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A256693 From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives denominator of b(n). 11

%I

%S 1,5,5,25,5,25,5,125,25,25,5,125,5,25,25,625,5,125,5,125,25,25,5,625,

%T 25,25,125,125,5,125,5,15625,25,25,25,625,5,25,25,625,5,125,5,125,125,

%U 25,5,3125,25,125,25,125,5,625,25,625,25,25,5,625,5,25,125,78125,25,125,5,125,25,125,5,3125,5,25,125,125,25,125,5,3125,625,25,5,625,25,25,25,625,5,625,25,125,25,25,25,78125,5,125,125,625

%N From fifth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is zeta function; sequence gives denominator of b(n).

%C Dirichlet g.f. of A256692(n)/A256693(n) is (zeta (x))^(1/5).

%C Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...

%H Wolfgang Hintze, <a href="/A256693/b256693.txt">Table of n, a(n) for n = 1..500</a>

%F with k = 5;

%F zeta(x)^(1/k) = Sum_{n>=1} b(n)/n^x;

%F c(1,n)=b(n); c(k,n) = Sum_{d|n} c(1,d)*c(k-1,n/d), k>1;

%F Then solve c(k,n) = 1 for b(m);

%F a(n) = denominator(b(n)).

%e b(1), b(2), ... =

%e 1, 1/5, 1/5, 3/25, 1/5, 1/25, 1/5, 11/125, 3/25, 1/25, 1/5, 3/125, 1/5, 1/25, 1/25, 44/625, 1/5, 3/125, 1/5, 3/125, 1/25, 1/25, 1/5, 11/625

%t k = 5;

%t c[1, n_] = b[n];

%t c[k_, n_] := DivisorSum[n, c[1, #1]*c[k - 1, n/#1] & ]

%t nn = 100; eqs = Table[c[k, n] == 1, {n, 1, nn}];

%t sol = Solve[Join[{b[1] == 1}, eqs], Table[b[i], {i, 1, nn}], Reals];

%t t = Table[b[n], {n, 1, nn}] /. sol[[1]];

%t num = Numerator[t] (* A256692 *)

%t den = Denominator[t] (* A256693 *)

%Y Cf. A046643/A046644 (k=2), A256688/A256689 (k=3), A256690/A256691 (k=4), A256692/A256693 (k=5).

%K nonn,frac,mult

%O 1,2

%A _Wolfgang Hintze_, Apr 08 2015

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Last modified August 13 17:37 EDT 2022. Contains 356107 sequences. (Running on oeis4.)