A056612 a(n) = gcd(n!, n!*(1 + 1/2 + 1/3 + ... + 1/n)).

1, 1, 1, 2, 2, 36, 36, 144, 144, 1440, 1440, 17280, 17280, 241920, 3628800, 29030400, 29030400, 1567641600, 1567641600, 156764160000, 9876142080000, 217275125760000, 217275125760000, 1738201006080000, 1738201006080000

Offset: 1

Author

Leroy Quet, Aug 08 2000

Keywords

• easy
• nonn

Comments

The first difference between this sequence and A131657 occurs for n = 20, while the first difference between this sequence and A131658 occurs for n = 21. - Christian Krattenthaler, Sep 30 2007

Examples

			a(4) = gcd(4!, 4!*(1 + 1/2 + 1/3 + 1/4)) = gcd(24, 50) = 2.
a(4) = gcd(A000254(5), A000254(4)) = gcd(5!*(1 + 1/2 + 1/3 + 1/4 + 1/5), 4!*(1 + 1/2 + 1/3 + 1/4)) = gcd(274, 50) = 2. - _Petros Hadjicostas_, May 18 2020
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..535

Crossrefs

Cf. A000142, A001008, A002805.

Cf. A007757, A131657, A131658.

Cf. A334958 (similar sequence for the alternative harmonic series).

Programs

Mathematica

Table[GCD[#, # Total@ Map[1/# &, Range@ n]] &[n!], {n, 25}] (* Michael De Vlieger, Sep 23 2017 *)
a[n_] := n!/Denominator@ HarmonicNumber@ n; Array[a, 25] (* Robert G. Wilson v, Jun 30 2018 *)

PARI

a(n) = gcd(n!, n!*sum(k=1, n, 1/k)); \\ Michel Marcus, Jul 14 2018

PARI

a(n) = gcd(stirling(n+1, 2, 1), n!); \\ Michel Marcus, May 20 2020

Formula

a(n) = A000142(n)/A002805(n) = A000254(n)/A001008(n). - Franz Vrabec, Sep 13 2005

a(n) = gcd(A000254(n+1), A000254(n)). - Petros Hadjicostas, May 18 2020

a(n) = gcd(Stirling1(n+1, 2), n!). - Michel Marcus, May 20 2020