A260610 Suprafactorials: Product of first n hyperfactorials divided by the product of the first n superfactorials.
1, 1, 2, 18, 1728, 4320000, 699840000000, 18525482136000000000, 204051433560311070720000000000, 2399547398533110254947300351672320000000000, 77759951835586717141477466390085274435584000000000000000000, 18011357710498321908881994832212360081640749122627567616000000000000000000000000
Offset: 0
Examples
a(3) = (Hyperfactorial(3)/Superfactorial(3)) * (Hyperfactorial(2)/Superfactorial(2)) * (Hyperfactorial(1)/Superfactorial(1)) * (Hyperfactorial(0)/Superfactorial(0)) = ((3^3 * 2^2 * 1^1)/(3! * 2! * 1!)) * ((2^2 * 1^1)/(2!*1!)) * (1^1/1!) * 1 = ((27 * 4)/(6 * 2)) * (4/2) * 1 = (108/12) * (4/2) = 9 * 2 = 18.
Links
- Matthew Campbell, Table of n, a(n) for n = 0..24
Programs
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Mathematica
Table[Product[Hyperfactorial[n]/BarnesG[n+2], {n, 0, m}], {m, 0, 12}] Table[BarnesG[n+2]^(n-1) / Product[BarnesG[k]^3, {k, 1, n + 1}], {n, 0, 12}] (* Vaclav Kotesovec, Nov 19 2023 *) -
PARI
a001142(n) = prod(m=1, n, binomial(n, m)); a(n) = prod(k=0, n, a001142(k)); \\ Michel Marcus, Aug 06 2015
Formula
a(n) = Product_{k=0..n} A001142(k).
a(n) = Product_{k=0..n} hyperfactorial(k)/superfactorial(k).
a(n) = Product_{i=1..n} (Product_{j=1..i} binomial(i,j)). - Pedro Caceres, Apr 13 2019
From Vaclav Kotesovec, Nov 19 2023: (Start)
a(n) = BarnesG(n+2)^(n-1) / Product_{k=1..n+1} BarnesG(k)^3.
a(n) ~ A^(2*n + 5/2) * exp(n^3/6 + 7*n^2/8 + 5*n/6 - 3*zeta(3)/(8*Pi^2) - 1/8) / ((2*Pi)^(n^2/4 + 3*n/4 + 1/2) * n^(n^2/4 + 7*n/12 + 7/24)), where A is the Glaisher-Kinkelin constant A074962. (End)
Comments