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A290250
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Smallest (prime) number a(n) > 2 such that Sum_{k=1..a(n)} k!^(2*n) is divisible by a(n).
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0
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1248829, 13, 1091, 13, 41, 37, 463, 13, 23, 13, 1667, 37, 23, 13, 41, 13, 139
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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Table of n, a(n) for n=1..17.
Eric Weisstein's World of Mathematics, Factorial Sums
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EXAMPLE
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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
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MATHEMATICA
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Table[Module[{sum = 1, fac = 1, k = 2}, While[! Divisible[sum += (fac *= k)^(2 n), k], k++]; k], {n, 17}]
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CROSSREFS
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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
Adjacent sequences: A290247 A290248 A290249 * A290251 A290252 A290253
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KEYWORD
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nonn,more,hard
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AUTHOR
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Eric W. Weisstein, Jul 24 2017
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STATUS
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approved
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