%I #15 Nov 25 2016 18:57:08
%S 1,1,0,1,1,-1,1,2,0,-2,1,3,2,-2,-3,1,4,5,0,-5,-4,1,5,9,5,-5,-9,-5,1,6,
%T 14,14,0,-14,-14,-6,1,7,20,28,14,-14,-28,-20,-7,1,8,27,48,42,0,-42,
%U -48,-27,-8
%N Triangle T(n,k) obtained by taking differences of consecutive pairs of row elements of Pascal's triangle A007318.
%C Row sums are 1,1,1,1,1,1 with g.f. 1/(1-x). Can also be obtained from triangle A080232 by taking sums of pairs of consecutive row elements.
%C Mirror image of triangle in A156644. - _Philippe Deléham_, Feb 14 2009
%F T(n, k) = if(k>n, 0, binomial(n, k)-binomial(n, k-1)).
%e Rows are {1}, {1,0}, {1,1,-1}, {1,2,0,-2}, {1,3,2,-2,-3}, ...
%t Table[Binomial[n, k] - Binomial[n, k - 1], {n, 0, 9}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Nov 24 2016 *)
%o (PARI) {T(n, k) = if( n<0 || k>n, 0, binomial(n, k) - binomial(n, k-1))}; /* _Michael Somos_, Nov 25 2016 */
%Y Cf. A007318, A080232.
%K easy,sign,tabl
%O 0,8
%A _Paul Barry_, Feb 10 2003
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