A092910 -id:A092910 - 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A. J. van der Poorten, An introduction to continued fractions, Unpublished.
A. J. van der Poorten, An introduction to continued fractions, Unpublished [Cached copy]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)

A088431 Half of the (n+1)-st component of the continued fraction expansion of Sum_{k>=0} 1/2^(2^k).

Original entry on oeis.org

2, 1, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 3, 2, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 2, 2, 1, 3, 2, 1, 2, 2, 3, 2, 1, 3, 2, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 2, 3, 1, 2, 3, 2, 2, 1, 2, 3, 1, 2, 2, 3, 2, 1, 3, 2, 1, 2, 3, 2, 2, 1, 3, 2, 1, 2, 2, 3, 2, 1, 2, 3

Offset: 1

Benoit Cloitre, Nov 08 2003

nonn

To construct the sequence use the rule: a(1)=2, then a(a(1) + a(2) + ... + a(n) + 1) = 2 and fill in any undefined places with the sequence 1,3,1,3,1,3,1,3,1,3,1,3,....
This sequence appears to be the sequence of run lengths of the regular paperfolding sequence A014577, i.e., the latter starts as follows: 2 zeros, 1 one, 2 zeros, 2 ones, etc. - Dimitri Hendriks, May 06 2010
Hendriks' conjecture is proved in Bunder, Bates, and Arnold (2024).  Also see Shallit (2024). - Jeffrey Shallit, Mar 10 2025

			Example to illustrate the comment: a(a(1)+1)=a(3)=2 and a(2) is undefined. The rule requires a(2)=1. Next, a(a(1)+a(2)+1)=a(4)=2, a(a(1)+a(2)+a(3)+1)=a(6)=2 and a(5) is undefined. The rule now requires a(5)=3.

Antti Karttunen, Table of n, a(n) for n = 1..8192
Martin Bunder, Bruce Bates, and Stephen Arnold, The summed paperfolding sequence, Bull. Austral. Math. Soc. 110 (2024), 189-198.
Kevin Ryde, Iterations of the Dragon Curve, see index "TurnRun", with a(n) = TurnRun(n-1).
Jeffrey Shallit, Runs in Paperfolding Sequences, arXiv:2412.17930 [math.CO], 2024. See p. 10.

Cf. A007400, A014577, A088435, A092910.

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]]]; Array[a[# + 1]/2 &, 98] (* after Jean-François Alcover at A007400 *)

(define (A088431 n) (* 1/2 (A007400 (+ 1 n)))) ;; Code for A007400 given under that entry. - Antti Karttunen, Aug 12 2017

a(n) = (1/2)*A007400(n+1); a(a(1) + a(2) + ... + a(n) + 1) = 2.

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

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A088431 Half of the (n+1)-st component of the continued fraction expansion of Sum_{k>=0} 1/2^(2^k).

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