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A168510 Products across consecutive rows of the denominators of the Leibniz harmonic triangle (A003506). 2
1, 4, 54, 2304, 300000, 116640000, 133413966000, 444110104166400, 4267295479315169280, 117595223746560000000000, 9245836018244425723200000000, 2065215715357207851951980544000000 (list; graph; refs; listen; history; text; internal format)
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: A003955 A182264 A094154 * A125531 A095209 A107101

Adjacent sequences:  A168507 A168508 A168509 * A168511 A168512 A168513

KEYWORD

easy,nonn

AUTHOR

Harlan J. Brothers, Nov 27 2009

STATUS

approved

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Last modified May 24 10:13 EDT 2018. Contains 304521 sequences. (Running on oeis4.)