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Numbers k such that (1!)^2 + (2!)^2 + (3!)^2 + ... + (k!)^2 is prime.
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%I #35 Oct 12 2023 15:37:51

%S 2,3,4,5,7,8,10,18,21,42,51,91,133,177,182,310,3175,9566,32841

%N Numbers k such that (1!)^2 + (2!)^2 + (3!)^2 + ... + (k!)^2 is prime.

%C All k <= 310 yield provable primes.

%C Write the sum as S(2,k)-1, where S(m,k) = Sum_{i=0..k} (i!)^m. Let p=1248829. Because p divides S(2,p-1)-1, p divides S(2,k)-1 for all k >= p-1. Hence there are no primes for k >= p-1.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FactorialSums.html">Factorial Sums</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/IntegerSequencePrimes.html">Integer Sequence Primes</a>

%p L:= [seq((i!)^2, i=1..1000)]:

%p S:= ListTools:-PartialSums(L):

%p select(t -> isprime(S[t]), [$1..1000]); # _Robert Israel_, Jul 17 2017

%t Select[Range[200], PrimeQ[Total[Range[#]!^2]] &]

%t Module[{nn=350,tt},tt=Accumulate[(Range[nn]!)^2];Position[tt,_?PrimeQ]]//Flatten (* The program generates the first 16 terms of the sequence. *) (* _Harvey P. Dale_, Oct 12 2023 *)

%o (PARI) is(n)=ispseudoprime(sum(k=1,n,k!^2)) \\ _Charles R Greathouse IV_, Apr 14 2015

%Y Cf. A100288 (primes of the form (1!)^2 + (2!)^2 + (3!)^2 +...+ (k!)^2).

%Y Cf. A061062 ((0!)^2 + (1!)^2 + (2!)^2 + (3!)^2 +...+ (n!)^2).

%Y Cf. A289947 (k!^6), A290014 (k!^10).

%Y Cf. also A104344.

%K nonn,fini,more

%O 1,1

%A _T. D. Noe_, Nov 11 2004 and Dec 11 2004

%E a(18) from _T. D. Noe_, Feb 15 2006

%E a(19) from _Serge Batalov_, Jul 29 2017