A129719 - OEIS

%I #11 Nov 17 2019 01:49:31

%S 1,1,1,2,1,2,2,1,4,3,1,4,5,3,1,8,8,4,1,8,12,9,4,1,16,20,13,5,1,16,28,

%T 25,14,5,1,32,48,38,19,6,1,32,64,66,44,20,6,1,64,112,104,63,26,7,1,64,

%U 144,168,129,70,27,7,1,128,256,272,192,96,34,8,1,128,320,416,360,225,104,35

%N Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 0's in odd positions (0 <= k <= ceiling(n/2)). A Fibonacci binary word is a binary word having no 00 subword.

%C Row n has 1+ceiling(n/2) terms.

%F G.f.: G(t,z) = (1+z)(1+tz-tz^2)/(1-(2+t)z^2+tz^4). The trivariate generating function H(t,s,z), where t marks number of 0's in odd position and s marks number of 0's in even position, is given by H(t,s,z) = (1+(1+t)z-tsz^3)/(1-(1+t+s)z^2+tsz^4).

%F Row sums are the Fibonacci numbers (A000045).

%F T(2n,k) = T(2n-1,k) + T(2n-2,k) (n >= 1).

%F T(2n,k) = A129721(2n,k).

%F Sum_{k=0..ceiling(n/2)} k*T(n,k) = A129720(n).

%e T(6,2)=4 because we have 110101, 011101, 010110 and 010111.

%e Triangle starts:

%e   1;

%e   1, 1;

%e   2, 1;

%e   2, 2, 1;

%e   4, 3, 1;

%e   4, 5, 3, 1;

%e   8, 8, 4, 1;

%p G:=(1+z)*(1+t*z-t*z^2)/(1-(2+t)*z^2+t*z^4): Gser:=simplify(series(G,z=0,20)): for n from 0 to 17 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 Cf. A000045, A129720, A129721.

%K nonn,tabf

%O 0,4

%A _Emeric Deutsch_, May 13 2007