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 A270441 Numbers n such that n^3+1 divides n!. 4
 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 (list; graph; refs; listen; history; text; internal format)
 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])

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Last modified June 14 09:06 EDT 2021. Contains 345018 sequences. (Running on oeis4.)