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From third root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is zeta function; sequence gives numerator of b(n).
10

%I #36 Apr 16 2015 15:25:37

%S 1,1,1,2,1,1,1,14,2,1,1,2,1,1,1,35,1,2,1,2,1,1,1,14,2,1,14,2,1,1,1,91,

%T 1,1,1,4,1,1,1,14,1,1,1,2,2,1,1,35,2,2,1,2,1,14,1,14,1,1,1,2,1,1,2,

%U 728,1,1,1,2,1,1,1,28,1,1,2,2,1,1,1,35,35,1,1,2,1,1,1,14,1,2,1,2,1,1,1,91,1,2,2,4

%N From third root of Riemann zeta function: form Dirichlet series Sum b(n)/n^x whose cube is zeta function; sequence gives numerator of b(n).

%C Dirichlet g.f. of A256688(n)/A256689(n) is (zeta (x))^(1/3).

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

%H Wolfgang Hintze, <a href="/A256688/b256688.txt">Table of n, a(n) for n = 1..500</a>

%F with k = 3;

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

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

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

%F a(n) = numerator(b(n)).

%e b(1), b(2), ... = 1, 1/3, 1/3, 2/9, 1/3, 1/9, 1/3, 14/81, 2/9, 1/9, 1/3, 2/27, 1/3, 1/9, 1/9, 35/243, ...

%t k = 3;

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

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

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

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

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

%t num = Numerator[t] (* A256688 *)

%t den = Denominator[t] (* A256689 *)

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

%K nonn,frac,mult

%O 1,4

%A _Wolfgang Hintze_, Apr 08 2015

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Last modified September 23 23:02 EDT 2024. Contains 376185 sequences. (Running on oeis4.)