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A253206 Coefficients of the Dirichlet series for zeta(5s)/zeta(s). 2
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 (list; graph; refs; listen; history; text; internal format)
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: A080323 A157657 A069158 * A133639 A323154 A060038

Adjacent sequences:  A253203 A253204 A253205 * A253207 A253208 A253209

KEYWORD

sign,mult

AUTHOR

Wolfgang Hintze, Mar 25 2015

STATUS

approved

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Last modified August 10 02:28 EDT 2020. Contains 336367 sequences. (Running on oeis4.)