A256691 - OEIS

%I #24 Apr 16 2015 15:26:27

%S 1,4,4,32,4,16,4,128,32,16,4,128,4,16,16,2048,4,128,4,128,16,16,4,512,

%T 32,16,128,128,4,64,4,8192,16,16,16,1024,4,16,16,512,4,64,4,128,128,

%U 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

%N 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).

%C Dirichlet g.f. of A256690(n)/A256691(n) is (zeta (x))^(1/4).

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

%H Wolfgang Hintze, <a href="/A256691/b256691.txt">Table of n, a(n) for n = 1..500</a>

%F with k = 4;

%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) = denominator(b(n)).

%e 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, ...

%t k = 4;

%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] (* A256690 *)

%t den = Denominator[t] (* A256691 *)

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

%K nonn,frac,mult

%O 1,2

%A _Wolfgang Hintze_, Apr 08 2015