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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
1, 5, 5, 25, 5, 25, 5, 125, 25, 25, 5, 125, 5, 25, 25, 625, 5, 125, 5, 125, 25, 25, 5, 625, 25, 25, 125, 125, 5, 125, 5, 15625, 25, 25, 25, 625, 5, 25, 25, 625, 5, 125, 5, 125, 125, 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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

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

LINKS

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

FORMULA

with k = 5;

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

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

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

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

EXAMPLE

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

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

MATHEMATICA

k = 5;

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

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

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

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

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

num = Numerator[t] (* A256692 *)

den = Denominator[t] (* A256693 *)

CROSSREFS

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

Sequence in context: A262117 A145764 A165826 * A255458 A256135 A227076

Adjacent sequences:  A256690 A256691 A256692 * A256694 A256695 A256696

KEYWORD

nonn,frac,mult

AUTHOR

Wolfgang Hintze, Apr 08 2015

STATUS

approved

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Last modified January 21 02:59 EST 2019. Contains 319344 sequences. (Running on oeis4.)