A112466 Riordan array ((1+2*x)/(1+x), x/(1+x)).
1, 1, 1, -1, 0, 1, 1, -1, -1, 1, -1, 2, 0, -2, 1, 1, -3, 2, 2, -3, 1, -1, 4, -5, 0, 5, -4, 1, 1, -5, 9, -5, -5, 9, -5, 1, -1, 6, -14, 14, 0, -14, 14, -6, 1, 1, -7, 20, -28, 14, 14, -28, 20, -7, 1, -1, 8, -27, 48, -42, 0, 42, -48, 27, -8, 1, 1, -9, 35, -75, 90, -42, -42, 90, -75, 35, -9, 1, -1, 10, -44, 110, -165, 132, 0, -132, 165, -110, 44, -10, 1
Offset: 0
Examples
Triangle starts 1; 1, 1; -1, 0, 1; 1, -1, -1, 1; -1, 2, 0, -2, 1; 1, -3, 2, 2, -3, 1; -1, 4, -5, 0, 5, -4, 1; From _Paul Barry_, Apr 08 2011: (Start) Production matrix begins 1, 1; -2, -1, 1; 2, 0, -1, 1; -2, 0, 0, -1, 1; 2, 0, 0, 0, -1, 1; -2, 0, 0, 0, 0, -1, 1; 2, 0, 0, 0, 0, 0, -1, 1; (End)
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150, flattened)
- Paul Barry, A Note on Riordan Arrays with Catalan Halves, arXiv:1912.01124 [math.CO], 2019.
- Emeric Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Mathematics, 34 (2005) pp. 101-122.
Crossrefs
Programs
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Magma
A112466:= func< n,k | n eq 0 select 1 else (-1)^(n+k)*(Binomial(n,k) - 2*Binomial(n-1,k)) >; [A112466(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Apr 30 2025
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Maple
seq(seq( (-1)^(n-k)*(2*binomial(n-1, k-1)-binomial(n, k)), k=0..n), n=0..10); # G. C. Greubel, Feb 19 2020
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Mathematica
{1}~Join~Table[(Binomial[n, n - k] - 2 Binomial[n - 1, n - k - 1])*(-1)^(n - k), {n, 12}, {k, 0, n}] // Flatten (* Michael De Vlieger, Feb 18 2020 *)
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PARI
T(n,k) = (-1)^(n-k)*(binomial(n, n-k) - 2*binomial(n-1, n-k-1)); \\ Michel Marcus, Feb 19 2020
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SageMath
def A112466(n,k): return 1 if (n==0) else (-1)^(n+k)*(binomial(n,k) - 2*binomial(n-1,k)) print(flatten([[A112466(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 30 2025
Formula
Number triangle: T(n,k) = (-1)^(n-k)*(C(n, n-k) - 2*C(n-1, n-k-1)), with T(0,0) = 1.
T(2*n, n) = 0 (main diagonal).
Sum_{k=0..n} T(n, k) = 0 + [n=0] + 2*[n=1] (row sums).
Sum_{k=0..floor(n/2)} T(n-k, k) = (-1)^(n+1)*Fibonacci(n-2) (diagonal sums).
Sum_{k=0..n} T(n,k)*x^k = (x+1)*(x-1)^(n-1), for n >= 1. - Philippe Deléham, Oct 03 2005
T(0,0) = T(1,0) = T(1,1) = 1, T(n,k) = 0 if n < 0 or if n < k, T(n,k) = T(n-1,k-1) - T(n-1,k) for n > 1. - Philippe Deléham, Nov 26 2006
G.f.: (1+2*x)/(1+x-x*y). - R. J. Mathar, Aug 11 2015
From G. C. Greubel, Apr 30 2025: (Start)
T(2*n+1, 2*n+1-k) = T(2*n+1, k) (symmetric odd n rows).
T(2*n, 2*n-k) = (-1)*T(2*n, k) (antisymmetric even n rows).
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n) (signed row sums).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = (-1)^n*A057079(n+2) (signed diagonal sums). (End)
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