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A110511 Riordan array (1/(1+x), x(1-x)/(1+x)^2).

%I A110511 #11 Aug 29 2017 22:50:52
%S A110511 1,-1,1,1,-4,1,-1,9,-7,1,1,-16,26,-10,1,-1,25,-70,52,-13,1,1,-36,155,
%T A110511 -190,87,-16,1,-1,49,-301,553,-403,131,-19,1,1,-64,532,-1372,1462,
%U A110511 -736,184,-22,1,-1,81,-876,3024,-4446,3206,-1216,246,-25,1,1,-100,1365,-6084,11826,-11584,6190,-1870,317,-28,1,-1,121,-2035
%N A110511 Riordan array (1/(1+x), x(1-x)/(1+x)^2).
%C A110511 Inverse of number triangle A110506. Row sums are A110512. Diagonal sums are A110513. Product of (1/(1+x), x/(1+x)) (inverse binomial transform matrix) and (1, x(1-2x)) (A110509).
%H A110511 G. C. Greubel, <a href="/A110511/b110511.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F A110511 Number triangle: T(n, k) = Sum_{j=0..n} (-1)^(n-j)*C(n, j)*(-2)^(j-k)*C(k, j-k).
%F A110511 T(n, k) = Sum_{j=0..n} Sum_{i=0..k} C(k, i)*C(n+k-i-j-1, n-k-i-j)*(-1)^(n-k).
%F A110511 T(n,k) = T(n-1,k-1) - 2*T(n-1,k) - T(n-2,k) - T(n-2,k-1), T(0,0)=1, T(1,0)=-1, T(1,1)=1, T(n,k)=0 if k < 0 or if k > n. - _Philippe Deléham_, Jan 12 2014
%e A110511 Rows begin
%e A110511    1;
%e A110511   -1,   1;
%e A110511    1,  -4,   1;
%e A110511   -1,   9,  -7,   1;
%e A110511    1, -16,  26, -10,   1;
%e A110511   -1,  25, -70,  52, -13,   1;
%t A110511 T[n_, k_] := Sum[(-1)^(n - j)*Binomial[n, j]*(-2)^(j - k)*Binomial[k, j - k], {j, 0, n}]; Table[T[n, k], {n, 0, 20}, {k, 0, n}] // Flatten (* _G. C. Greubel_, Aug 29 2017 *)
%o A110511 (PARI) for(n=0,20, for(k=0,n, print1(sum(j=0,n, (-1)^(n-j)*binomial(n, j)*(-2)^(j-k)*binomial(k, j-k)), ", "))) \\ _G. C. Greubel_, Aug 29 2017
%K A110511 easy,sign,tabl
%O A110511 0,5
%A A110511 _Paul Barry_, Jul 24 2005