A122433 Riordan array ((1 + x)^2, x/(1 + x)).

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, -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

Offset: 0

Author

Paul Barry, Sep 04 2006

Keywords

• easy
• sign
• tabl

Examples

			Triangle begins
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, -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, 0, -1, 7, -21, 35, -35, 21, -7, 1
		

Links

• Igor Victorovich Statsenko, Riordan generalizations of binomial coefficients, Innovation science No 9-2, State Ufa, Aeterna Publishing House, 2024, pp. 10-13. In Russian.

Crossrefs

Row sums are binomial(3, n).

Diagonal sums are A122434.

Product of A007318 and A122432.

Cf. A007318.

Programs

Maple

C := proc(n, k) if 0 <= k and k <= n then
factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if; end proc:
for n from 0 to 10 do
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);
end do; # Peter Bala, Mar 21 2018
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

Formula

Inverse is Riordan array ((1 - x)^2, x/(1 - x)).

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

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, Sep 23 2024

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