OFFSET
1,1
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
EXAMPLE
a(4) = 77 because 77!^(77 + 4) divides (77^2)! and 77 is the smallest integer m, m > 1, with this property.
PROG
(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
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Farideh Firoozbakht, Aug 24 2014
EXTENSIONS
a(9)-a(13) from Jens Kruse Andersen, Aug 31 2014
a(14)-a(27) from Hiroaki Yamanouchi, Sep 29 2014
STATUS
approved