A129706 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k inversions (n>=0, 0<=k<=floor(n(n+1)/6)). A Fibonacci binary word is a binary word having no 00 subword.

1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 4, 2, 1, 2, 2, 2, 4, 4, 4, 2, 1, 2, 2, 2, 4, 4, 6, 6, 4, 2, 2, 2, 2, 2, 4, 4, 6, 8, 8, 6, 6, 4, 2, 1, 2, 2, 2, 4, 4, 6, 8, 10, 10, 10, 10, 8, 6, 4, 2, 1, 2, 2, 2, 4, 4, 6, 8, 10, 12, 14, 14, 14, 14, 12, 10, 8, 4, 2, 2, 2, 2, 2, 4, 4, 6, 8, 10, 12, 16, 18, 18, 20

Offset: 0

Author

Emeric Deutsch, May 12 2007

Keywords

• nonn
• tabf

Comments

Row n has 1+floor(n(n+1)/6) terms. Row sums are the Fibonacci numbers (A000045). Sum(k*T(n,k), k>=0)=A129707(n).

Examples

			T(5,3)=4 because we have 11101, 10101, 01110 and 01010.
Triangle starts:
1;
2;
2,1;
2,2,1;
2,2,2,2;
2,2,2,4,2,1;
2,2,2,4,4,4,2,1;
		

Crossrefs

Cf. A000045, A129707.

Programs

Maple

Q[0]:=1: Q[1]:=1+x: for n from 2 to 12 do Q[n]:=expand(x*Q[n-1]+t*x*subs(x=t*x,Q[n-2])) od: for n from 0 to 15 do P[n]:=sort(subs(x=1,Q[n])) od: for n from 0 to 12 do seq(coeff(P[n],t,j),j=0..floor(n*(n+1)/6)) od; # yields sequence in triangular form

Formula

G.f.=G(t,z)=H(t,1,z), where H(t,x,z)=1+z+xzH(t,x,z)+txz^2*H(t,tx,z). 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)=xQ[n-1](t,x)+txQ[n-2](t,tx) for n>=2.