OFFSET
0,13
COMMENTS
Inverse is A112466. Note that C(n,k) = Sum_{j = 0..n-k} C(j+k-1, j).
LINKS
Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
E. Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Mathematics, 34 (2005) pp. 101-122.
FORMULA
Number triangle T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*C(j+k-1, j).
T(2*n, n) = A072547(n) (main diagonal). - Paul Barry, Apr 08 2011
From Reinhard Zumkeller, Jan 03 2014: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n, with T(n, 0) = (-1)^n and T(n, n) = 1.
T(n, k) = A108561(n, n-k). (End)
T(n, k) = T(n-1, k-1) + 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 11 2014
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(-1 + x + x^2/2! + x^3/3!) = -1 + 2*x^2/2! + 6*x^3/3! + 13*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
EXAMPLE
Triangle starts
1;
-1, 1;
1, 0, 1;
-1, 1, 1, 1;
1, 0, 2, 2, 1;
-1, 1, 2, 4, 3, 1;
1, 0, 3, 6, 7, 4, 1;
-1, 1, 3, 9, 13, 11, 5, 1;
1, 0, 4, 12, 22, 24, 16, 6, 1;
Production matrix begins
-1, 1;
0, 1, 1;
0, 0, 1, 1;
0, 0, 0, 1, 1;
0, 0, 0, 0, 1, 1;
0, 0, 0, 0, 0, 1, 1;
0, 0, 0, 0, 0, 0, 1, 1;
0, 0, 0, 0, 0, 0, 0, 1, 1; - Paul Barry, Apr 08 2011
MATHEMATICA
T[n_, k_]:= Sum[Binomial[j+k-1, j]*(-1)^(n-k-j), {j, 0, n-k}];
Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* Jean-François Alcover, Jul 23 2018 *)
PROG
(Haskell)
a112465 n k = a112465_tabl !! n !! k
a112465_row n = a112465_tabl !! n
a112465_tabl = iterate f [1] where
f xs'@(x:xs) = zipWith (+) ([-x] ++ xs ++ [0]) ([0] ++ xs')
-- Reinhard Zumkeller, Jan 03 2014
(Magma)
A112465:= func< n, k | (-1)^(n+k)*(&+[(-1)^j*Binomial(j+k-1, j): j in [0..n-k]]) >;
[A112465(n, k): k in [0..n], n in [0..13]]; // G. C. Greubel, Apr 18 2025
(SageMath)
def A112465(n, k): return (-1)^(n+k)*sum((-1)^j*binomial(j+k-1, j) for j in range(n-k+1))
print(flatten([[A112465(n, k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 18 2025
CROSSREFS
KEYWORD
AUTHOR
Paul Barry, Sep 06 2005
STATUS
approved
