A168510 Products across consecutive rows of the denominators of the Leibniz harmonic triangle (A003506).

1, 4, 54, 2304, 300000, 116640000, 133413966000, 444110104166400, 4267295479315169280, 117595223746560000000000, 9245836018244425723200000000, 2065215715357207851951980544000000

Offset

1,2

Comments

As in A001142, lim_{n->inf} (a(n)a(n+2))/a(n+1)^2 = e, demonstrating an underlying relation between A003506 and Pascal's triangle A007318. Unlike A001142, in this case the function is asymptotic from above.

Links

Table of n, a(n) for n=1..12.

A. Bogomolny, Cut The Knot: Leibniz and Pascal Triangles

H. J. Brothers, Pascal's prism, The Mathematical Gazette, 96 (July 2012), 213-220.

Formula

a(n) = n!*Product_{k=1..n} k^(2k-n-1).

a(n) = Product_{j=1..n} Product_{k=2..j} ((1-1/k)^-k).

a(1) = 1; a(n) = a(n-1)*Product_{k=2..n} ((1-1/k)^-k).

a(n) ~ A^2 * exp(n^2/2 - 1/12) * n^(n/2 + 1/6) / (2*Pi)^(n/2), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Oct 22 2017

a(n) = Product_{k=0..n-1} (n-k)^(n-2k). - Peter Munn, Mar 07 2018

Example

For n=3, row 3 of A003506 = {3, 6, 3} and a(3)=54.
a(5) = 5^5 * 4^3 * 3^1 * 2^-1 * 1^-3 = 5^5 * 3 * 2^5 = 300000. - Peter Munn, Mar 07 2018

Mathematica

Table[n! Product[k^(2 k - n - 1), {k, 1, n}], {n, 1, 12}]
Table[Product[Product[(1 - 1/k)^-k, {k, 2, j}], {j, 1, n}], {n, 1, 12}]
(* or *)
a[1] = 1; a[n_] := a[n - 1] Product[(1 - 1/k)^-k, {k, 2, n}]

Crossrefs

Cf. A003506, A001142, A007318. For n >= 1, a(n) = n!*A001142(n).

Sequence in context: A355126 A357511 A094154 * A324235 A125531 A095209

Adjacent sequences: A168507 A168508 A168509 * A168511 A168512 A168513

Keyword

easy,nonn

Author

Harlan J. Brothers, Nov 27 2009

Status

approved