[go: up one dir, main page]

login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A082476
a(n) = Sum_{d|n} mu(d)^2*tau(d)^2.
6
1, 5, 5, 5, 5, 25, 5, 5, 5, 25, 5, 25, 5, 25, 25, 5, 5, 25, 5, 25, 25, 25, 5, 25, 5, 25, 5, 25, 5, 125, 5, 5, 25, 25, 25, 25, 5, 25, 25, 25, 5, 125, 5, 25, 25, 25, 5, 25, 5, 25, 25, 25, 5, 25, 25, 25, 25, 25, 5, 125, 5, 25, 25, 5, 25, 125, 5, 25, 25, 125, 5, 25, 5, 25, 25, 25, 25, 125
OFFSET
1,2
COMMENTS
More generally : sum(d|n, mu(d)^2*tau(d)^m) = (2^m+1)^omega(n).
FORMULA
a(n) = 5^omega(n); multiplicative with a(p^e)=5.
a(n) = abs(sum(d|n, mu(d)*tau_3(d^2))), where tau_3 is A007425. - Enrique Pérez Herrero, Mar 29 2010
a(n) = tau_5(rad(n)) = A061200(A007947(n)). - Enrique Pérez Herrero, Jun 24 2010
a(n) = A000351(A001221(n)). - Antti Karttunen, Jul 26 2017
From Vaclav Kotesovec, Feb 28 2023: (Start)
Dirichlet g.f.: Product_{primes p} (1 + 5/(p^s - 1)).
Dirichlet g.f.: zeta(s)^5 * Product_{primes p} (1 - 10/p^(2*s) + 20/p^(3*s) - 15/p^(4*s) + 4/p^(5*s)), (with a product that converges for s=1). (End)
MATHEMATICA
tau[1, n_] := 1; SetAttributes[tau, Listable];
tau[k_, n_] := Plus @@ (tau[k - 1, Divisors[n]]) /; k > 1;
A082476[n_] := Abs[DivisorSum[n, MoebiusMu[ # ]*tau[3, #^2] &]]; (* Enrique Pérez Herrero, Mar 29 2010 *)
(* or more easy *)
A082476[n_] := 5^PrimeNu[n] (* Enrique Pérez Herrero, Mar 29 2010 *)
PROG
(PARI) a(n)=5^omega(n)
(PARI) for(n=1, 100, print1(direuler(p=2, n, (4*X+1)/(1-X))[n], ", ")) \\ Vaclav Kotesovec, Feb 28 2023
KEYWORD
mult,nonn
AUTHOR
Benoit Cloitre, Apr 27 2003
STATUS
approved