A135838 Triangle read by rows: T(n,k) = 2^floor(n/2)*binomial(n-1,k-1).

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, 48, 8, 16, 112, 336, 560, 560, 336, 112, 16, 16, 128, 448, 896, 1120, 896, 448, 128, 16, 32, 288, 1152, 2688, 4032, 4032, 2688, 1152, 288, 32

Offset

1,2

Links

Gheorghe Coserea, Rows n = 1..100, flattened

Formula

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.

Sum_{k=1..n} T(n, k) = A094015(n-1).

From G. C. Greubel, Feb 07 2022: (Start)

T(n, n-k) = T(n, k).

T(n, 1) = A016116(n).

T(n, 2) = 2*A093968(n-1).

T(2*n-1, n) = A059304(n-1).

T(2*n, n) = 2*A069720(n). (End)

Example

First few rows of the triangle are:
  1;
  2,  2;
  2,  4,  2;
  4, 12, 12,  4;
  4, 16, 24, 16,  4;
  8, 40, 80, 80, 40, 8;
  ...

Maple

A135838 := proc(n, k)
    2^floor(n/2)*binomial(n-1, k-1) ;
end proc:
seq(seq( A135838(n, k), k=1..n), n=1..10) ; # R. J. Mathar, Aug 15 2022

Mathematica

T[n_, k_]:= 2^Floor[n/2]*Binomial[n-1, k-1];
Table[T[n, k], {n, 12}, {k, n}] //Flatten (* G. C. Greubel, Feb 07 2022 *)

Prog

(PARI)
A(n, k) = 2^(n\2)*binomial(n-1, k-1);
concat(vector(10, n, vector(n, k, A(n, k))))  \\ Gheorghe Coserea, May 18 2016
(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

Crossrefs

Cf. A016116, A059304, A069720, A093968, A094015 (row sums), A135837.

Sequence in context: A286849 A098069 A173738 * A279966 A349483 A114349

Adjacent sequences: A135835 A135836 A135837 * A135839 A135840 A135841

Keyword

nonn,tabl,easy

Author

Gary W. Adamson, Dec 01 2007

Status

approved