cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A338198 Triangle read by rows, T(n,k) = ((k+1)*2^(n-k)-(k-2)*(-1)^(n-k))/3 for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 2, 3, 2, 1, 6, 5, 4, 3, 1, 10, 11, 8, 5, 4, 1, 22, 21, 16, 11, 6, 5, 1, 42, 43, 32, 21, 14, 7, 6, 1, 86, 85, 64, 43, 26, 17, 8, 7, 1, 170, 171, 128, 85, 54, 31, 20, 9, 8, 1, 342, 341, 256, 171, 106, 65, 36, 23, 10, 9, 1, 682, 683, 512, 341, 214, 127, 76, 41, 26, 11, 10, 1
Offset: 0

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Author

Werner Schulte, Oct 15 2020

Keywords

Comments

This triangle is related to the Jacobsthal numbers (A001045).

Examples

			The triangle T(n,k) for 0 <= k <= n starts:
n\k :    0     1     2    3    4    5    6   7   8   9
======================================================
  0 :    1
  1 :    0     1
  2 :    2     1     1
  3 :    2     3     2    1
  4 :    6     5     4    3    1
  5 :   10    11     8    5    4    1
  6 :   22    21    16   11    6    5    1
  7 :   42    43    32   21   14    7    6   1
  8 :   86    85    64   43   26   17    8   7   1
  9 :  170   171   128   85   54   31   20   9   8   1
etc.
		

Crossrefs

For columns k = 0 to 8 see A078008, A001045, A000079, A001045, A084214, A014551, A083595, A083582, A259713 respectively.

Programs

  • Mathematica
    Table[((k + 1)*2^(n - k) - (k - 2)*(-1)^(n - k))/3, {n, 0, 11}, {k, 0, n}] // Flatten (* Michael De Vlieger, Oct 15 2020 *)

Formula

T(n,n) = 1 for n >= 0; T(n,n-1) = n-1 for n > 0.
T(n,k) = T(n-1,k) + 2 * T(n-2,k) for 0 <= k <= n-2.
T(n,k) = 2 * T(n-1,k) - (k-2) * (-1)^(n-k) for 0 <= k < n.
T(n,k) = T(n+1-k,1) + (k-1) * T(n-k,1) for 0 <= k < n.
T(n+1,k) * T(n-1,k) - T(n,k+1) * T(n,k-1) = T(n-k,1)^2 for 0 < k < n.
Row sums are A083579(n+1) for n >= 0.
G.f. of column k >= 0: (1+(k-1)*t) * t^k / (1-t-2*t^2).
G.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = (1 - (1+x)*t + 2*x*t^2) / ((1 - x*t)^2 * (1 - t - 2*t^2)).
Conjecture: Let M(n,k) be the matrix inverse of T(n,k), seen as a matrix. Then M(i,j) = 0 if j < 0 or j > i, M(n,n) = 1 for n >= 0, M(n,n-1) = 1-n for n > 0, and M(n,k) = (-1)^(n-k) * (k^2-2) * (n-2)! / k! for 0 <= k <= n-2.