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%I #19 Mar 11 2016 06:04:54
%S 1,4,9,33,56,203,441,1596,2585,9353,20304,73461,124033,448756,974169,
%T 3524577,5702888,20633243,44791065,162055596,273617239,989956471,
%U 2149017696,7775219067,12591974497,45558191716,98898651657
%N Sum_{i=0..2n} (C(2n,i) mod 2)*Fibonacci(i+2) = Sum_{i=0..n} (C(n,i) mod 2)*Fibonacci(2i+2).
%C The history of 1-D CA Rule 90 starting from the seed pattern 1 interpreted as Zeckendorffian expansion.
%C Also, product of distinct terms of A001566 and appropriate Fibonacci or Lucas numbers: a(n) = FL(n+2)Product(L(2^i)^bit(n,i),i=0..) Here L(2^i) = A001566 and FL(n) = n-th Fibonacci number if n even, n-th Lucas number if n odd. bit(n,i) is the i-th digit (0 or 1) in the binary expansion of n, with the least significant digit being bit(n,0).
%H Antti 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.
%e 1 = Fib(2) = 1;
%e 101 = Fib(4) + Fib(2) = 3 + 1 = 4;
%e 10001 = Fib(6) + Fib(2) = 8 + 1 = 9;
%e 1010101 = Fib(8) + Fib(6) + Fib(4) + Fib(2) = 21 + 8 + 3 + 1 = 33; etc.
%t Table[Sum[Mod[Binomial[2n, i], 2] Fibonacci[i + 2], {i, 0, 2n}], {n, 0, 19}] (* _Alonso del Arte_, Apr 27 2014 *)
%Y a(n) = A022290(A038183(n)) = A022290(A048723(5, n)) = A003622(A051656(n)) = A075148(n, 2)*A050613(n). Third row of A050609, third column of A050610.
%Y Cf. A054433.
%K easy,nonn
%O 0,2
%A _Antti Karttunen_, Jul 13 1999
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