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A064237
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Numbers k such that k! + 1 is divisible by a square.
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6
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OFFSET
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1,1
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COMMENTS
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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
Comment from Francois BRUNAULT, Nov 23 2008: 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.
It is very likely that 229 and 562 are the next two terms, but this has not yet been proved. - Nov 29 2008
Contains A007540(n)-1 for all n. That sequence is conjectured to be infinite. - Robert Israel, Jul 04 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
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LINKS
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EXAMPLE
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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.
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). - Thomas Richard, Aug 31 2021
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MAPLE
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remove(t -> numtheory:-issqrfree(t!+1), [$1..50]); # Robert Israel, Jul 04 2016
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MATHEMATICA
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Flatten[Position[MoebiusMu[Range[30]!+1], 0]]; (* T. D. Noe, Mar 01 2006, Nov 21 2008 *)
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PROG
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(PARI) lista(nn) = for(n=1, nn, if(!issquarefree(n!+1), print1(n, ", "))); \\ Altug Alkan, Mar 08 2016
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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