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Partial sums of the Jordan function J_2(k), for 1 <= k <= n.
2

%I #18 Nov 24 2018 05:05:41

%S 0,1,4,12,24,48,72,120,168,240,312,432,528,696,840,1032,1224,1512,

%T 1728,2088,2376,2760,3120,3648,4032,4632,5136,5784,6360,7200,7776,

%U 8736,9504,10464,11328,12480,13344,14712,15792,17136,18288,19968,21120,22968,24408

%N Partial sums of the Jordan function J_2(k), for 1 <= k <= n.

%C In general, for m >= 1, Sum_{k=1..n} J_m(k) = Sum_{k=1..n} mu(k) * (Bernoulli(m+1, 1+floor(n/k)) - Bernoulli(m+1, 0)) / (m+1), where mu(k) is the Moebius function and Bernoulli(n,x) are the Bernoulli polynomials.

%C In general, for m >= 1, Sum_{k=1..n} J_m(k) ~ n^(m+1) / ((m+1) * zeta(m+1)).

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Jordan%27s_totient_function">Jordan's totient function</a>

%F a(n) ~ n^3 / (3*zeta(3)).

%F a(n) = Sum_{k=1..n} A007434(k).

%F a(n) = Sum_{k=1..n} mu(k) * Bernoulli(3, 1+floor(n/k)) / 3, where mu(k) is the Moebius function and Bernoulli(n,x) are the Bernoulli polynomials.

%t a[n_]:= Sum[MoebiusMu[k]*BernoulliB[3,1+Floor[n/k]]/3, {k,1,n}]; Array[a, 50, 0] (* _Stefano Spezia_, Nov 21 2018 *)

%o (PARI) a(n) = sum(k=1, n, moebius(k) * ((n\k)^3/3 + (n\k)^2/2 + (n\k)/6));

%Y Cf. A002088, A007434, A008683.

%K nonn,easy

%O 0,3

%A _Daniel Suteu_, Nov 20 2018