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

1,1

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

Cf. A118742, A120416, A269665.

Sequence in context: A160961 A260805 A267781 * A256374 A286512 A087166

Adjacent sequences: A270438 A270439 A270440 * A270442 A270443 A270444

Keyword

nonn

Author

José Hernández, Mar 17 2016

Status

approved