A270441 Numbers n such that n^3+1 divides n!.

17, 31, 50, 68, 69, 75, 80, 101, 103, 122, 147, 155, 159, 160, 164, 170, 173, 179, 182, 212, 230, 231, 236, 257, 263, 264, 274, 278, 293, 302, 325, 327, 335, 353, 362, 373, 374, 381, 394, 407, 411, 424, 431, 437, 440, 451, 459, 467, 471, 472, 485, 491, 495, 500

Offset: 1

Author

José Hernández, Mar 17 2016

Keywords

• nonn

Comments

There exist infinitely many natural numbers n such that n^3+1 divides n!, because for k > 0, (3*k+1)^2 + 1 and 16*k^4 + 1 are terms. (Edited by Jinyuan Wang, Feb 05 2019)

There are 1738 members up to 10^4, 19912 up to 10^5, 216921 up to 10^6, 2299173 up to 10^7, and 23960698 up to 10^8. Perhaps the asymptotic density is 1 - log 2 = 30.68...%. - Charles R Greathouse IV, Apr 05 2016 (Edited by Jinyuan Wang, Feb 06 2019)

Examples

			a(1) = 17 because 17 is the least natural number n such that n^3+1 | n!.
		

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A118742, A120416, A269665.

Programs

Maple

A270441:=n->`if`(n! mod (n^3+1) = 0, n, NULL): seq(A270441(n), n=1..800); # Wesley Ivan Hurt, Apr 02 2016

Mathematica

For[n = 1, n <= 500, n++, If[Mod[n!, n^3 + 1] == 0, Print[n]]]
Select[Range@ 500, Divisible[#!, #^3 + 1] &] (* Michael De Vlieger, Mar 17 2016 *)

PARI

isok(n) = (n! % (n^3+1)) == 0; \\ Michel Marcus, Mar 17 2016

PARI

my(f=1);for(n=2,10^3,f*=n;if(f%(n^3+1)==0,print1(n,", "))); \\ Joerg Arndt, Apr 03 2016

PARI

valp(n,p)=my(s); while(n>=p, s += n\=p); s
is(n)=if(isprime(n+1), return(0)); my(f=factor(n^2-n+1)); for(i=1,#f~, if(valp(n,f[i,1])Charles R Greathouse IV, Apr 04 2016

Python

from math import factorial
for n in range(2,1000):
    if(factorial(n)%(n**3+1)==0):print(n)
# Soumil Mandal, Apr 03 2016