A129708 - OEIS

%I #6 May 03 2017 08:42:32

%S 1,2,3,4,1,6,2,9,3,1,13,6,2,19,11,3,1,28,18,7,2,41,30,14,3,1,60,50,24,

%T 8,2,88,81,43,17,3,1,129,130,77,30,9,2,189,208,132,57,20,3,1,277,330,

%U 224,108,36,10,2,406,520,379,193,72,23,3,1,595,816,633,342,143,42,11,2,872

%N Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 010 subwords (n>=0, 0<=k<=floor((n-1)/2)). A Fibonacci binary word is a binary word having no 00 subword.

%C Row n has floor((n+1)/2) terms (n>=1). Row sums are the Fibonacci numbers (A000045). T(n,0)=A000930(n+2). Sum(k*T(n,k), k>=0)=A001629(n-1).

%H G. C. Greubel, <a href="/A129708/b129708.txt">Table of n, a(n) for the first 100 rows, flattened</a>

%F G.f.: G(t,z) = (1+z+z^2-tz^2)/(1-z-tz^2+tz^3-z^3). Row generating polynomials P[n] are given by P[n](t)=Q[n](t,1), where Q[0]=1, Q[1]=1+x, Q[n](t,x)=Q[n-1](t,1)+xQ[n-2](t,t) for n>=2.

%e T(7,2)=3 because we have 1101010, 1010101 and 0101011.

%e Triangle starts:

%e 1;

%e 2;

%e 3;

%e 4,1;

%e 6,2;

%e 9,3,1;

%e 13,6,2;

%e 19,11,3,1;

%p Q[0]:=1: Q[1]:=1+x: for n from 2 to 30 do Q[n]:=expand(subs(x=1,Q[n-1])+x*subs(x=t,Q[n-2])) od: for n from 0 to 18 do P[n]:=subs(x=1,Q[n]) od; 1; for n from 1 to 18 do seq(coeff(P[n],t,j),j=0..floor((n-1)/2)) od; # yields sequence in triangular form

%t Flatten[CoefficientList[CoefficientList[Series[(1 + z + z^2 - t*z^2)/(1 - z - t*z^2 + t*z^3 - z^3), {z, 0, 20}, {t, 0, 20}], z], t]] (* _G. C. Greubel_, May 02 2017 *)

%Y Cf. A000045, A000930, A001629.

%K nonn,tabf

%O 0,2

%A _Emeric Deutsch_, May 12 2007