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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
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LINKS
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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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KEYWORD
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nonn,more,hard
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AUTHOR
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STATUS
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approved
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