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%I #11 Oct 27 2018 21:57:02
%S 1,1,3,3,7,7,21,21,47,47,141,141,329,329,987,987,2207,2207,6621,6621,
%T 15449,15449,46347,46347,103729,103729,311187,311187,726103,726103,
%U 2178309,2178309,4870847,4870847,14612541,14612541,34095929,34095929
%N Products of distinct terms of 1 and rest from A001566: a(n) = Product_{i=0..floor(log_2(n+1))} L(2^i)^bit(n,i).
%C Used to produce the rows of A050609.
%C Also Sum(((C(2((n+((n+1) mod 2)) mod (2^floor(log_2(n)))),i) mod 2)*F(n+((n+1) mod 2)-i)),i=0..2((n+((n+1) mod 2)) mod (2^floor(log_2(n))))) or Sum(((C(2((n-(n mod 2)) mod (2^floor(log_2(n)))),i) mod 2)*L(n-(n mod 2)-i)),i=0..2((n-(n mod 2)) mod (2^floor(log_2(n))))) for all n > 1. Here F(n) and L(n) are n-th Fibonacci (A000045) and Lucas (A000032) numbers respectively.
%H A. Karttunen, <a href="http://www.fq.math.ca/Papers1/42-1/quartkarttunen01_2004.pdf">On Pascal's Triangle Modulo 2 in Fibonacci Representation</a>, Fibonacci Quarterly, 42 (2004), 38-46.
%p with(combinat); A050613 := n -> product('luc(2^i)^bit_i(n,i)','i'=0..floor_log_2(n+1));
%p luc := n -> (fibonacci(n-1)+fibonacci(n+1));
%p bit_i := (n,i) -> `mod`(floor(n/(2^i)),2);
%p floor_log_2 := proc(n) local nn,i; nn := n; for i from -1 to n do if(0 = nn) then RETURN(i); fi; nn := floor(nn/2); od; end;
%Y Bisection: A050614.
%K nonn
%O 0,3
%A _Antti Karttunen_, Dec 02 1999
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