A073088 -id:A073088 - cp's OEIS frontend

A007400 Continued fraction for Sum_{n>=0} 1/2^(2^n) = 0.8164215090218931...

0, 1, 4, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 4, 6, 4, 2, 6, 4, 2, 4, 6, 4, 4, 2, 6, 4

Offset: 0

Simon Plouffe, Jeffrey Shallit

			0.816421509021893143708079737... = 0 + 1/(1 + 1/(4 + 1/(2 + 1/(4 + ...))))

M. Kmošek, Rozwinieçie Niektórych Liczb Niewymiernych na Ulamki Lancuchowe (Continued Fraction Expansion of Some Irrational Numbers), Master's thesis, Uniwersytet Warszawski, 1979.

Harry J. Smith, Table of n, a(n) for n = 0..20000
Henry Cohn, Symmetry and specializability in continued fractions, Acta Arithmetica, volume 75, number 4, 1996, pages 297-320 (PDF). Also arXiv:math/0008221 [math.NT].
W. F. Lunnon, Q-D Tables and Zero-Squares, Manuscript, Jan. 1974. (Annotated scanned copy)
R. M. MacGregor, Generalizing the notion of a periodic sequence, American Math. Monthly 87 (1980), 90-102. (Annotated scanned copy)
B. Salvy, Email to N. J. A. Sloane, May 1994
Ratpoly info, May 1994
Jeffrey Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), no. 2, 209-217.
Jeffrey O. Shallit, Simple continued fractions for some irrational numbers, J. Number Theory 11 (1979), no. 2, 209-217.
A. J. van der Poorten, An introduction to continued fractions, Unpublished.
A. J. van der Poorten, An introduction to continued fractions, Unpublished [Cached copy]
G. Xiao, Contfrac
Index entries for continued fractions for constants

Cf. A007404 (decimal), A073088 (partial sums), A073414/A073415 (convergents), A088431 (half), A089267, A092910.

a:= proc(n) option remember; local n8, n16;
    n8:= n mod 8;
    if n8 = 0 or n8 = 3 then return 2
    elif n8 = 4 or n8 = 7 then return 4
    elif n8 = 1 then return procname((n+1)/2)
    elif n8 = 2 then return procname((n+2)/2)
    fi;
    n16:= n mod 16;
    if n16 = 5 or n16 = 14 then return 4
    elif n16 = 6 or n16 = 13 then return 6
    fi
end proc:
a(0):= 0: a(1):= 1: a(2):= 4:
map(a, [$0..1000]); # Robert Israel, Jun 14 2016

a[n_] := a[n] = Which[n < 3, {0, 1, 4}[[n+1]], Mod[n, 8] == 1, a[(n+1)/2], Mod[n, 8] == 2, a[(n+2)/2], True, {2, 0, 0, 2, 4, 4, 6, 4, 2, 0, 0, 2, 4, 6, 4, 4}[[Mod[n, 16]+1]]]; Table[a[n], {n, 0, 98}] (* Jean-François Alcover, Nov 29 2013, after Ralf Stephan *)

a(n)=if(n<3,[0,1,4][n+1],if(n%8==1,a((n+1)/2),if(n%8==2,a((n+2)/2),[2,0,0,2,4,4,6,4,2,0,0,2,4,6,4,4][(n%16)+1]))) /* Ralf Stephan */

a(n)=contfrac(suminf(n=0,1/2^(2^n)))[n+1]

{ allocatemem(932245000); default(realprecision, 26000); x=suminf(n=0, 1/2^(2^n)); x=contfrac(x); for (n=1, 20001, write("b007400.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 07 2009

(define (A007400 n) (cond ((<= n 1) n) ((= 2 n) 4) (else (case (modulo n 8) ((0 3) 2) ((4 7) 4) ((1) (A007400 (/ (+ 1 n) 2))) ((2) (A007400 (/ (+ 2 n) 2))) (else (case (modulo n 16) ((5 14) 4) ((6 13) 6))))))) ;; (After Ralf Stephan's recurrence) - Antti Karttunen, Aug 12 2017

From Ralf Stephan, May 17 2005: (Start)
a(0)=0, a(1)=1, a(2)=4; for n > 2:
a(8k)    = a(8k+3)   = 2;
a(8k+4)  = a(8k+7)   = a(16k+5) = a(16k+14) = 4;
a(16k+6) = a(16k+13) = 6;
a(8k+1)  = a(4k+1);
a(8k+2)  = a(4k+2). (End)

A073089 a(n) = (1/2)*(4n - 3 - Sum_{k=1..n} A007400(k)).

Original entry on oeis.org

0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1

Offset: 1

Benoit Cloitre, Aug 18 2002

From Joerg Arndt, Oct 28 2013: (Start)
Sequence is (essentially) obtained by complementing every other term of A014577.
Turns (by 90 degrees) of a curve similar to the Heighway dragon which can be rendered as follows: [Init] Set n=0 and direction=0. [Draw] Draw a unit line (in the current direction). Turn left/right if a(n) is zero/nonzero respectively. [Next] Set n=n+1 and goto (draw).
See the linked pdf files for two renderings of the curve. (End)

			From _Paul D. Hanna_, Oct 19 2012: (Start)
Let F(x) = x + 1/x + 1/x^3 + 1/x^7 + 1/x^15 + 1/x^31 +...+ 1/x^(2^n-1) +...
then F(x) = x + 1/(x + 1/(-x + 1/(-x + 1/(-x + 1/(x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(x + 1/(x + 1/(-x + 1/(-x + 1/(x + 1/(-x + 1/(-x + 1/(-x + 1/(x +...+ 1/((-1)^a(n)*x +...)))))))))))))))))))))),
a continued fraction in which the partial quotients equal (-1)^a(n)*x.  (End)

Antti Karttunen, Table of n, a(n) for n = 1..65537
Joerg Arndt, pdf rendering of the curve described in comment, alternate rendering of the curve.
Index entries for characteristic functions

Cf. A007400, A073088 (the sum part here), A123725.

a(n)=if(n<2,0,if(n%8==1,a((n+1)/2),[1,-1,0,1,1,1,0,0,1,-1,0,1,1,0,0,0][(n%16)+1])) \\ Ralf Stephan

/* Using the Continued Fraction, Print 2^N terms of this sequence: */
{N=10;CF=contfrac(x+sum(n=1,N,1/x^(2^n-1)),2^N);for(n=1,2^N,print1((1-CF[n]/x)/2,", "))} \\ Paul D. Hanna, Oct 19 2012

a(n) = { if ( n<=1, return(0)); n-=1; my(v=2^valuation(n,2) ); return( (0==bitand(n, v<<1)) != (v%2) ); } \\ Joerg Arndt, Oct 28 2013

Recurrence: a(1) = a(4n+2) = a(8n+7) = a(16n+13) = 0, a(4n) = a(8n+3) = a(16n+5) = 1, a(8n+1) = a(4n+1).
G.f.: The following series has a simple continued fraction expansion:
x + Sum_{n>=1} 1/x^(2^n-1) = [x; x, -x, -x, -x, x, ..., (-1)^a(n)*x, ...]. - Paul D. Hanna, Oct 19 2012
a(n) = A014577(n-2) + A056594(n).  Conjecture: a(n) = (1 + (-1)^A057661(n - 1))/2 for all n > 1. - Velin Yanev, Feb 01 2021

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A007400 Continued fraction for Sum_{n>=0} 1/2^(2^n) = 0.8164215090218931...

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A073089 a(n) = (1/2)*(4n - 3 - Sum_{k=1..n} A007400(k)).

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