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%I #33 Sep 04 2017 12:32:24
%S 1248829,13,1091,13,41,37,463,13,23,13,1667,37,23,13,41,13,139
%N Smallest (prime) number a(n) > 2 such that Sum_{k=1..a(n)} k!^(2*n) is divisible by a(n).
%C If a(i) exists, then the number of primes in the sequence {Sum_{k=1..n} k!^(2*i)}_n is finite. This follows since all subsequent terms in the sum involve adding (1*2*...*a(i)*...)^(2*i) to the previous term, both of which are divisible by a(i).
%C The terms from a(19) to a(36) are 46147, 13, 587, 13, 107, 23, 41, 13, 163, 13, 43, 37, 23, 13, 397, 13, 23, 433, and the terms from a(38) to a(50) are 13, 419, 13, 9199, 23, 2129, 13, 41, 13, 2358661, 37, 409, 13. If they exist, a(18) > 25*10^6 and a(37) > 14*10^6. - _Giovanni Resta_, Jul 27 2017
%C a(37) = 17424871; a(18) > 5*10^7 - _Mark Rodenkirch_, Sep 04 2017
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FactorialSums.html">Factorial Sums</a>
%e sum(k=1..1248829, k!^2) = 14+ million-digit number which is divisible by 1248829
%e sum(k=1..13, k!^4) = 1503614384819523432725006336630745933089, which is divisible by 13
%e sum(k=1..1091, k!^6) = 17055-digit number which is divisible by 1091
%t Table[Module[{sum = 1, fac = 1, k = 2}, While[! Divisible[sum += (fac *= k)^(2 n), k], k++]; k], {n, 17}]
%Y Cf. A100289 (n such that Sum_{k=1..n} k!^2 is prime), A289945 (k!^4), A289946 (k!^6), A290014 (k!^10).
%K nonn,more,hard
%O 1,1
%A _Eric W. Weisstein_, Jul 24 2017
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