A120416 Numbers k such that k^2+1 divides k!.

18, 21, 38, 43, 47, 57, 68, 70, 72, 73, 83, 99, 111, 117, 119, 123, 128, 132, 133, 142, 157, 172, 173, 174, 182, 185, 191, 192, 193, 200, 211, 212, 216, 233, 237, 239, 242, 251, 253, 255, 265, 268, 273, 278, 293, 294, 302, 305, 307, 313, 319, 322, 327, 336

Offset: 1

Author

R. J. Mathar, Jul 07 2006

Keywords

• easy
• nonn

Comments

Let Product_j (p_j)^(e_j) be the prime factorization of n^2+1. Then n is in the sequence if and only if for each j, e_j <= Sum_{k>=1} floor(n/(p_j)^k). - Robert Israel, Nov 11 2016

There exist infinitely many natural numbers n such that n^2+1 divides n!, because 2*(5*k-2)^2 is a term for k > 0. - Jinyuan Wang, Feb 06 2019

There are 2082 terms up to 10^4, 22792 up to 10^5, 242421 up to 10^6, 2523043 up to 10^7. Perhaps the asymptotic density is 1 - log(2) = 30.68...%. - Jinyuan Wang, Feb 09 2019

Examples

			For the first number, k=18: 18^2+1=325 divides 18!=6402373705728000.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Hojoo Lee, Problem A 26, Problems in elementary number theory, 2003.

Crossrefs

Cf. A002522, A000142, A118742, A270441.

Programs

Maple

select(t -> t! mod (t^2+1)=0, [$1..1000]); # Robert Israel, Nov 11 2016

Mathematica

Select[Range@ 336, Divisible[#!, #^2 + 1] &] (* Jinyuan Wang, Feb 06 2019 *)

PARI

isok(n) = (n! % (n^2+1) == 0) \\ Michel Marcus, Jul 23 2013

PARI

valp(n, p)=my(s); while(n>=p, s += n\=p); s
is(n)=my(f=factor(n^2+1)); for(i=1, #f~, if(valp(n, f[i, 1])Jinyuan Wang, Feb 06 2019

Formula

k such that A002522(k) | A000142(k).