%I #38 Jan 17 2024 09:07:56
%S 1,1023,59048,1047552,9765624,60406104,282475248,1072693248,
%T 3486725352,9990233352,25937424600,61855850496,137858491848,
%U 288972178704,576640565952,1098437885952,2015993900448,3566920035096,6131066257800
%N Jordan function J_10(n).
%C a(n) is divisible by 264 = (2^3)*3*11 = A006863(5), except for n = 1, 2, 3 or 11. See Lugo. - _Peter Bala_, Jan 13 2024
%D L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 199, #3.
%H Robert Israel, <a href="/A069095/b069095.txt">Table of n, a(n) for n = 1..10000</a>
%H Michael Lugo, <a href="http://godplaysdice.blogspot.com/2008/05/little-number-theory-problem.html">A little number theory problem</a> (2008)
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Jordan%27s_totient_function">Jordan's totient function</a>.
%F a(n) = Sum_{d|n} d^10*mu(n/d).
%F Multiplicative with a(p^e) = p^(10e)-p^(10(e-1)).
%F Dirichlet generating function: zeta(s-10)/zeta(s). - _Ralf Stephan_, Jul 04 2013
%F a(n) = n^10*Product_{distinct primes p dividing n} (1-1/p^10). - _Tom Edgar_, Jan 09 2015
%F Sum_{k=1..n} a(k) ~ n^11 / (11*zeta(11)). - _Vaclav Kotesovec_, Feb 07 2019
%F From _Amiram Eldar_, Oct 12 2020: (Start)
%F lim_{n->oo} (1/n) * Sum_{k=1..n} a(k)/k^10 = 1/zeta(11).
%F Sum_{n>=1} 1/a(n) = Product_{p prime} (1 + p^10/(p^10-1)^2) = 1.0009955309... (End)
%p f:= n -> n^10*mul(1-1/p^10, p=numtheory:-factorset(n)):
%p map(f, [$1..30]); # _Robert Israel_, Jan 09 2015
%t JordanJ[n_, k_] := DivisorSum[n, #^k*MoebiusMu[n/#] &]; f[n_] := JordanJ[n, 10]; Array[f, 21]
%t f[p_, e_] := p^(10*e) - p^(10*(e-1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Oct 12 2020 *)
%o (PARI) for(n=1,100,print1(sumdiv(n,d,d^10*moebius(n/d)),","))
%Y Cf. A059379 and A059380 (triangle of values of J_k(n)), A000010 (J_1), A007434 (J-2), A059376 (J_3), A059377 (J_4), A059378 (J_5), A069091 - A069094 (J_6 through J_9).
%Y Cf. A013669.
%K easy,nonn,mult
%O 1,2
%A _Benoit Cloitre_, Apr 05 2002