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I discussed the prime zeta function at some length in this math.SE answer. In particular, the infinite Möbius inversion $$P(s)=\sum_{k=1}^\infty \frac{\mu(k)}{k}\log\zeta(ks)$$ is the actual computational formula used, as recommended in Fröberg's paper. (It is also noted there that numerical evaluation becomes more difficult at values near the imaginary ...

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You made a few mistakes. jk[0] should be 0 in your code and your function tr is wrong. Corrected version: t[0] = 0; t[1] = 0; t[2] = 1; t[n_] := t[n] = LengthWhile[Range[1, 11], Divisible[n, #1] &] + 1 Sum[Nest[t, m, 3], {m, 1, 2006}] 1171 10x faster version: t[n_] := t[n] = Module[{i = 1}, While[MemberQ[Divisors[n], i], i++]; i];

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This works for all numbers which are not multiples of $\text{lcm}(1, \dots, 20) = 232792560$: smallest[n_] := LengthWhile[Range[3, 20], Divisible[n, #] &] + 3 If you use $50$ instead of $20$, you get $3099044504245996706400$ as the forbidden number, which might be more acceptable to you. You could compile this to get something which might be faster. ...

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Unfortunately, Mathematica's capabilities for manipulating Dirichlet series are not very extensive at the moment. In particular, there isn't anything built-in that does the inverse of DirichletTransform[]. (This MO question is apropos.) Here is a workaround. It seems you're trying to generate the coefficients for $\zeta(s)^2$. Recall that $\zeta(s)$ is the ...

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f[0] := 0 f[1] := 0 f[2] := 1 f[x_] := Module[{j = 1}, While[Mod[x, j] == 0, j++]; j]; bt[x_] := Nest[f, x, 3] Total[bt /@ Range[2006]] yields 1171

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Not sure if this is what you want. a = Sum[1/n^s, {n, 1, 6}]; a2 = (a /. Power[n_, Times[-1, s]] :> t[n])^2 // Expand; expr = (a2 /. {t[n_]^2 :> t[n^2], t[n_]*t[m_] :> t[n*m]}) /. t[n_] :> HoldForm[1/n^s] Simplify[a^2 == ReleaseHold[expr]] (* True *) EDIT : Perhaps this is closer expr2 = (a2 /. {t[n_]^2 :> t[n^2], t[n_]*t[m_] ...

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