A115091 -id:A115091 - cp's OEIS frontend

A259230 a(n) = smallest k such that (A115091(n)-k)! == -1 (mod A115091(n)^2).

1, 6, 1, 24, 64, 1, 384

Offset: 1

Felix Fröhlich, Nov 08 2015

The values of m in A115091.

A115091(n) is in A007540 iff a(n) = 1.

			a(2) = 6, because 6 is the smallest k such that (A115091(2)-k)! == -1 (mod A115091(2)^2), which yields the congruence (11-6)! == -1 (mod 11^2).

Cf. A007540, A115091.

t = Select[Prime@ Range@ 120, AnyTrue[Range@ #, Function[m, Divisible[m! + 1, #^2]]] &]; Table[k = 1; While[Mod[(t[[n]] - k)!, t[[n]]^2] != t[[n]]^2 - 1, k++]; k, {n, 7}] (* Michael De Vlieger, Nov 10 2015, Version 10 *)

forprime(p=1, , for(k=1, p-1, if(Mod((p-k)!, p^2)==-1, print1(k, ", "); break({1}))))

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

Original entry on oeis.org

4, 5, 7, 12, 23

Offset: 1

Vladeta Jovovic, Sep 22 2001

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

			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)

Hisanori Mishima, Factorizations of m!+1

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

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

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

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

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

cp's OEIS Frontend

A259230 a(n) = smallest k such that (A115091(n)-k)! == -1 (mod A115091(n)^2).

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