A253206 - OEIS

%I #59 Jun 14 2020 08:10:36

%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,

%T -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,

%U 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

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

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

%C Multiplicative with a(p^d) = 1 if d == 0 mod 5, -1 if d == 1 mod 5, 0 otherwise. - _Robert Israel_, Mar 27 2015

%C Inverse Mobius transform gives the characteristic function of 5th powers. - _R. J. Mathar_, Jun 05 2020

%H Wolfgang Hintze, <a href="/A253206/b253206.txt">Table of n, a(n) for n = 1..500</a>

%H StackExchange, <a href="http://mathematica.stackexchange.com/questions/77711/what-are-the-terms-of-the-sequence-generated-by-zeta3s-zetas/77731">Question 77711</a>

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

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

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

%p g:= [1,-1,0,0,0]:

%p [seq(mul(g[x[2] mod 5 + 1], x = ifactors(n)[2]),n=1..100)]; # _Robert Israel_, Mar 27 2015

%t (* Number theoretical approach *)

%t a[k_,n_]:=Plus@@(MoebiusMu[n/#^k]&/@Select[(Divisors[n])^(1/k),IntegerQ[#]&])

%t nn=200; Table[a[5,n],{n,1,nn} (* _Wolfgang Hintze_, Mar 27 2015 *)

%t (* Comparison of "Dirichlet"-powers n^-x *)

%t fqZeta[k_, nn_] := Module[{z, d, x, g, eqs, sol, t, p},

%t z[x_, p_] := Sum[1/n^x, {n, 1, p}]; d[x_, p_] := Sum[f[n]/n^x, {n, 1, p}];

%t g[k] = Expand[z[k*x, nn] - z[x, nn]*d[x, nn]] /. (a_)^((c_)*(b_)) ->

%t Simplify[a^b]^c; eqs[k] =

%t Table[Simplify[0 == (1/m^(-x))*Plus @@ Cases[g[k], (_.)/m^x]], {m, 2, nn}];

%t sol[k] = Solve[Join[{f[1] == 1}, eqs[k]]][[1]];

%t t[k] = Table[f[n], {n, 1, nn}] /. sol[k]]

%t fqZeta[5,200]

%t (* Using G.f. (beyond nn = 73 numerical problems appear) *)

%t nn = 73; f[x_] := Sum[a[n]*(x^n/(1 - x^n)), {n, 1, nn}];

%t sol = SolveAlways[0 == Series[f[x] - Sum[x^n^5, {n, 1, nn}], {x, 0, nn}], x];

%t Flatten[Table[a[n], {n, 1, nn}] /. sol]

%o (PARI) a(n) = sumdiv(n, d, if (ispower(d, 5), moebius(n/d), 0)); \\ _Michel Marcus_, Mar 27 2015

%o (PARI) for(n=1, 100, print1(direuler(p=2, n, (1-X)/(1-X^5))[n], ", ")) \\ _Vaclav Kotesovec_, Jun 14 2020

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

%K sign,mult

%O 1

%A _Wolfgang Hintze_, Mar 25 2015