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

Author

Simon Plouffe, Jeffrey Shallit

Keywords

• nonn
• cofr
• easy

Examples

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

References

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

Links

• 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• A. J. van der Poorten, An introduction to continued fractions, Unpublished.
• A. J. van der Poorten, An introduction to continued fractions, Unpublished [Cached copy]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

Crossrefs

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

Programs

Maple

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

Mathematica

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 *)

PARI

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 */

PARI

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

PARI

{ 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

Scheme

(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

Formula

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)