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

0,3

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A. J. van der Poorten, An introduction to continued fractions, Unpublished.

A. J. van der Poorten, An introduction to continued fractions, Unpublished [Cached copy]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

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)

Example

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

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A103859 A253808 A346171 * A057696 A057697 A019921

Adjacent sequences: A007397 A007398 A007399 * A007401 A007402 A007403

Keyword

nonn,cofr,easy

Author

Simon Plouffe, Jeffrey Shallit

Status

approved