A119810 Partial quotients of the continued fraction of the constant defined by binary sums involving Beatty sequences: c = Sum_{n>=1} 1/2^A049472(n) = Sum_{n>=1} A001951(n)/2^n.
2, 3, 10, 132, 131104, 2199023259648, 633825300114114700748888473600, 883423532389192164791648750371459257913741948437810659652423818057613312
Offset: 1
Examples
c = 2.32258852258806773012144068278798408011950250800432925665718... The partial quotients start: a(1) = 2^1; a(2) = 2^1 + 2^0; a(3) = 2^3 + 2^1; a(4) = 2^7 + 2^2; a(5) = 2^17 + 2^5; a(6) = 2^41 + 2^12; and continue as a(n) = 2^A001333(n-1) + 2^A000129(n-2) where A001333(n) = ( (1+sqrt(2))^n + (1-sqrt(2))^n )/2; A000129(n) = ( (1+sqrt(2))^n - (1-sqrt(2))^n )/(2*sqrt(2)).
Links
- W. W. Adams and J. L. Davison, A remarkable class of continued fractions, Proc. Amer. Math. Soc. 65 (1977), 194-198.
- P. G. Anderson, T. C. Brown, P. J.-S. Shiue, A simple proof of a remarkable continued fraction identity, Proc. Amer. Math. Soc. 123 (1995), 2005-2009.
- J. L. Davison, A series and its associated continued fraction, Proc. Amer. Math. Soc., 63 (1977), 29-32.
Programs
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Mathematica
(* b = A001333 *) b[0] = 1; b[1] = 1; b[n_] := b[n] = 2 b[n-1] + b[n-2]; a[1] = 2; a[n_] := 2^b[n-1] + 2^Fibonacci[n-2, 2]; Array[a, 10] (* Jean-François Alcover, May 04 2017 *)
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PARI
{a(n)=if(n==1,2,2^round(((1+sqrt(2))^(n-1)+(1-sqrt(2))^(n-1))/2) +2^round(((1+sqrt(2))^(n-2)-(1-sqrt(2))^(n-2))/(2*sqrt(2))))}
Comments