A379718 The second Jordan totient function applied to the cubefull numbers: a(n) = A007434(A036966(n)).

1, 48, 192, 648, 768, 3072, 5832, 15000, 12288, 31104, 52488, 49152, 115248, 124416, 196608, 375000, 279936, 472392, 497664, 720000, 786432, 1119744, 1756920, 1990656, 2519424, 2880000, 3145728, 4251528, 4798248, 5647152, 4478976, 5531904, 9375000, 9720000, 7962624

Offset: 1

Author

Amiram Eldar, Dec 31 2024

Keywords

• nonn
• easy

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007434, A013661, A036966, A371412 (analogous with J_1 = phi), A379715, A379716, A379717.

Programs

Mathematica

f[p_, e_] := (p^2-1) * p^(2*e-2); j2[1] = 1; j2[n_] := Times @@ f @@@ FactorInteger[n]; Join[{1}, j2 /@ Select[Range[20000], AllTrue[Last /@ FactorInteger[#], #1 > 2 &] &]]

PARI

j2(f) = prod(i = 1, #f~, (f[i,1]^2 - 1) * f[i,1]^(2*f[i,2] - 2));
list(lim) = {my(f); print1(1, ", "); for(k = 2, lim, f = factor(k); if(vecmin(f[, 2]) > 2, print1(j2(f), ", "))); }

Formula

Sum_{n>=1} 1/a(n) = zeta(2)^2 * Product_{p prime} (1 - 2/p^2 + 1/p^4 + 1/p^6) = 1.02964361441212748276... .

In general, Sum_{m cubefull} 1/J_k(m) = zeta(k)^2 * Product_{p prime} (1 - 2/p^k + 1/p^(2*k) + 1/p^(3*k)), for k >= 2, where J_k is the k-th Jordan totient function.

In general, Sum_{m k-full} 1/J_2(m) = zeta(2)^2 * Product_{p prime} (1 - 2/p^2 + 1/p^4 + 1/p^(2*k)), for k >= 2.