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Asymmetricity-index for Zeckendorf-expansion A014417(n) of n.
2

%I #2 Mar 31 2012 14:02:28

%S 0,0,1,1,0,1,0,2,1,0,2,1,0,1,0,2,2,1,2,1,3,1,0,2,2,1,1,0,2,2,1,3,1,0,

%T 1,0,2,2,1,2,1,3,2,1,3,3,2,2,1,3,1,0,3,2,4,1,0,2,2,1,2,1,3,1,0,2,2,1,

%U 2,1,3,3,2,1,0,2,2,1,3,1,0,3,2,4,2,1,3,1,0,1,0,2,2,1,2,1,3,2,1,3,3,2,2,1,3

%N Asymmetricity-index for Zeckendorf-expansion A014417(n) of n.

%C Least number of flips of "fibits" (changing either 0 to 1 or 1 to 0 in Zeckendorf-expansion A014417(n)) so that a palindrome is produced.

%e The integers 0 and 1 look as '0' and '1' also in Fibonacci-representation,

%e and being palindromes, a(0) and a(1) = 0.

%e 2 has Fibonacci-representation '10', which needs a flip of other 'fibit',

%e that it would become a palindrome, thus a(2) = 1. Similarly 3 has representation

%e '100', so flipping for example the least significant fibit, we get '101',

%e thus a(3)=1 as well. 7 (= F(3)+F(5)) has representation '1010', which needs

%e two flips to produce a palindrome, thus a(7)=2. Here F(n) = A000045(n).

%Y a(n) = A037888(A003714(n)). A094202 gives the positions of zeros. Cf. also A095732.

%K base,nonn

%O 0,8

%A _Antti Karttunen_, Jun 05 2004