cp's OEIS Frontend

A129718 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k runs of 1's (n >= 0, 0 <= k <= floor((n+1)/2)). A Fibonacci binary word is a binary word having no 00 subword. A run of 1's is a maximal subword of the form 11..1.

%I A129718 #11 Dec 29 2023 10:55:39
%S A129718 1,1,1,0,3,0,4,1,0,4,4,0,4,8,1,0,4,12,5,0,4,16,13,1,0,4,20,25,6,0,4,
%T A129718 24,41,19,1,0,4,28,61,44,7,0,4,32,85,85,26,1,0,4,36,113,146,70,8,0,4,
%U A129718 40,145,231,155,34,1,0,4,44,181,344,301,104,9,0,4,48,221,489,532,259,43,1
%N A129718 Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k runs of 1's (n >= 0, 0 <= k <= floor((n+1)/2)). A Fibonacci binary word is a binary word having no 00 subword. A run of 1's is a maximal subword of the form 11..1.
%C A129718 Row n has 1+floor((n+1)/2) terms.
%C A129718 Row sums are the Fibonacci numbers (A000045).
%C A129718 T(n,k) = A129717(n,k-1) (since in each word the number of runs of 1's = 1 + the number of 101's).
%C A129718 Sum_{k=0..floor((n+1)/2)} k*T(n,k) = A055244(n) (n >= 1).
%F A129718 G.f. = G(t,z) = (1+z)(1-z+tz)/(1-z-tz^2).
%F A129718 T(n,k) = binomial(n-k,k-1) + 2*binomial(n-k-1,k-1) + binomial(n-k-2,k-1) for n >= 4 and 0 <= k < floor((n+1)/2).
%e A129718 T(6,3)=5 because we have 110101, 101101, 101010, 101011 and 010101.
%e A129718 Triangle starts:
%e A129718   1;
%e A129718   1,  1;
%e A129718   0,  3;
%e A129718   0,  4,  1;
%e A129718   0,  4,  4;
%e A129718   0,  4,  8,  1;
%e A129718   0,  4, 12,  5;
%p A129718 G:=(1+z)*(1-z+t*z)/(1-z-t*z^2): Gser:=simplify(series(G,z=0,21)): for n from 0 to 18 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 17 do seq(coeff(P[n],t,j),j=0..ceil(n/2)) od; # yields sequence in triangular form
%Y A129718 Cf. A000045, A129717, A055244.
%K A129718 nonn,tabf
%O A129718 0,5
%A A129718 _Emeric Deutsch_, May 12 2007