cp's OEIS Frontend

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.

%I A174736 #6 Jul 30 2012 18:11:22
%S A174736 2,8,2048,8796093022208,
%T A174736 2993155353253689176481146537402947624255349848014848
%N 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.
%C A174736 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 A174736 Proof :
%C A174736 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 A174736 Graphic representation :
%C A174736 0....................^
%C A174736 1............l.................l
%C A174736 2............^.................^
%C A174736 3.......^........^........^........^
%C A174736 4.....l...l... l...l... l...l... l...l
%C A174736 5.....l...l... l...l... l...l... l...l
%C A174736 6.....l...l... l...l... l...l... l...l
%C A174736 7.....l...l... l...l... l...l... l...l
%C A174736 8.....^...^... ^...^... ^...^... ^...^
%C A174736 9....^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
%C A174736 10..^^^^^^^^ ^^^^^^^^ ^^^^^^^^ ^^^^^^^^
%D A174736 J. E. Hutchinson, Fractal and self-similarity, Ind. U. Math. J. 30 (1981), 713-747.
%F A174736 a(2^2k) = 2^A007583.
%e A174736 k=0, a(1) = 2 ; k=1, a(4) = 8 ; k=2, a(16) = 2048 ; k=3, a(64) = 2^43 = 8796093022208.
%p A174736 for n from 0 to 5 do: p:=(2*4^n + 1)/3:q:=2^p:print(q):od:
%K A174736 nonn
%O A174736 1,1
%A A174736 _Michel Lagneau_, Mar 28 2010