A338997 Number of (i,j,k) in {1,2,...,n}^3 such that gcd(n,i) = gcd(n,j) = gcd(n,k).
1, 2, 9, 10, 65, 18, 217, 74, 225, 130, 1001, 90, 1729, 434, 585, 586, 4097, 450, 5833, 650, 1953, 2002, 10649, 666, 8065, 3458, 6057, 2170, 21953, 1170, 27001, 4682, 9009, 8194, 14105, 2250, 46657, 11666, 15561, 4810, 64001, 3906, 74089, 10010, 14625, 21298, 97337, 5274, 74305, 16130
Offset: 1
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
a[n_] := DivisorSum[n, EulerPhi[#]^3 &]; Array[a, 100] (* Amiram Eldar, Dec 31 2020 *)
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PARI
a(n)=sumdiv(n,d,eulerphi(d)^3)
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PARI
a(n) = sum(k=1, n, eulerphi(n/gcd(k, n))^2); \\ Seiichi Manyama, Mar 13 2021
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PARI
my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, eulerphi(k)^3*x^k/(1-x^k))) \\ Seiichi Manyama, Mar 13 2021
Formula
a(n) = Sum_{d|n} phi(d)^3.
From Seiichi Manyama, Mar 13 2021: (Start)
a(n) = Sum_{k=1..n} phi(n/gcd(k, n))^2.
G.f.: Sum_{k>=1} phi(k)^3 * x^k/(1 - x^k). (End)
From Amiram Eldar, Nov 15 2022: (Start)
Multiplicative with a(p^e) = 1 + ((p-1)^2 (p^(3*e)-1))/(p^2 + p + 1).
Sum_{k=1..n} a(k) ~ c * n^4, where c = (Pi^4/360) * Product_{p prime} (1 - 3/p^2 + 3/p^3 - 1/p^4) = 0.09123656748... . (End)