A064237 Numbers k such that k! + 1 is divisible by a square.

4, 5, 7, 12, 23

Offset: 1

Author

Vladeta Jovovic, Sep 22 2001

Keywords

• nonn
• hard
• more

Comments

229 is another term because 613^2 divides 229!+1. See A115091 for primes whose square divides m!+1 for some m. An examination of the factorizations of m!+1 for m<=100 found no additional squares. - T. D. Noe, Mar 01 2006

562 is also a term because 562!+1 is divisible by 563^2. - Vladeta Jovovic, Mar 30 2004

A web search reveals that for 1 <= k <= 228 there are 82 values of k for which k! + 1 has not been completely factored (the smallest is k=103), so showing that 229 and 562 are indeed the next two terms will be a huge task. I checked that k!+1 is not divisible by p^2 for k <= 1000 and prime p < 10^8. - Francois Brunault, Nov 23 2008

It is very likely that 229 and 562 are the next two terms, but this has not yet been proved. - Francois Brunault, Nov 29 2008

Contains A007540(n)-1 for all n. That sequence is conjectured to be infinite. - Robert Israel, Jul 04 2016

This sequence includes A146968 (solutions of Brocard's problem). - Salvador Cerdá, Mar 08 2016

If k > 562 and k! + 1 is divisible by p^2 where p is prime, then either k > 10000 or p > 2038074743 (the hundred millionth prime). - Jason Zimba, Oct 21 2021

Examples

			4 is in the sequence because 4! + 1 = 5^2.
5 is in the sequence because 5! + 1 = 11^2.
6 is not in the sequence because 6! + 1 = 721
7 is in the sequence because 7! + 1 = 71^2.
12 is in the sequence because 12! + 1 = 13^2 * 2834329.
23 is a term because 23!+1 = 47^2*79*148139754736864591.
From _Thomas Richard_, Aug 31 2021: (Start)
229 and 562 are terms because
229!+1 = 613^2 * 38669 * 1685231 * 3011917759 * (417-digit composite)
562!+1 = 563^2 * 64467346976659839517037 * 112870688711507255213769871 * 63753966393108716329397432599379239 * (1214-digit prime). (End)
		

Links

• Hisanori Mishima, Factorizations of m!+1

Crossrefs

Cf. A007540 (Wilson primes), A115091, A146968, A038507, A085692.

Programs

Maple

remove(t -> numtheory:-issqrfree(t!+1), [$1..50]); # Robert Israel, Jul 04 2016

Mathematica

Flatten[Position[MoebiusMu[Range[30]!+1], 0]]; (* T. D. Noe, Mar 01 2006, Nov 21 2008 *)

PARI

lista(nn) = for(n=1, nn, if(!issquarefree(n!+1), print1(n, ", "))); \\ Altug Alkan, Mar 08 2016

Extensions

Example corrected by T. D. Noe, Nov 26 2008