|
%I #19 Aug 15 2015 13:05:02
%S 2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,
%T 16,32,32,64,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32,64,8,16,16,32,16,
%U 32,32,64,16,32,32,64,32,64,64,128,4,8,8,16,8,16,16,32,8,16,16,32,16,32
%N a(n) = 2^(wt(n)+1), where wt() = A000120().
%C Denominator of Zeta'(-2n).
%C If Gould's sequence A001316 is written as a triangle, this is what the rows converge to. In other words, let S_0 = [2], and construct S_{n+1} by following S_n with 2*S_n. Then this is S_{oo}. - _N. J. A. Sloane_, May 30 2009
%C In A160464 the coefficients of the ES1 matrix are defined. This matrix led to the discovery that the successive differences of the ES1[1-2*m,n] coefficients for m = 1, 2, 3, ..., are equal to the values of Zeta'(-2n), see also A094665 and A160468. - _Johannes W. Meijer_, May 24 2009
%H J. Sondow and E. W. Weisstein, <a href="http://mathworld.wolfram.com/RiemannZetaFunction.html">MathWorld: Riemann Zeta Function</a>
%F For n>=0, a(n) = 2 * A001316(n). - _N. J. A. Sloane_, May 30 2009
%F For n>0, a(n) = 4 * A048896(n). - _Peter Luschny_, May 02 2009
%F a(0) = 2; for n>0, write n = 2^i + j where 0 <= j < 2^i; then a(n) = 2*a(j).
%F a((2*n+1)*2^p-1) = 2^(p+1) * A001316(n), p >= 0. - _Johannes W. Meijer_, Jan 28 2013
%e -zeta(3)/(4*Pi^2), (3*zeta(5))/(4*Pi^4), (-45*zeta(7))/(8*Pi^6), (315*zeta(9))/(4*Pi^8), (-14175*zeta(11))/(8*Pi^10), ...
%p S := [2]; S := [op(S), op(2*S)]; # repeat ad infinitum! - _N. J. A. Sloane_, May 30 2009
%p a := n -> 2^(add(i,i=convert(n,base,2))+1); # _Peter Luschny_, May 02 2009
%t Denominator[(2*n)!/2^(2*n + 1)]
%Y Cf. A001316, A117972, A160464, A094665, A160468, A220466.
%K nonn,frac
%O 0,1
%A _Eric W. Weisstein_, Apr 06 2006
%E Entry revised by _N. J. A. Sloane_, May 30 2009
|
