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%I #21 Aug 15 2022 04:25:41
%S 1,2,2,2,4,2,4,12,12,4,4,16,24,16,4,8,40,80,80,40,8,8,48,120,160,120,
%T 48,8,16,112,336,560,560,336,112,16,16,128,448,896,1120,896,448,128,
%U 16,32,288,1152,2688,4032,4032,2688,1152,288,32
%N Triangle read by rows: T(n,k) = 2^floor(n/2)*binomial(n-1,k-1).
%H Gheorghe Coserea, <a href="/A135838/b135838.txt">Rows n = 1..100, flattened</a>
%F M * Pascal's triangle as infinite lower triangular matrices, where M = a triangle with (1, 2, 2, 4, 4, 8, 8, 16, 16, ...) in the main diagonal and the rest zeros.
%F Sum_{k=1..n} T(n, k) = A094015(n-1).
%F From _G. C. Greubel_, Feb 07 2022: (Start)
%F T(n, n-k) = T(n, k).
%F T(n, 1) = A016116(n).
%F T(n, 2) = 2*A093968(n-1).
%F T(2*n-1, n) = A059304(n-1).
%F T(2*n, n) = 2*A069720(n). (End)
%e First few rows of the triangle are:
%e 1;
%e 2, 2;
%e 2, 4, 2;
%e 4, 12, 12, 4;
%e 4, 16, 24, 16, 4;
%e 8, 40, 80, 80, 40, 8;
%e ...
%p A135838 := proc(n,k)
%p 2^floor(n/2)*binomial(n-1,k-1) ;
%p end proc:
%p seq(seq( A135838(n,k),k=1..n),n=1..10) ; # _R. J. Mathar_, Aug 15 2022
%t T[n_, k_]:= 2^Floor[n/2]*Binomial[n-1, k-1];
%t Table[T[n, k], {n,12}, {k,n}] //Flatten (* _G. C. Greubel_, Feb 07 2022 *)
%o (PARI)
%o A(n,k) = 2^(n\2)*binomial(n-1,k-1);
%o concat(vector(10, n, vector(n, k, A(n,k)))) \\ _Gheorghe Coserea_, May 18 2016
%o (Sage) flatten([[2^(n//2)*binomial(n-1, k-1) for k in (1..n)] for n in (1..12)]) # _G. C. Greubel_, Feb 07 2022
%Y Cf. A016116, A059304, A069720, A093968, A094015 (row sums), A135837.
%K nonn,tabl,easy
%O 1,2
%A _Gary W. Adamson_, Dec 01 2007
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