A244443 Smallest integer m > 1 such that m!^(m + n) divides (m^2)!.

2, 6, 15, 77, 185, 187, 475, 3820, 4043, 4090, 11231, 30589, 57023, 126815, 131055, 983032, 983033, 2617339, 4046839, 11534206, 11534207, 65011702, 66777087, 368279551, 469745405, 973061887, 1064828671

Offset: 1

Author

Farideh Firoozbakht, Aug 24 2014

Keywords

• nonn
• more
• hard

Comments

The constraint m > 1 is necessary because (1^2)! = 1.

The motivation for this sequence came from comments on the sequence A246048 by M. F. Hasler.

The integer (3820^2)!/(3820!)^3828 related to a(8) has 52166326 digits, so it isn't easy to find more terms.

a(28) > 1.5 * 10^9. - Hiroaki Yamanouchi, Sep 29 2014

Examples

			a(4) = 77 because 77!^(77 + 4) divides (77^2)! and 77 is the smallest integer m, m > 1, with this property.
		

Crossrefs

Cf. A096126, A096127, A246048.

Programs

PARI

for(n=1, 7, m=2; while((m^2)!%(m!^(m+n)), m++); print1(m", ")) \\ Jens Kruse Andersen, Aug 31 2014

PARI

n=f=1; for(m=2, 5000, f*=m; s=m^2; forprime(p=2, m, e=0; b=p; while(b<=s, e+=s\b; b*=p); if(valuation(f,p)*(m+n)>e, next(2))); print1(m", "); n++) \\ Faster program. Jens Kruse Andersen, Aug 31 2014

Extensions

a(9)-a(13) from Jens Kruse Andersen, Aug 31 2014

a(14)-a(27) from Hiroaki Yamanouchi, Sep 29 2014