|
|
A257101
|
|
From fifth root of the inverse of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose fifth power is 1/zeta; sequence gives numerator of b(n).
|
|
4
|
|
|
1, -1, -1, -2, -1, 1, -1, -6, -2, 1, -1, 2, -1, 1, 1, -21, -1, 2, -1, 2, 1, 1, -1, 6, -2, 1, -6, 2, -1, -1, -1, -399, 1, 1, 1, 4, -1, 1, 1, 6, -1, -1, -1, 2, 2, 1, -1, 21, -2, 2, 1, 2, -1, 6, 1, 6, 1, 1, -1, -2, -1, 1, 2, -1596, 1, -1, -1, 2, 1, -1, -1, 12, -1, 1, 2, 2, 1, -1, -1, 21, -21, 1, -1, -2, 1, 1, 1, 6, -1, -2, 1, 2, 1, 1, 1, 399, -1, 2, 2, 4
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
COMMENTS
|
Denominator is the same as for Dirichlet g.f. (zeta(x))^(+1/5).
Formula holds for general Dirichlet g.f. zeta(x)^(-1/k) with k = 1, 2, ...
|
|
LINKS
|
|
|
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) = mu(n) for b(m);
a(n) = numerator(b(n)).
|
|
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]==MoebiusMu[n], {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]];
|
|
CROSSREFS
|
|
|
KEYWORD
|
sign
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|