A379717 The second Jordan totient function applied to the cubefree numbers.
1, 3, 8, 12, 24, 24, 48, 72, 72, 120, 96, 168, 144, 192, 288, 216, 360, 288, 384, 360, 528, 600, 504, 576, 840, 576, 960, 960, 864, 1152, 864, 1368, 1080, 1344, 1680, 1152, 1848, 1440, 1728, 1584, 2208, 2352, 1800, 2304, 2016, 2808, 2880, 2880, 2520, 3480, 2304
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
f[p_, e_] := (p^2-1) * p^(2*e-2); j2[1] = 1; j2[n_] := Times @@ f @@@ FactorInteger[n]; cubeFreeQ[n_] := Max[FactorInteger[n][[;;, 2]]] < 3; j2 /@ Select[Range[100], cubeFreeQ]
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PARI
j2(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i,1]^2 - 1) * f[i,1]^(2*f[i,2] - 2));} iscubefree(n) = if(n == 1, 1, vecmax(factor(n)[, 2]) < 3); list(lim) = apply(j2, select(iscubefree, vector(lim, i, i)));
Formula
Sum_{n>=1} 1/a(n) = zeta(2) * zeta(4) / zeta(8) = 35 / (2*Pi^2) = 1.77312071374091100026... .
In general, Sum_{m cubefree} 1/J_k(m) = zeta(k) * zeta(2*k) / zeta(4*k), for k >= 2, where J_k is the k-th Jordan totient function.
In general, Sum_{m k-free} 1/J_2(m) = zeta(2)^2 * Product_{p prime} (1 - 1/p^2 + 1/p^4 - 1/p^(2*k)), for k >= 2.
Sum_{k=1..n} a(k) ~ c * n^3 / 3, where c = zeta(3)^3 * Product_{p prime} (1 - 2/p^3 + 1/p^5) = 1.23061243656940899916... . - Amiram Eldar, Jan 03 2025
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