A125693 Riordan array ((1-x)/(1-3*x), x*(1-x)/(1-3*x)).
1, 2, 1, 6, 4, 1, 18, 16, 6, 1, 54, 60, 30, 8, 1, 162, 216, 134, 48, 10, 1, 486, 756, 558, 248, 70, 12, 1, 1458, 2592, 2214, 1168, 410, 96, 14, 1, 4374, 8748, 8478, 5160, 2150, 628, 126, 16, 1, 13122, 29160, 31590, 21744, 10442, 3624, 910, 160, 18, 1
Offset: 0
Examples
Triangle begins
1;
2, 1;
6, 4, 1;
18, 16, 6, 1;
54, 60, 30, 8, 1;
162, 216, 134, 48, 10, 1;
Links
- G. C. Greubel, Rows n=0..100 of triangle, flattened
Programs
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GAP
Flat(List([0..10], n-> List([0..n], k-> Sum([0..n], j-> (-1)^j*3^(n-k-j)*Binomial(k+1,j)*Binomial(n-j, n-k-j) )))); # G. C. Greubel, Oct 28 2019
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Magma
T:= func< n,k | &+[(-1)^j*3^(n-k-j)*Binomial(k+1,j)*Binomial(n-j, n-k-j): j in [0..n]] >; [T(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Oct 28 2019
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Maple
seq(seq( add( (-1)^j*3^(n-k-j)*binomial(k+1,j)*binomial(n-j, n-k-j), j=0..n), k=0..n), n=0..10); # G. C. Greubel, Oct 28 2019
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Mathematica
T[0, 0]=1; T[1, 0]=2; T[1, 1]=1; T[n_, k_]/; 0<=k<=n:= T[n, k]= 3T[n-1, k] + T[n-1, k-1] - T[n-2, k-1]; T[, ]=0; Table[T[n, k], {n, 0, 9}, {k, 0, n}] (* Jean-François Alcover, Jun 13 2019 *) T[n_, k_]:= Sum[(-1)^j*3^(n-k-j)*Binomial[k+1,j]*Binomial[n-j,n-k-j], {j, 0, n}]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 28 2019 *)
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PARI
T(n,k) = sum(j=0,n, (-1)^j*3^(n-k-j)*binomial(k+1,j)*binomial(n-j, n-k-j)); for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 28 2019
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Sage
[[sum( (-1)^j*3^(n-k-j)*binomial(k+1,j)*binomial(n-j, n-k-j) for j in (0..n) ) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Oct 28 2019
Formula
Number triangle T(n,k) = Sum_{j=0..k+1} C(k+1,j)*C(n-j,n-k-j)* (-1)^j * 3^(n-k-j).
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - T(n-2,k-1), T(0,0)=1, T(1,0)=2, T(1,1)=1, T(n,k)=0 if k>n or if kPhilippe Deléham, Jan 08 2013
Comments