%I #18 Jul 20 2017 07:19:35
%S 1,2,3,6,13,26,50,96,184,350,661,1242,2324,4332,8047,14902,27521,
%T 50700,93191,170942,312974,572030,1043852,1902044,3461067,6289972,
%U 11417576,20702328,37498589,67856074,122677727,221599538,399962369,721333090
%N a(n) = the number of occurrences of 1 in all compositions of n without 2's = # of occurrences of the integer k in compositions of n+k-1 without 2's (k > 2).
%H P. Chinn and S. Heubach, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL6/Heubach/heubach5.html">Integer Sequences Related to Compositions without 2's</a>, J. Integer Seqs., Vol. 6 no. 2, 2003, Article 03.2.3.
%H J. J. Madden, <a href="http://arxiv.org/abs/1707.04351">A generating function for the distribution of runs in binary words</a>, arXiv:1707.04351, Theorem 1.1, r=2, k=1.
%F a(n) = c(0)c(n-1) + c(1)c(n-2) + c(2)c(n-3) + ... + c(n-1)c(0), where c(i) is given by sequence A005251; generating function = (x(1-x)^2)/(1-2x+x^2-x^3)^2
%F a(n) = Sum_{k=1..floor((n+2)/3)} k*binomial(n-k+1, 2*k-1). - _Vladeta Jovovic_, Apr 10 2004
%e a(4)=6 since the compositions of 4 that do not contain a 2 are 1+1+1+1, 1+3, 3+1 and 4, for a total of 6 1's. Also there are 6 occurrences of 5 in the compositions of 8 (= 4+5-1): 1+1+1+5, 1+1+5+1, 1+5+1+1, 5+1+1+1, 5+3 and 3+5 (only compositions without 2's that contain a 5 are listed).
%t Rest[CoefficientList[ Normal[Series[x(1 - x)^2/((1 - 2x + x^2 - x^3)^2), {x, 0, 50}]], x]]
%Y Cf. A005251.
%K easy,nonn
%O 1,2
%A Silvia Heubach (sheubac(AT)calstatela.edu), Jan 23 2003