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

1,1

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

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Hojoo Lee, Problem A 26, Problems in elementary number theory, 2003.

Formula

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

Example

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

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 *)

Prog

(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])<f[i, 2], return(0))); 1 \\ Jinyuan Wang, Feb 06 2019

Crossrefs

Cf. A002522, A000142, A118742, A270441.

Sequence in context: A088341 A105145 A244603 * A293751 A090891 A303298

Adjacent sequences: A120413 A120414 A120415 * A120417 A120418 A120419

Keyword

easy,nonn

Author

R. J. Mathar, Jul 07 2006

Status

approved