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a(1) = 1, a(n+1) = numerator of the continued fraction [1; 2, 4, 8, ..., 2^n].
6

%I #18 Dec 28 2016 08:22:23

%S 1,3,13,107,1725,55307,3541373,453351051,116061410429,59423895490699,

%T 60850185043886205,124621238393774438539,510448653311085144141949,

%U 4181595492545647894585284747,68511261060316548415970449436797

%N a(1) = 1, a(n+1) = numerator of the continued fraction [1; 2, 4, 8, ..., 2^n].

%H Harvey P. Dale, <a href="/A061377/b061377.txt">Table of n, a(n) for n = 1..82</a>

%F a(n) = 2^(n-1)*a(n-1) + a(n-2). - _Orson R. L. Peters_, Dec 28 2016

%F 0 = a(n)*(-2*a(n+2)) + a(n+1)*(+a(n+1) - a(n+3)) + a(n+2)*(+2*a(n+2)) if n>0. - _Michael Somos_, Dec 28 2016

%e G.f. = x + 3*x^2 + 13*x^3 + 107*x^4 + 1725*x^5 + 55307*x^6 + 3541373*x^7 + ...

%e a(3) = 13, the numerator of 1 + 1/(2 + 1/4) = 13/9.

%p with(numtheory); f := n->numer(cfrac([seq (2^i,i=0..n)])); for n from 0 to 25 do printf("%d,",f(n)) od;

%t Module[{nn=20,c},c=2^Range[0,nn];Table[Numerator[ FromContinuedFraction[ Take[ c,n]]],{n,nn}]] (* _Harvey P. Dale_, Jun 04 2014 *)

%o (PARI) {a(n) = if( n<1, 0, n<3, 2*n-1, 2^(n-1)*a(n-1) + a(n-2))}; /* _Michael Somos_, Dec 28 2016 */

%Y Denominators are sequence A015473.

%K nonn,easy,frac

%O 1,2

%A _Amarnath Murthy_, May 02 2001

%E More terms from Larry Reeves (larryr(AT)acm.org) and Winston C. Yang (winston(AT)cs.wisc.edu), May 15 2001