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A120416 Numbers k such that k^2+1 divides k!. 7
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified July 28 23:07 EDT 2021. Contains 346340 sequences. (Running on oeis4.)