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%I #22 Sep 08 2022 08:45:24
%S 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,
%T 42,70,140,84,56,8,1,8,56,112,280,224,224,64,16,1,9,72,168,504,504,
%U 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
%N Triangle read by rows: T(n, k) = 2^floor((k-1)/2)*binomial(n-1, k-1).
%C Row sums are the Pell sequence A000129.
%C Right border = inverse binomial transform of the Pell sequence: (A016116).
%C This triangle = difference terms of columns from an array generated from binomial transforms of (1,0,0,0...); (1,1,0,0,0...); (1,1,2,2...); (1,1,2,2,4,...); where (1, 1, 2, 2, 4, 4,...) = A016116, the inverse binomial transform of the Pell sequence A000129.
%C Triangle read by rows, iterates of X * [1,0,0,0,...] where X = an infinite bidiagonal matrix with (1,1,1,...) in the main diagonal and (1,2,1,2,1,2,...) in the subdiagonal, with the rest zeros. - _Gary W. Adamson_, May 10 2008
%C This sequence is jointly generated with A135837 as a triangular array of coefficients of polynomials u(n,x): initially, u(1,x) = v(1,x) = 1; for n>1, u(n,x) = u(n-1,x) + x*v(n-1) and v(n,x) = 2*x*u(n-1,x) + v(n-1,x). See the Mathematica section. - _Clark Kimberling_, Feb 26 2012
%H G. C. Greubel, <a href="/A117919/b117919.txt">Rows n = 1..50 of the triangle, flattened</a>
%F From _G. C. Greubel_, Oct 23 2021:
%F T(n, k) = 2^floor((k-1)/2)*binomial(n-1, k-1).
%F Sum_{k=0..n} T(n, k) = A000129(n). (End)
%e First few rows of the generating array are:
%e 1, 1, 1, 1, 1, ...
%e 1, 2, 3, 4, 5, ...
%e 1, 2, 5, 10, 17, ...
%e 1, 2, 5, 12, 25, ...
%e 1, 2, 5, 12, 29, ...
%e ...
%e Taking difference terms of the columns, we get this triangle. First few rows are:
%e 1;
%e 1, 1;
%e 1, 2, 2;
%e 1, 3, 6, 2;
%e 1, 4, 12, 8, 4;
%e 1, 5, 20, 20, 20, 4;
%e 1, 6, 30, 40, 60, 24, 8;
%e 1, 7, 42, 70, 140, 84, 56, 8;
%e ...
%t (* First program *)
%t u[1, x_]:= 1; v[1, x_]:= 1; z = 13;
%t u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
%t v[n_, x_]:= 2*x*u[n-1, x] + v[n-1, x];
%t Table[Expand[u[n, x]], {n, 1, z/2}]
%t Table[Expand[v[n, x]], {n, 1, z/2}]
%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
%t TableForm[cu]
%t Flatten[%] (* A117919 *)
%t Table[Expand[v[n, x]], {n, 1, z}]
%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
%t TableForm[cv]
%t Flatten[%] (* A135837 *)
%t (* Second program *)
%t Table[2^Floor[(k-1)/2]*Binomial[n-1, k-1], {n,15}, {k,n}]//Flatten (* _G. C. Greubel_, Oct 23 2021 *)
%o (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
%o (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
%Y Cf. A000129, A016116, A135837.
%K nonn,tabl
%O 1,5
%A _Gary W. Adamson_, Apr 02 2006
%E Name changed and more terms added by _G. C. Greubel_, Oct 23 2021
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