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A256693
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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).
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11
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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
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OFFSET
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1,2
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COMMENTS
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Formula holds for general Dirichlet g.f. zeta(x)^(1/k) with k = 1, 2, ...
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LINKS
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FORMULA
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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)).
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EXAMPLE
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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
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MATHEMATICA
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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]];
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CROSSREFS
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KEYWORD
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nonn,frac,mult
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AUTHOR
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STATUS
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approved
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