%I #14 Sep 23 2024 13:10:55
%S 1,2,1,1,1,1,0,0,0,1,0,0,0,-1,1,0,0,0,1,-2,1,0,0,0,-1,3,-3,1,0,0,0,1,
%T -4,6,-4,1,0,0,0,-1,5,-10,10,-5,1,0,0,0,1,-6,15,-20,15,-6,1,0,0
%N Riordan array ((1 + x)^2, x/(1 + x)).
%H Igor Victorovich Statsenko, <a href="https://aeterna-ufa.ru/sbornik/IN-2024-09-2.pdf#page=10">Riordan generalizations of binomial coefficients</a>, Innovation science No 9-2, State Ufa, Aeterna Publishing House, 2024, pp. 10-13. In Russian.
%F Inverse is Riordan array ((1 - x)^2, x/(1 - x)).
%F T(n, k) = (-1)^(n + k)*(C(n, n-k) - 3*C(n-1, n-k-1) + 3*C(n-2, n-k-2) - C(n-3, n-k-3)), where C(n, k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0. - _Peter Bala_, Mar 21 2018
%F T(n, k) = Sum_{i=0..n-k} binomial(i+3,3)*binomial(n+1,n-k-i)*(-1)^(n+k+i). - _Igor Victorovich Statsenko_, Sept 23 2024
%F T(m, n, k) = (-1)^(k + n)*binomial(n + 1, n - k)*hypergeom([m, k - n], [k + 2], 1) for m = 4. - _Peter Luschny_, Sep 23 2024
%e Triangle begins
%e 1,
%e 2, 1,
%e 1, 1, 1,
%e 0, 0, 0, 1,
%e 0, 0, 0, -1, 1,
%e 0, 0, 0, 1, -2, 1,
%e 0, 0, 0, -1, 3, -3, 1,
%e 0, 0, 0, 1, -4, 6, -4, 1,
%e 0, 0, 0, -1, 5, -10, 10, -5, 1,
%e 0, 0, 0, 1, -6, 15, -20, 15, -6, 1,
%e 0, 0, 0, -1, 7, -21, 35, -35, 21, -7, 1
%p C := proc(n, k) if 0 <= k and k <= n then
%p factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if; end proc:
%p for n from 0 to 10 do
%p seq((-1)^(n+k)*(C(n, n-k)-3*C(n-1, n-k-1)+3*C(n-2, n-k-2)-C(n-3, n-k-3)), k = 0..n);
%p end do; # _Peter Bala_, Mar 21 2018
%p T := (m, n, k) -> (-1)^(k + n)*binomial(n + 1, n - k)*hypergeom([m, k - n], [k + 2], 1); for n from 0 to 9 do seq(simplify(T(4, n, k)), k=0..n) od; # _Peter Luschny_, Sep 23 2024
%Y Row sums are binomial(3, n).
%Y Diagonal sums are A122434.
%Y Product of A007318 and A122432.
%Y Cf. A007318.
%K easy,sign,tabl
%O 0,2
%A _Paul Barry_, Sep 04 2006