A118742 -id:A118742 - cp's OEIS frontend

A120416 Numbers k such that k^2+1 divides k!.

18, 21, 38, 43, 47, 57, 68, 70, 72, 73, 83, 99, 111, 117, 119, 123, 128, 132, 133, 142, 157, 172, 173, 174, 182, 185, 191, 192, 193, 200, 211, 212, 216, 233, 237, 239, 242, 251, 253, 255, 265, 268, 273, 278, 293, 294, 302, 305, 307, 313, 319, 322, 327, 336

Offset: 1

R. J. Mathar, Jul 07 2006

Let Product_j (p_j)^(e_j) be the prime factorization of n^2+1. Then n is in the sequence if and only if for each j, e_j <= Sum_{k>=1} floor(n/(p_j)^k). - Robert Israel, Nov 11 2016
There exist infinitely many natural numbers n such that n^2+1 divides n!, because 2*(5*k-2)^2 is a term for k > 0. - Jinyuan Wang, Feb 06 2019
There are 2082 terms up to 10^4, 22792 up to 10^5, 242421 up to 10^6, 2523043 up to 10^7. Perhaps the asymptotic density is 1 - log(2) = 30.68...%. - Jinyuan Wang, Feb 09 2019

			For the first number, k=18: 18^2+1=325 divides 18!=6402373705728000.

Robert Israel, Table of n, a(n) for n = 1..10000
Hojoo Lee, Problem A 26, Problems in elementary number theory, 2003.

Cf. A002522, A000142, A118742, A270441.

select(t -> t! mod (t^2+1)=0, [$1..1000]); # Robert Israel, Nov 11 2016

Select[Range@ 336, Divisible[#!, #^2 + 1] &] (* Jinyuan Wang, Feb 06 2019 *)

isok(n) = (n! % (n^2+1) == 0) \\ Michel Marcus, Jul 23 2013

valp(n, p)=my(s); while(n>=p, s += n\=p); s
is(n)=my(f=factor(n^2+1)); for(i=1, #f~, if(valp(n, f[i, 1])Jinyuan Wang, Feb 06 2019

k such that A002522(k) | A000142(k).

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

Original entry on oeis.org

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

José Hernández, Mar 17 2016

nonn

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)

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

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

Cf. A118742, A120416, A269665.

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

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 *)

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

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

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

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

A166460 Numbers k such that k + (-1)^k is not prime.

Original entry on oeis.org

0, 1, 5, 7, 8, 9, 11, 13, 14, 15, 17, 19, 20, 21, 23, 24, 25, 26, 27, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 59, 61, 62, 63, 64, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 80, 81, 83, 84, 85, 86, 87, 89, 90, 91, 92, 93, 94

Offset: 1

Juri-Stepan Gerasimov, Oct 14 2009

This is the complement of A068499 (except that both include 1 as a term).
From Don Reble, Aug 31 2021: (Start)
Proof for all k except 0, 1, 3 with cases
(i)   If k is odd and >=5, then k+1 = 2*x, 2 < x < k, k! = k*...*x*...*2*1
        A068499: k+1 divides k!            : absent
        A166460: k-1 is even and composite : present
(ii)  If k is even and k+1 is prime,
        A068499: k+1 does not divide k!    : present
        A166460: k+1 is prime              : absent
(iii) If k is even and k+1 = p^2 is the square of a (odd) prime, then k+1 >= 3p, k > 2p.
        A068499: k! = k*...*2p*...*p*...*1;
                 k+1 divides k!            : absent
        A166460: k+1 is composite          : present
(iv)  If k is even and k+1 is composite but not the square of a prime, then there are two distinct factors x*y = k+1:
                 3 <= x < y = (k+1)/x < k.
        A068499: k! = k*...*y*...*x*...*1:
                 k+1 divides k!            : absent
        A166460: k+1 is composite          : present
(End)

			0 + (-1)^0 = 1 is not prime, which adds 0 to the sequence.
5 + (-1)^5 = 4 is not prime, which adds 5 to the sequence.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A002808, A072668, A141468, A060462, A118742, A068499.

Select[Range[0, 94], ! PrimeQ[# + (-1)^#] &] (* Michael De Vlieger, Sep 08 2021 *)

from sympy import composite
def A166460(n): return composite(n-1)-1 if n>2 else n-1 # Chai Wah Wu, Aug 27 2024

a(n) = A002808(n-1)-1 for n>2. - Chai Wah Wu, Aug 27 2024

0 added by R. J. Mathar, Oct 21 2009

A269665 For n>=0, let A_n be the set of natural numbers k such that (k^n + 1) | k!. If A is nonempty, then a(n) is the least element of A_n; otherwise a(n) = 0.

Original entry on oeis.org

2, 5, 18, 17, 1600, 984, 2888, 460747, 99271723, 792174, 32917926

Offset: 0

José Hernández, Mar 02 2016

a(n) is the smallest k such that (k^n + 1) | k! if it exists, otherwise a(n) = 0.

			For n=2, a(2) is equal to 18 because k=18 is the least natural number k such that (k^2+1)|k! (see A120416).

Cf. A118742, A120416, A270441.

For[k = 0, k < 11, k++, x = 0; r = 0; n = 1; While[x != 1, If[Mod[n!, n^k + 1] != 0, x = 0, x = 1; r = n]; n++]; Print[r]]
Table[SelectFirst[Range[10^4], Divisible[#!, #^n + 1] &], {n, 0, 6}] (* Michael De Vlieger, Mar 04 2016, Version 10 *)

a(n) = {my(k = 1); while (k! % (k^n+1), k++); k;} \\ Michel Marcus, Mar 03 2016

a(7)-a(9) from Hiroaki Yamanouchi, Apr 04 2016

a(10) from Giovanni Resta, Apr 20 2016

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A120416 Numbers k such that k^2+1 divides k!.

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

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A166460 Numbers k such that k + (-1)^k is not prime.

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A269665 For n>=0, let A_n be the set of natural numbers k such that (k^n + 1) | k!. If A is nonempty, then a(n) is the least element of A_n; otherwise a(n) = 0.

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