A290250 Smallest (prime) number a(n) > 2 such that Sum_{k=1..a(n)} k!^(2*n) is divisible by a(n).

1248829, 13, 1091, 13, 41, 37, 463, 13, 23, 13, 1667, 37, 23, 13, 41, 13, 139

Offset: 1

Author

Eric W. Weisstein, Jul 24 2017

Keywords

• nonn
• more
• hard

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

Examples

			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
		

Links

• Eric Weisstein's World of Mathematics, Factorial Sums

Crossrefs

Cf. A100289 (n such that Sum_{k=1..n} k!^2 is prime), A289945 (k!^4), A289946 (k!^6), A290014 (k!^10).

Programs

Mathematica

Table[Module[{sum = 1, fac = 1, k = 2}, While[! Divisible[sum += (fac *= k)^(2 n), k], k++]; k], {n, 17}]