OFFSET
1,2
COMMENTS
In general, if g.f.: 1/(1-x) * Sum_{j>=1} phi(j) * x^j/(1 - x^j)^k, where k > 2 and phi is the Euler totient function (A000010), then a(n) ~ zeta(k-1) * n^k / (k! * zeta(k)).
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..10000
FORMULA
G.f.: 1/(1-x) * Sum_{k>=1} phi(k) * x^k/(1 - x^k)^4, where phi is the Euler totient function (A000010).
a(n) = Sum_{k=1..n} Sum_{d|k} phi(k/d) * d*(d+1)*(d+2)/6.
a(n) ~ 15 * zeta(3) * n^4 / (4*Pi^4).
a(n) = Sum_{k=1..n} phi(k) * binomial(floor(n/k)+3,4). - Seiichi Manyama, Sep 13 2024
MATHEMATICA
Table[Sum[Sum[Sum[Sum[GCD[i, j, k, m], {i, 1, j}], {j, 1, k}], {k, 1, m}], {m, 1, n}], {n, 1, 100}]
nmax = 100; Rest[CoefficientList[Series[1/(1-x) * Sum[EulerPhi[k]*x^k/(1 - x^k)^4, {k, 1, nmax}], {x, 0, nmax}], x]]
Accumulate[Table[Sum[EulerPhi[n/d] * d*(d+1)*(d+2)/6, {d, Divisors[n]}], {n, 1, 100}]] (* faster *)
PROG
(PARI) a(n) = sum(i=1, n, sum(j=i, n, sum(k=j, n, sum(m=k, n, gcd([i, j, k, m]))))); \\ Michel Marcus, Jun 06 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Vaclav Kotesovec, Jun 05 2021
STATUS
approved