%I A119900
%S A119900 1,0,2,0,1,3,0,0,4,4,0,0,1,10,5,0,0,0,6,20,6,0,0,0,1,21,35,7,0,0,0,0,8,
%T A119900 56,56,8,0,0,0,0,1,36,126,84,9,0,0,0,0,0,10,120,252,120,10,0,0,0,0,0,1,
%U A119900 55,330,462,165,11,0,0,0,0,0,0,12,220,792,792,220,12,0,0,0,0,0,0,1,78
%N A119900 Triangle read by rows: T(n,k) is the number of binary words of length 
               n with k strictly increasing runs (0<=k<=n; for example, the binary 
               word 1/0/01/01/1/1/01 has 7 strictly increasing runs).
%C A119900 Sum of entries in row n is 2^n (A000079). Sum of entries in column k 
               is A001906(k+1) (the even indexed Fibonacci numbers). Row n contains 
               1+floor(n/2) nonzero terms. Sum(k*T(n,k),k=0..n)=(3n+1)*2^(n-2)=A066373(n+1) 
               for n>=1. T(n,k)=A034867(n,n-k).
%C A119900 Triangle T(n,k), 0<=k<=n, read by rows, given by [0,1/2,-1/2,0,0,0,0,
               0,0,...] DELTA [2,-1/2,1/2,0,0,0,0,0,0,...] where DELTA is the operator 
               defined in A084938. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), 
               Dec 02 2008]
%F A119900 T(n,k)=binom(n+1,2k-n). G.f.=1/[1-2tz-t(1-t)z^2].
%e A119900 T(5,3)=6 because we have 0/01/01, 01/0/01, 01/01/0, 01/1/01, 01/01/1 
               and 1/01/01 (the runs are separated by /).
%e A119900 Triangle starts:
%e A119900 1;
%e A119900 0,2;
%e A119900 0,1,3;
%e A119900 0,0,4,4;
%e A119900 0,0,1,10,5;
%e A119900 0,0,0,6,20,6;
%p A119900 T:=(n,k)->binomial(n+1,2*k-n): for n from 0 to 12 do seq(T(n,k),k=0..n) 
               od; # yields sequence in triangular form
%Y A119900 Cf. A000079, A001906, A066373, A034867.
%Y A119900 Cf. A098158 [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Dec 02 2008]
%Y A119900 Sequence in context: A136493 A132213 A154312 this_sequence A141097 A096335 
               A129503
%Y A119900 Adjacent sequences: A119897 A119898 A119899 this_sequence A119901 A119902 
               A119903
%K A119900 nonn,tabf
%O A119900 0,3
%A A119900 Emeric Deutsch (deutsch(AT)duke.poly.edu), May 27 2006