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Numerator of Sum[ 1/k^p, {k,1,p-1} ], where p = Prime[n].
1

%I #5 Jan 02 2024 14:12:55

%S 1,9,257875,940908897061,26038773205374138944970092886340352227,

%T 5706439637514064062030256049808675747470805004854626598761,

%U 3819751175863358416058062379293843331497647520922258560223903226691067255782388923965399403291707829

%N Numerator of Sum[ 1/k^p, {k,1,p-1} ], where p = Prime[n].

%C p^3 divides a(n) for n>2. A119722[n] = a(n)/p^3, p=Prime[n].

%C Numerators of Sum[ 1/k^n, {k,1,n-1} ] are listed in A120347(n) = {1, 9, 1393, 257875, 47463376609, 940908897061, ...}.

%F a(n) = Numerator[ Sum[ 1/k^Prime[n], {k,1,Prime[n]-1} ]]. a(n) = Numerator[ Zeta[p] - Zeta[p,p] ], for p = Prime[n].

%F a(n) = A120347[ Prime[n] ].

%t Table[Numerator[Sum[1/k^Prime[n],{k,1,Prime[n]-1}]],{n,1,8}]

%Y Cf. A119722.

%Y Cf. A120347.

%K frac,nonn

%O 1,2

%A _Alexander Adamchuk_, Aug 16 2006, Oct 31 2006