132300, 4233600, 61122600, 3911846400, 104886381600, 13425456844800, 674943865596000, 172785629592576000, 16407885372638760000, 8400837310791045120000, 1515727634953623371280000, 1552105098192510332190720000

Offset: 0

Author

Amarnath Murthy, Aug 07 2002

Keywords

• nonn

Comments

The sum of the terms pertaining to the above product is the (n+1)-th Fibonacci number: 1 + 5 + 6 + 1 = 13.

n divides A073617(n+1) for n>=1; see the Mathematica section. [Clark Kimberling, Feb 29 2012]

Examples

			For n=6 the diagonal is 1,5,6,1 and product of the terms = 30 hence a(6) = 30.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..118

Crossrefs

Cf. A000045, A073618, A007685, A208649, A000984.

Programs

Maple

a:= n-> mul(binomial(n-i, i), i=0..floor(n/2)):
seq(a(n), n=0..21);  # Alois P. Heinz, Nov 27 2023

Mathematica

p[n_] := Product[Binomial[n + 1 - k, k], {k, 1, Floor[(n + 1)/2]}]
Table[p[n], {n, 1, 20}]   (* A073617(n+1) *)
Table[p[n]/n, {n, 1, 20}] (* A208649 *)
(* Clark Kimberling, Feb 29 2012 *)
(* Second program *)
Join[{1}, Table[If[EvenQ[n], 2^(3/2 - n/4) * Sqrt[BarnesG[n]] * Gamma[n] / (n*BarnesG[n/2]^2 * Gamma[n/2]^(7/2)), Glaisher^3 * 2^((-10 + 3*n + 6*n^2)/12) * BarnesG[n/2]^2 * Gamma[n/2]^(5/2) / (E^(1/4) * Pi^(1/4 + n/2) * Sqrt[BarnesG[n]] * Gamma[n])], {n, 1, 25}]] (* Vaclav Kotesovec, Jun 10 2025 *)

Formula

a(n) = Product_{k=0..floor(n/2)} binomial(n-k,k).

a(2n+1)/a(2n-1) = binomial(2n,n); a(2n)/a(2n-2) = (1/2)*binomial(2n,n); (a(2n+1)*a(2n-2))/(a(2n)*a(2n-1)) = 2. - John Molokach, Sep 09 2013

a(n) ~ A^(3/2) * 2^(n*(n+1)/4 - 1/6 + (-1)^n/4) * exp(n/4 - 1/8) / (n^((n+1)/4 + (-1)^n/8) * Pi^(n/4 + 3/8 + (-1)^n/8)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Jun 10 2025

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 22 2003