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A290250
Smallest (prime) number a(n) > 2 such that Sum_{k=1..a(n)} k!^(2*n) is divisible by a(n).
0
1248829, 13, 1091, 13, 41, 37, 463, 13, 23, 13, 1667, 37, 23, 13, 41, 13, 139
OFFSET
1,1
COMMENTS
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).
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
a(37) = 17424871; a(18) > 5*10^7 - Mark Rodenkirch, Sep 04 2017
LINKS
Eric Weisstein's World of Mathematics, Factorial Sums
EXAMPLE
sum(k=1..1248829, k!^2) = 14+ million-digit number which is divisible by 1248829
sum(k=1..13, k!^4) = 1503614384819523432725006336630745933089, which is divisible by 13
sum(k=1..1091, k!^6) = 17055-digit number which is divisible by 1091
MATHEMATICA
Table[Module[{sum = 1, fac = 1, k = 2}, While[! Divisible[sum += (fac *= k)^(2 n), k], k++]; k], {n, 17}]
CROSSREFS
Cf. A100289 (n such that Sum_{k=1..n} k!^2 is prime), A289945 (k!^4), A289946 (k!^6), A290014 (k!^10).
Sequence in context: A206019 A210084 A212489 * A251156 A210132 A251118
KEYWORD
nonn,more,hard
AUTHOR
Eric W. Weisstein, Jul 24 2017
STATUS
approved