A099567 Riordan array (1/(1-x-x^3), 1/(1-x)).
1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 3, 5, 6, 4, 1, 4, 8, 11, 10, 5, 1, 6, 12, 19, 21, 15, 6, 1, 9, 18, 31, 40, 36, 21, 7, 1, 13, 27, 49, 71, 76, 57, 28, 8, 1, 19, 40, 76, 120, 147, 133, 85, 36, 9, 1, 28, 59, 116, 196, 267, 280, 218, 121, 45, 10, 1, 41, 87, 175, 312, 463, 547, 498, 339, 166, 55, 11, 1
Offset: 0
Examples
Rows begin: 1; 1, 1; 1, 2, 1; 2, 3, 3, 1; 3, 5, 6, 4, 1; 4, 8, 11, 10, 5, 1; 6, 12, 19, 21, 15, 6, 1; 9, 18, 31, 40, 36, 21, 7, 1; 13, 27, 49, 71, 76, 57, 28, 8, 1; 19, 40, 76, 120, 147, 133, 85, 36, 9, 1; 28, 59, 116, 196, 267, 280, 218, 121, 45, 10, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Crossrefs
Programs
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Magma
T:= func< n,k | (&+[Binomial(n-2*j, k+j): j in [0..Floor(n/3)]]) >; [[T(n,k): k in [0..n]]: n in [0..15]]; // G. C. Greubel, Jul 27 2022
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Mathematica
T[n_, 0]:=T[n,0]=HypergeometricPFQ[{(1-n)/3,(2-n)/3,-n/3}, {(1-n)/2,-n/2}, -27/4]; T[n_, k_]:= T[n,k]= If[k==n, 1, T[n-1,k-1] +T[n-1,k]]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 28 2017 *)
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SageMath
@CachedFunction def A099567(n, k): return sum( binomial(n-2*j, k+j) for j in (0..(n//3)) ) flatten([[A099567(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 27 2022
Formula
Number triangle T(n, k) = Sum_{j=0..floor(n/3)} binomial(n-2*j, k+j).
Columns have g.f. (1/(1-x-x^3))*(x/(1-x))^k.
Sum_{k=0..n} T(n, k) = A099568(n).
T(n,0) = A000930(n), T(n,n) = 1, T(n,k) = T(n-1,k-1) + T(n-1,k) for 0Philippe Deléham, Dec 29 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(2 + 3*x + 3*x^2/2! + x^3/3!) = 2 + 5*x + 11*x^2/2! + 21*x^3/3! + 36*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
From G. C. Greubel, Jul 27 2022: (Start)
T(n, n-1) = n, for n >= 1.
T(n, n-2) = A000217(n-1), for n >= 2.
T(n, n-3) = A050407(n+1), for n >= 3.
T(2*n, n) = A144904(n+1), for n >= 1. (End)
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