A253206 Coefficients of the Dirichlet series for zeta(5s)/zeta(s).

1, -1, -1, 0, -1, 1, -1, 0, 0, 1, -1, 0, -1, 1, 1, 0, -1, 0, -1, 0, 1, 1, -1, 0, 0, 1, 0, 0, -1, -1, -1, 1, 1, 1, 1, 0, -1, 1, 1, 0, -1, -1, -1, 0, 0, 1, -1, 0, 0, 0, 1, 0, -1, 0, 1, 0, 1, 1, -1, 0, -1, 1, 0, -1, 1, -1, -1, 0, 1, -1, -1, 0, -1, 1, 0, 0, 1, -1, -1, 0, 0, 1, -1, 0, 1, 1, 1, 0, -1, 0, 1, 0, 1, 1, 1, -1, -1

Offset

1

Comments

First formula works also for zeta(k*s)/zeta(s), with k>5, thus generating a whole class of sequences.

Multiplicative with a(p^d) = 1 if d == 0 mod 5, -1 if d == 1 mod 5, 0 otherwise. - Robert Israel, Mar 27 2015

Inverse Mobius transform gives the characteristic function of 5th powers. - R. J. Mathar, Jun 05 2020

Links

Wolfgang Hintze, Table of n, a(n) for n = 1..500

StackExchange, Question 77711

Formula

a(k,n) = Sum_{d^k divides n} MoebiusMu[n/d^k], k = 5 gives this sequence. - Wolfgang Hintze, Mar 27 2015

Dirichlet G.f.: defining relation: zeta(5s)/zeta(s) = Sum_{n>=1} a(n)/n^s.

G.f.: Sum_{n>=1} a(n)*x^n/(1 - x^n) = Sum_{n>=1} x^(n^5).

Maple

g:= [1, -1, 0, 0, 0]:
[seq(mul(g[x[2] mod 5 + 1], x = ifactors(n)[2]), n=1..100)]; # Robert Israel, Mar 27 2015

Mathematica

(* Number theoretical approach *)
a[k_, n_]:=Plus@@(MoebiusMu[n/#^k]&/@Select[(Divisors[n])^(1/k), IntegerQ[#]&])
nn=200; Table[a[5, n], {n, 1, nn} (* Wolfgang Hintze, Mar 27 2015 *)
(* Comparison of "Dirichlet"-powers n^-x *)
fqZeta[k_, nn_] := Module[{z, d, x, g, eqs, sol, t, p},
z[x_, p_] := Sum[1/n^x, {n, 1, p}]; d[x_, p_] := Sum[f[n]/n^x, {n, 1, p}];
g[k] = Expand[z[k*x, nn] - z[x, nn]*d[x, nn]] /. (a_)^((c_)*(b_)) ->
Simplify[a^b]^c; eqs[k] =
Table[Simplify[0 == (1/m^(-x))*Plus @@ Cases[g[k], (_.)/m^x]], {m, 2, nn}];
sol[k] = Solve[Join[{f[1] == 1}, eqs[k]]][[1]];
t[k] = Table[f[n], {n, 1, nn}] /. sol[k]]
fqZeta[5, 200]
(* Using G.f. (beyond nn = 73 numerical problems appear) *)
nn = 73; f[x_] := Sum[a[n]*(x^n/(1 - x^n)), {n, 1, nn}];
sol = SolveAlways[0 == Series[f[x] - Sum[x^n^5, {n, 1, nn}], {x, 0, nn}], x];
Flatten[Table[a[n], {n, 1, nn}] /. sol]

Prog

(PARI) a(n) = sumdiv(n, d, if (ispower(d, 5), moebius(n/d), 0)); \\ Michel Marcus, Mar 27 2015
(PARI) for(n=1, 100, print1(direuler(p=2, n, (1-X)/(1-X^5))[n], ", ")) \\ Vaclav Kotesovec, Jun 14 2020

Crossrefs

Cf. A219009 (k=4), A210826 (k=3), A008836 (k=2).

Sequence in context: A359155 A069158 A353559 * A133639 A323154 A060038

Adjacent sequences: A253203 A253204 A253205 * A253207 A253208 A253209

Keyword

sign,mult

Author

Wolfgang Hintze, Mar 25 2015

Status

approved