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%I #37 Feb 06 2019 03:27:51
%S 17,31,50,68,69,75,80,101,103,122,147,155,159,160,164,170,173,179,182,
%T 212,230,231,236,257,263,264,274,278,293,302,325,327,335,353,362,373,
%U 374,381,394,407,411,424,431,437,440,451,459,467,471,472,485,491,495,500
%N Numbers n such that n^3+1 divides n!.
%C 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)
%C 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)
%H Charles R Greathouse IV, <a href="/A270441/b270441.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 17 because 17 is the least natural number n such that n^3+1 | n!.
%p A270441:=n->`if`(n! mod (n^3+1) = 0, n, NULL): seq(A270441(n), n=1..800); # _Wesley Ivan Hurt_, Apr 02 2016
%t For[n = 1, n <= 500, n++, If[Mod[n!, n^3 + 1] == 0, Print[n]]]
%t Select[Range@ 500, Divisible[#!, #^3 + 1] &] (* _Michael De Vlieger_, Mar 17 2016 *)
%o (PARI) isok(n) = (n! % (n^3+1)) == 0; \\ _Michel Marcus_, Mar 17 2016
%o (PARI) my(f=1);for(n=2,10^3,f*=n;if(f%(n^3+1)==0,print1(n,", "))); \\ _Joerg Arndt_, Apr 03 2016
%o (PARI) valp(n,p)=my(s); while(n>=p, s += n\=p); s
%o 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
%o (Python)
%o from math import factorial
%o for n in range(2,1000):
%o if(factorial(n)%(n**3+1)==0):print(n)
%o # _Soumil Mandal_, Apr 03 2016
%Y Cf. A118742, A120416, A269665.
%K nonn
%O 1,1
%A _José Hernández_, Mar 17 2016
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