A141689 Average of Eulerian numbers (A008292) and Pascal's triangle (A007318): t(n,m) = (A008292(n,m) + A007318(n,m))/2.

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 36, 15, 1, 1, 31, 156, 156, 31, 1, 1, 63, 603, 1218, 603, 63, 1, 1, 127, 2157, 7827, 7827, 2157, 127, 1, 1, 255, 7318, 44145, 78130, 44145, 7318, 255, 1, 1, 511, 23938, 227638, 655240, 655240, 227638, 23938, 511, 1

Offset

1,5

Comments

Row sums are: {1, 2, 5, 16, 68, 376, 2552, 20224, 181568, 1814656, ...}.

If Pascal's triangle and the Eulerian numbers are both fundamental arrays, then there should be a combinatorial set "between" them.

Links

G. C. Greubel, Rows n=1..100 of triangle, flattened

Example

{1},
{1, 1},
{1, 3, 1},
{1, 7, 7, 1},
{1, 15, 36, 15, 1},
{1, 31, 156, 156, 31, 1},
{1, 63, 603, 1218, 603, 63, 1},
{1, 127, 2157, 7827, 7827, 2157, 127, 1},
{1, 255, 7318, 44145, 78130, 44145, 7318, 255, 1},
{1, 511, 23938, 227638, 655240, 655240, 227638, 23938, 511, 1}

Mathematica

Table[Table[(Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}] + Binomial[n - 1, k])/2, {k, 0, n - 1}], {n, 1, 10}]; Flatten[%]

Crossrefs

Cf. A008292, A007318.

Sequence in context: A184173 A359985 A022166 * A058669 A057004 A059328

Adjacent sequences: A141686 A141687 A141688 * A141690 A141691 A141692

Keyword

nonn,tabl

Author

Roger L. Bagula, Sep 09 2008

Extensions

Edited by N. J. A. Sloane, Dec 13 2008

Status

approved