A117919 Triangle read by rows: T(n, k) = 2^floor((k-1)/2)*binomial(n-1, k-1).
1, 1, 1, 1, 2, 2, 1, 3, 6, 2, 1, 4, 12, 8, 4, 1, 5, 20, 20, 20, 4, 1, 6, 30, 40, 60, 24, 8, 1, 7, 42, 70, 140, 84, 56, 8, 1, 8, 56, 112, 280, 224, 224, 64, 16, 1, 9, 72, 168, 504, 504, 672, 288, 144, 16, 1, 10, 90, 240, 840, 1008, 1680, 960, 720, 160, 32, 1, 11, 110, 330, 1320, 1848, 3696, 2640, 2640, 880, 352, 32
Offset: 1
Examples
First few rows of the generating array are: 1, 1, 1, 1, 1, ... 1, 2, 3, 4, 5, ... 1, 2, 5, 10, 17, ... 1, 2, 5, 12, 25, ... 1, 2, 5, 12, 29, ... ... Taking difference terms of the columns, we get this triangle. First few rows are: 1; 1, 1; 1, 2, 2; 1, 3, 6, 2; 1, 4, 12, 8, 4; 1, 5, 20, 20, 20, 4; 1, 6, 30, 40, 60, 24, 8; 1, 7, 42, 70, 140, 84, 56, 8; ...
Links
- G. C. Greubel, Rows n = 1..50 of the triangle, flattened
Programs
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Magma
[2^Floor((k-1)/2)*Binomial(n-1, k-1): k in [1..n], n in [1..15]]; // G. C. Greubel, Oct 23 2021
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Mathematica
(* First program *) u[1, x_]:= 1; v[1, x_]:= 1; z = 13; u[n_, x_]:= u[n-1, x] + x*v[n-1, x]; v[n_, x_]:= 2*x*u[n-1, x] + v[n-1, x]; Table[Expand[u[n, x]], {n, 1, z/2}] Table[Expand[v[n, x]], {n, 1, z/2}] cu = Table[CoefficientList[u[n, x], x], {n, 1, z}]; TableForm[cu] Flatten[%] (* A117919 *) Table[Expand[v[n, x]], {n, 1, z}] cv = Table[CoefficientList[v[n, x], x], {n, 1, z}]; TableForm[cv] Flatten[%] (* A135837 *) (* Second program *) Table[2^Floor[(k-1)/2]*Binomial[n-1, k-1], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Oct 23 2021 *)
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Sage
flatten([[2^((k-1)//2)*binomial(n-1,k-1) for k in (1..n)] for n in (1..15)]) # G. C. Greubel, Oct 23 2021
Formula
From G. C. Greubel, Oct 23 2021: (Start)
T(n, k) = 2^floor((k-1)/2)*binomial(n-1, k-1).
Sum_{k=0..n} T(n, k) = A000129(n). (End)
Extensions
Name changed and more terms added by G. C. Greubel, Oct 23 2021
Comments