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Numerators of the convergents to the continued fraction for the constant A119812 defined by binary sums involving Beatty sequences: c = Sum_{n>=1} A049472(n)/2^n = Sum_{n>=1} 1/2^A001951(n).
2

%I #3 Mar 30 2012 18:36:57

%S 0,1,6,109,112494,1887350536045,543991754934632523092182415214,

%T 758213844806172103575972149363453352380811718063209070444420739586832237

%N Numerators of the convergents to the continued fraction for the constant A119812 defined by binary sums involving Beatty sequences: c = Sum_{n>=1} A049472(n)/2^n = Sum_{n>=1} 1/2^A001951(n).

%C The number of digits in these numerators are (beginning at n=2): [1,1,3,6,13,30,72,174,420,1013,2444,5901,14245,34391,83027,...].

%e c = 0.858267656461002055792260308433375148664905190083506778667684867..

%e Convergents begin:

%e [0/1, 1/1, 6/7, 109/127, 112494/131071, 1887350536045/2199023255551,..]

%e where the denominators of the convergents equal [2^A001333(n-1)-1]:

%e [1,1,7,127,131071,2199023255551,633825300114114700748351602687,...]

%e and A001333 is numerators of continued fraction convergents to sqrt(2).

%o (PARI) {a(n)=local(M=contfracpnqn(vector(n,k,if(k==1,0,if(k==2,1, 4^round(((1+sqrt(2))^(k-2)+(1-sqrt(2))^(k-2))/(2*sqrt(2))) +if(k==3,2,2^round(((1+sqrt(2))^(k-3)-(1-sqrt(2))^(k-3))/2))))))); return(M[1,1])}

%Y Cf. A119812 (constant), A119813 (continued fraction), A001333; A119809 (dual constant).

%K frac,nonn

%O 1,3

%A _Paul D. Hanna_, May 26 2006