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A126225
Least number k > 0 such that the numerator of Sum_{i=1..k} 1/prime(i)^n is a prime.
0
2, 2, 3, 2, 3, 5, 3, 11, 3, 22
OFFSET
1,1
COMMENTS
a(12) > 80, a(13) = 30, a(14) = 16, a(18) = 7, a(19) = 3. - J. Mulder (jasper.mulder(AT)planet.nl), Jan 25 2010
a(11) > 200, a(12) > 200. - Michel Marcus, May 27 2019
If they exist, a(11) > 1263; a(17) > 954; a(22) > 795; a(23) > 720; a(25) > 570; a(12) = 799, a(15) = 313, a(16) = 780, a(20) = 433, a(21) = 7, a(24) = 4, a(27) = 12, a(29) = 37. - J.W.L. (Jan) Eerland, Jan 26 2023
EXAMPLE
a(1) = 2 corresponds to A024451(2) = 5, a prime.
a(2) = 2 corresponds to A061015(2) = 13, a prime.
MATHEMATICA
a[n_] := Block[{i = 1, sum = 0}, While[True, sum += 1/Prime[i]^n; If[PrimeQ[Numerator@sum], Return[i]]; i++ ]] (* J. Mulder (jasper.mulder(AT)planet.nl), Jan 25 2010 *)
Table[y[x_, y_]:=Numerator[FullSimplify[Sum[1/Prime[m]^x, {m, 1, y}]]]; k=1; Monitor[Parallelize[While[True, If[PrimeQ[y[n, k]], Break[]]; k++]; k], k], {n, 1, 10}] (* J.W.L. (Jan) Eerland, Jan 25 2023 *)
PROG
(PARI) a(n) = {my(k=1, s=1/prime(k)^n); while (! isprime(numerator(s)), k++; s += 1/prime(k)^n); k; } \\ Michel Marcus, May 27 2019
CROSSREFS
Cf. A024451 (1/p), A061015 (1/p^2), A115963 (1/p^3).
Sequence in context: A037126 A080092 A164738 * A306997 A056160 A106245
KEYWORD
hard,more,nonn
AUTHOR
Alexander Adamchuk, Mar 08 2007
STATUS
approved