OFFSET
1,1
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)
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
EXAMPLE
a(1) = 17 because 17 is the least natural number n such that n^3+1 | n!.
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 *)
PROG
(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])<f[i, 2], return(0))); 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
CROSSREFS
KEYWORD
nonn
AUTHOR
José Hernández, Mar 17 2016
STATUS
approved