A105658 a(n) = (Product_{i=1..n} i^i) / denominator( Sum_{j=1..n} j*(j+1)/2 / (Product_{k=0..j-1} j!/k!) ).

1, 1, 1, 3, 2, 5, 3, 7, 4, 9, 5, 11, 6, 143, 7, 15, 104, 935, 9, 19, 10, 21, 11, 4025, 3900, 325, 3289, 27, 14, 29, 15, 31, 368, 33, 17, 35, 18, 185, 19, 39, 380, 451, 399, 215, 770, 45, 23, 29563, 24, 12397, 725, 51, 26, 1537, 837, 2365, 1036, 285, 377, 2537, 30

Offset: 0

Author

Jess E. Boling (tdbpeekitup(AT)yahoo.com), Apr 17 2005

Keywords

• nonn

Comments

Most of the time a(2n-1)=2n-1, but a(2n-1)!=2n-1 for 2n-1 = 13,17,23,25,37,41,43,47,49,53,55,57,59,61,63,...

Most of the time a(2n)=n, but a(2n)!=n for 2n = 16,24,26,32,40,42,44,50,54,56,58,64,84,86,96,100,102,104,...

Examples

			a(3) = 108/36 = 3.
		

Crossrefs

Cf. A002109 (hyperfactorial numbers).

Programs

Mathematica

f[n_] := Product[k^k, {k, 1, n}]/ Denominator[Sum[i(i + 1)/2/Product[i!/j!, {j, 0, i - 1}], {i, n}]]; Table[ f[n], {n, 0, 61}] (* Robert G. Wilson v, Apr 18 2005 *)

PARI

a(n) = prod(i=1, n, i^i) / denominator(sum(j=1, n, j*(j+1)/2 / prod(k=0, j-1, j!/k!))) \\ Jason Yuen, Jan 18 2025

Extensions

Edited by Robert G. Wilson v, Apr 18 2005

Name corrected by Jason Yuen, Jan 18 2025