%I #19 Jan 26 2023 16:14:30

%S 2,2,3,2,3,5,3,11,3,22

%N Least number k > 0 such that the numerator of Sum_{i=1..k} 1/prime(i)^n is a prime.

%C 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

%C a(11) > 200, a(12) > 200. - _Michel Marcus_, May 27 2019

%C 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

%e a(1) = 2 corresponds to A024451(2) = 5, a prime.

%e a(2) = 2 corresponds to A061015(2) = 13, a prime.

%t 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 *)

%t 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 *)

%o (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

%Y Cf. A024451 (1/p), A061015 (1/p^2), A115963 (1/p^3).

%K hard,more,nonn

%O 1,1

%A _Alexander Adamchuk_, Mar 08 2007