

A007400


Continued fraction for sum(n>=0, 1/2^(2^n) ) = 0.8164215090218931...


18



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
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OFFSET

0,3


LINKS

Harry J. Smith, Table of n, a(n) for n = 0..20000
H. Cohn, Symmetry and specializability in continued fractions
Jeffrey Shallit, Simple continued fractions for some irrational numbers. J. Number Theory 11 (1979), no. 2, 209217.
A. J. van der Poorten, An introduction to continued fractions
G. Xiao, Contfrac
Index entries for continued fractions for constants


FORMULA

Recurrence: a(0)=0, a(1)=1, a(2)=4, a(8n)=a(8n+3)=2, a(8n+4)=a(8n+7)=a(16n+5)=a(16n+14)=4, a(16n+6)=a(16n+13)=6, a(8n+1)=a(4n+1), a(8n+2)=a(4n+2).  Ralf Stephan, May 17 2005


EXAMPLE

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


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}] (* JeanFrançois 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", n1, " ", x[n])); } \\ Harry J. Smith, May 07 2009


CROSSREFS

Cf. A007404, A089267.
Sequence in context: A135282 A179411 A103859 * A057696 A057697 A019921
Adjacent sequences: A007397 A007398 A007399 * A007401 A007402 A007403


KEYWORD

nonn,cofr


AUTHOR

Simon Plouffe, Jeffrey Shallit


STATUS

approved



