A256691 From fourth root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fourth power is zeta function; sequence gives denominator of b(n).
1, 4, 4, 32, 4, 16, 4, 128, 32, 16, 4, 128, 4, 16, 16, 2048, 4, 128, 4, 128, 16, 16, 4, 512, 32, 16, 128, 128, 4, 64, 4, 8192, 16, 16, 16, 1024, 4, 16, 16, 512, 4, 64, 4, 128, 128, 16, 4, 8192, 32, 128, 16, 128, 4, 512, 16, 512, 16, 16, 4, 512, 4, 16, 128, 65536, 16, 64, 4, 128, 16, 64, 4, 4096, 4, 16, 128, 128, 16, 64, 4, 8192, 2048, 16, 4, 512, 16, 16, 16, 512, 4, 512, 16, 128, 16, 16, 16, 32768, 4, 128, 128, 1024
Offset: 1
Examples
b(1), b(2), ... = 1, 1/4, 1/4, 5/32, 1/4, 1/16, 1/4, 15/128, 5/32, 1/16, 1/4, 5/128, 1/4, 1/16, 1/16, 195/2048, ...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..500 from Wolfgang Hintze)
Crossrefs
Programs
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Mathematica
k = 4; 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] (* A256690 *) den = Denominator[t] (* A256691 *) -
PARI
for(n=1, 100, print1(denominator(direuler(p=2, n, 1/(1-X)^(1/4))[n]), ", ")) \\ Vaclav Kotesovec, May 04 2025
Formula
with k = 4;
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} A256690(j)/A256691(j) ~ n / (Gamma(1/4) * log(n)^(3/4)) * (1 + (3*(1 - gamma/4))/(4*log(n))), where gamma is the Euler-Mascheroni constant A001620 and Gamma() is the gamma function. - Vaclav Kotesovec, May 04 2025
Comments