%I #21 Jul 15 2017 12:18:20
%S 0,1,5,15,40,95,213,455,940,1890,3720,7194,13710,25805,48055,88665,
%T 162272,294865,532395,955795,1707110,3034836,5372400,9473700,16646700,
%U 29155225,50908793,88644915,153952120,266726195,461066385,795320159
%N Positions of those 1's that are followed by a 0, summed over all Fibonacci binary words of length n. A Fibonacci binary word is a binary word having no 00 subword.
%C a(n) = Sum(k*A119469(n+1,k),k>=0).
%C For n>1, a(n-1) is the n-th antidiagonal sum of A213777. [_Clark Kimberling_, Jun 21 2012]
%H M. Janjic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Janjic/janjic33.html">Hessenberg Matrices and Integer Sequences </a>, J. Int. Seq. 13 (2010) # 10.7.8.
%H A. F. Y. Zhao, <a href="http://www.emis.de/journals/JIS/VOL17/Zhao/zhao3.html">Pattern Popularity in Multiply Restricted Permutations</a>, Journal of Integer Sequences, 17 (2014), #14.10.3.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (3,0,-5,0,3,1).
%F G.f.: z^2*(1+2z)/(1-z-z^2)^3.
%F a(n) = A001628(n-1) + 2*A001628(n-2), n>1, a(0)=0, a(1)=1. [_Vladimir Kruchinin_, Apr 26 2011]
%e a(4)=15 because the Fibonacci binary words of length 4 are 1110, 1111, 1101, 1010, 1011, 0110, 0111, 0101 and the positions of those 1's that are followed by a 0 are 3, 2, 1, 3, 1, 3 and 2; their sum is 15.
%p G := z^2*(1+2*z)/(1-z-z^2)^3: Gser := series(G, z = 0, 38): seq(coeff(Gser, z, n), n = 1 .. 34);
%Y Cf. A119469.
%K nonn,easy
%O 1,3
%A _Emeric Deutsch_, Jan 04 2009