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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

%I #6 Jul 30 2012 18:11:22

%S 2,8,2048,8796093022208,

%T 2993155353253689176481146537402947624255349848014848

%N 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.

%C 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.

%C Proof :

%C 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.

%C Graphic representation :

%C 0....................^

%C 1............l.................l

%C 2............^.................^

%C 3.......^........^........^........^

%C 4.....l...l... l...l... l...l... l...l

%C 5.....l...l... l...l... l...l... l...l

%C 6.....l...l... l...l... l...l... l...l

%C 7.....l...l... l...l... l...l... l...l

%C 8.....^...^... ^...^... ^...^... ^...^

%C 9....^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^

%C 10..^^^^^^^^ ^^^^^^^^ ^^^^^^^^ ^^^^^^^^

%D J. E. Hutchinson, Fractal and self-similarity, Ind. U. Math. J. 30 (1981), 713-747.

%F a(2^2k) = 2^A007583.

%e k=0, a(1) = 2 ; k=1, a(4) = 8 ; k=2, a(16) = 2048 ; k=3, a(64) = 2^43 = 8796093022208.

%p for n from 0 to 5 do: p:=(2*4^n + 1)/3:q:=2^p:print(q):od:

%K nonn

%O 1,1

%A _Michel Lagneau_, Mar 28 2010

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Last modified June 22 03:16 EDT 2024. Contains 373561 sequences. (Running on oeis4.)