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A006464
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Continued fraction for Sum_{n>=0} 1/4^(2^n).
(Formerly M2512)
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5
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0, 3, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 4, 2, 6, 4, 2, 4, 4, 6, 4, 2, 6, 4, 4, 2, 4, 6, 2, 4, 6, 4, 2, 4, 4, 6, 2, 4, 6, 4, 4, 2, 4, 6, 4
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OFFSET
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0,2
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COMMENTS
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a(n)=A004200(n) if n=0; A004200(n)+1 if n>0 (according to case u=3, b=1 of Theorem 5 (of the reference) which states that: if B(u,infinity) = Sum_{n>=0} 1/u^(2^n) = [a0, a1, a2, ...] then B(u + b,infinity) = [a0, a1+b, a2+b, a3+b,... ] (u >= 3, b >= 0)).
The sum is equal to 0.316421509021893143708079...= A078585.
After computing the first 10^5 terms and dropping the first two (0 & 3), only the numbers 2, 4 & 6 occur. Further I found no two 0's in a row and no three 2's or three 1's in a row. - Robert G. Wilson v, Dec 01 2002
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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EXAMPLE
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0.316421509021893143708079737... = 0 + 1/(3 + 1/(6 + 1/(4 + 1/(4 + ...)))). - Harry J. Smith, May 11 2009
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MAPLE
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u := 4: v := 7: Buv := [u, 1, [0, u-1, u+1]]: for k from 2 to v do n := nops(Buv[3]): Buv := [u, Buv[2]+1, [seq(Buv[3][i], i=1..n-1), Buv[3][n]+1, Buv[3][n]-1, seq(Buv[3][n-i], i=1..n-2)]] od:seq(Buv[3][i], i=1..2^v); # first 2^v terms of A006464, Antonio G. Astudillo (aft_astudillo(AT)hotmail.com), Dec 02 2002
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MATHEMATICA
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ContinuedFraction[ N[ Sum[1/4^(2^n), {n, 0, Infinity}], 1000]]
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PROG
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(PARI) { allocatemem(932245000); default(realprecision, 25000); x=suminf(n=0, 1/4^(2^n)); x=contfrac(x); for (n=1, 20001, write("b006464.txt", n-1, " ", x[n])); } \\ Harry J. Smith, May 11 2009
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CROSSREFS
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KEYWORD
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nonn,cofr
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AUTHOR
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EXTENSIONS
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Better description and more terms from Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 19 2001
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STATUS
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approved
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