%I #20 Jan 27 2020 01:24:41

%S 1,0,1,0,1,1,0,0,1,1,0,0,1,2,1,0,0,0,1,2,1,0,0,0,1,3,3,1,0,0,0,0,1,3,

%T 3,1,0,0,0,0,1,4,6,4,1,0,0,0,0,0,1,4,6,4,1,0,0,0,0,0,1,5,10,10,5,1,0,

%U 0,0,0,0,0,1,5,10,10,5,1,0,0,0,0,0,0,1,6,15,20,15,6,1,0,0,0,0,0,0,0,1,6,15

%N Triangle read by rows: triangle of repeated stepped binomial coefficients.

%C Row sums are Sum_{k=0..n} binomial(floor(n/2),n-k) = (1,1,2,2,4,4,...). Diagonal sums have g.f. (1+x^2)/(1-x^3-x^4) (see A079398). Matrix inverse of the signed triangle (-1)^(n-k)T(n,k) is A103631. Matrix inverse of T(n,k) is the alternating signed version of A103631.

%C Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ....] DELTA [1, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - _Philippe Deléham_, Oct 08 2005

%F Number triangle T(n, k) = binomial(floor(n/2), n-k).

%F Sum_{n>=0} T(n, k) = A000045(k+2) = Fibonacci(k+2). - _Philippe Deléham_, Oct 08 2005

%F Sum_{k=0..n} T(n,k) = 2^floor(n/2) = A016116(n). - _Philippe Deléham_, Dec 03 2006

%F G.f.: (1+x*y)/(1-x^2*y-x^2*y^2). - _Philippe Deléham_, Nov 10 2013

%F T(n,k) = T(n-2,k-1) + T(n-2,k-2) for n > 2, T(0,0) = T(,1) = T(2,1) = T(2,2) = 1, T(1,0) = T(2,0) = 0, T(n,k) = 0 if k > n or if k < 0. - _Philippe Deléham_, Nov 10 2013

%e Triangle begins:

%e 1;

%e 0, 1;

%e 0, 1, 1;

%e 0, 0, 1, 1;

%e 0, 0, 1, 2, 1;

%e 0, 0, 0, 1, 2, 1;

%e 0, 0, 0, 1, 3, 3, 1;

%e 0, 0, 0, 0, 1, 3, 3, 1;

%e 0, 0, 0, 0, 1, 4, 6, 4, 1;

%e 0, 0, 0, 0, 0, 1, 4, 6, 4, 1;

%e 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1;

%e 0, 0, 0, 0, 0, 0, 1, 5, 10, 10, 5, 1;

%e 0, 0, 0, 0, 0, 0, 1, 6, 15, 20, 15, 6, 1; ...

%K easy,nonn,tabl

%O 0,14

%A _Paul Barry_, Feb 11 2005