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URL: https://oeis.org/A256693

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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).
13
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
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, ...
In general, for m > 0, if Dirichlet g.f. is zeta(s)^m, then Sum_{j=1..n} a(j) ~ n*log(n)^(m-1)/Gamma(m) * (1 + (m-1)*(m*gamma - 1)/log(n)), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)
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)).
Sum_{j=1..n} A256692(j)/A256693(j) ~ n / (Gamma(1/5) * log(n)^(4/5)) * (1 + (4*(1 - gamma/5))/(5*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025
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 *)
PROG
(PARI) for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-X)^(1/5))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025
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
KEYWORD
nonn,frac,mult
AUTHOR
Wolfgang Hintze, Apr 08 2015
STATUS
approved