A121574 Riordan array (1/(1-2*x), x*(1+x)/(1-2*x)).
1, 2, 1, 4, 5, 1, 8, 16, 8, 1, 16, 44, 37, 11, 1, 32, 112, 134, 67, 14, 1, 64, 272, 424, 305, 106, 17, 1, 128, 640, 1232, 1168, 584, 154, 20, 1, 256, 1472, 3376, 3992, 2641, 998, 211, 23, 1, 512, 3328, 8864, 12592, 10442, 5221, 1574, 277, 26, 1
Offset: 0
Examples
Triangle begins 1; 2, 1; 4, 5, 1; 8, 16, 8, 1; 16, 44, 37, 11, 1; 32, 112, 134, 67, 14, 1; 64, 272, 424, 305, 106, 17, 1;
Links
- G. C. Greubel, Rows n = 0..100 of triangle, flattened
- M. Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.
Programs
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GAP
T:=Flat(List([0..9],n->List([0..n],k->Sum([0..n-k],j->Binomial(k,j)*Binomial(n-j,k)*2^(n-k-j))))); # Muniru A Asiru, Nov 02 2018
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Magma
[[(&+[ Binomial(k, j)*Binomial(n-j, k)*2^(n-k-j): j in [0..(n-k)]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Nov 02 2018
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Maple
T:=(n,k)->add(binomial(k,j)*binomial(n-j,k)*2^(n-k-j),j=0..n-k): seq(seq(T(n,k),k=0..n),n=0..9); # Muniru A Asiru, Nov 02 2018
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Mathematica
Table[Sum[Binomial[k, j] Binomial[n-j, k] 2^(n-k-j), {j, 0, n-k}], {n, 0, 15}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 02 2018 *)
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PARI
for(n=0,10, for(k=0,n, print1(sum(j=0, n-k, binomial(k, j)* binomial(n-j, k)*2^(n-k-j)), ", "))) \\ G. C. Greubel, Nov 02 2018
Formula
Number array T(n,k) = Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(n-k-j).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) + T(n-2,k-1). - Philippe Deléham, Nov 10 2011
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = (x + 2)*R(n-1,x) + x*R(n-2,x) with R(1,x) = 1 and R(2,x) = x + 2. - Peter Bala, Feb 07 2024
Comments