A079662 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).
1, 2, 3, 6, 13, 26, 50, 96, 184, 350, 661, 1242, 2324, 4332, 8047, 14902, 27521, 50700, 93191, 170942, 312974, 572030, 1043852, 1902044, 3461067, 6289972, 11417576, 20702328, 37498589, 67856074, 122677727, 221599538, 399962369, 721333090
Offset: 1
Examples
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).
Links
- P. Chinn and S. Heubach, Integer Sequences Related to Compositions without 2's, J. Integer Seqs., Vol. 6 no. 2, 2003, Article 03.2.3.
- J. J. Madden, A generating function for the distribution of runs in binary words, arXiv:1707.04351 [math.CO], 2017, Theorem 1.1, r=2, k=1.
- Index entries for linear recurrences with constant coefficients, signature (4,-6,6,-5,2,-1).
Crossrefs
Cf. A005251.
Programs
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Mathematica
Rest[CoefficientList[ Normal[Series[x(1 - x)^2/((1 - 2x + x^2 - x^3)^2), {x, 0, 50}]], x]]
Formula
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
a(n) = Sum_{k=1..floor((n+2)/3)} k*binomial(n-k+1, 2*k-1). - Vladeta Jovovic, Apr 10 2004
a(n) = 4*a(n-1) - 6*a(n-2) + 6*a(n-3) - 5*a(n-4) + 2*a(n-5) - a(n-6) for n > 6. - Chai Wah Wu, Apr 15 2025