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 A174736 Number of nodes with index 2^2k in a dyadic tree build alternatively with the schema "l" between the index 2^2k and 2^(2k+1) - 1, and the schema "^" between the index 2^(2k+1) and 2^(2k+2) - 1. 0
 2, 8, 2048, 8796093022208, 2993155353253689176481146537402947624255349848014848 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS This tree represent the set of the interval [0,1], and the number of nodes of rank n is a(n). The number nodes with index 2^2k is a(2^2k) = (2^(2k+1) + 1)/3= 2^A007583. Proof : Let n(k) such that a(2^2k) = 2^n(k). By recurrence we have n(k) = 2^2k - 2^2k-1 + n(k-1) = 2^2k-1 + n(k-1). With n(1) = 3 = 2+1 we obtain : n(k) = 1 + 2(1 + 2^2 + ... + 2^(2k-2)) = (2^(2k+1) + 1)/3. Graphic representation : 0....................^ 1............l.................l 2............^.................^ 3.......^........^........^........^ 4.....l...l... l...l... l...l... l...l 5.....l...l... l...l... l...l... l...l 6.....l...l... l...l... l...l... l...l 7.....l...l... l...l... l...l... l...l 8.....^...^... ^...^... ^...^... ^...^ 9....^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 10..^^^^^^^^ ^^^^^^^^ ^^^^^^^^ ^^^^^^^^ REFERENCES J. E. Hutchinson, Fractal and self-similarity, Ind. U. Math. J. 30 (1981), 713-747. LINKS Table of n, a(n) for n=1..5. FORMULA a(2^2k) = 2^A007583. EXAMPLE k=0, a(1) = 2 ; k=1, a(4) = 8 ; k=2, a(16) = 2048 ; k=3, a(64) = 2^43 = 8796093022208. MAPLE for n from 0 to 5 do: p:=(2*4^n + 1)/3:q:=2^p:print(q):od: CROSSREFS Sequence in context: A073630 A027733 A054874 * A324567 A135238 A133376 Adjacent sequences: A174733 A174734 A174735 * A174737 A174738 A174739 KEYWORD nonn AUTHOR Michel Lagneau, Mar 28 2010 STATUS approved

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