

A331466


The number of common terms in the Zeckendorf and dual Zeckendorf representations of n.


2



0, 1, 1, 0, 2, 0, 1, 2, 0, 1, 1, 1, 3, 0, 1, 1, 0, 2, 1, 2, 3, 0, 1, 1, 1, 2, 0, 1, 2, 1, 2, 2, 2, 4, 0, 1, 1, 0, 2, 1, 2, 2, 0, 1, 1, 1, 3, 1, 2, 2, 1, 3, 2, 3, 4, 0, 1, 1, 1, 2, 0, 1, 2, 1, 2, 2, 2, 3, 0, 1, 1, 0, 2, 1, 2, 3, 1, 2, 2, 2, 3, 1, 2, 3, 2, 3, 3
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,5


COMMENTS

The indices of records are numbers of the form F(2*k  1)  1, for k > 0, where F(k) is the kth Fibonacci number. The corresponding record values are k  1 = 0, 1, 2, ...


LINKS



FORMULA



EXAMPLE

a(6) = 1 since the Zeckendorf representation of 6 is 1001 (i.e., F(2) + F(5)), its dual Zeckendorf representation is 111 (i.e., F(2) + F(3) + F(4)), and there is only one position with a common digit 1, corresponding to the one common summand F(2).


MATHEMATICA

m = 1000; zeck = Select[Range[0, m], BitAnd[#, 2 #] == 0 &]; dualZeck = Select[Range[0, m], SequenceCount[IntegerDigits[#, 2], {0, 0}] == 0 &]; DigitCount[BitAnd[zeck[[#]], dualZeck[[#]]] & /@ Range[Min[Length[zeck], Length[dualZeck]]], 2, 1]


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



