A178346 A symmetrical triangle between the Eulerian numbers, A008292, and MacMahon numbers, A060187, by linear combination with Pascal(A007318), Eulerian numbers ,and Narayana numbers,A001263,:k=3;t(n,m,k)=Binomial[n, m] - k*(Binomial[n, m]*Binomial[n + 1, m]/(m + 1)) + k*Eulerian[n + 1, m]

1, 1, 1, 1, 5, 1, 1, 18, 18, 1, 1, 52, 144, 52, 1, 1, 131, 766, 766, 131, 1, 1, 303, 3273, 6743, 3273, 303, 1, 1, 664, 12312, 45422, 45422, 12312, 664, 1, 1, 1406, 42844, 261230, 463348, 261230, 42844, 1406, 1, 1, 2913, 141936, 1358100, 3915312

Offset

0,5

Comments

Row sums are:

{1, 2, 7, 38, 250, 1796, 13897, 116798, 1074310, 10836524, 119575066,...}.

By solving the linear combination of Pascal,Narayana and Eulerian to give

the MacMahon at k=4, I got this functional set of symmetrical triangles.

By modulo two pattern the k=3 set is Narayana numbers like and the k=6 ,{1.8.1} level,

is a new even Sierpinski type.

Links

Table of n, a(n) for n=0..49.

Formula

k=3;

t(n,m,k)=Binomial[n, m] - k*(Binomial[n, m]*Binomial[n + 1, m]/(m + 1)) + k*Eulerian[n + 1, m]

Example

{1},
{1, 1},
{1, 5, 1},
{1, 18, 18, 1},
{1, 52, 144, 52, 1},
{1, 131, 766, 766, 131, 1},
{1, 303, 3273, 6743, 3273, 303, 1},
{1, 664, 12312, 45422, 45422, 12312, 664, 1},
{1, 1406, 42844, 261230, 463348, 261230, 42844, 1406, 1},
{1, 2913, 141936, 1358100, 3915312, 3915312, 1358100, 141936, 2913, 1},
{1, 5953, 455481, 6595734, 29172972, 47114784, 29172972, 6595734, 455481, 5953, 1}

Mathematica

<< DiscreteMath`Combinatorica`
t[n_, m_, k_] = Binomial[n, m] - k*(Binomial[n, m]* Binomial[n + 1, m]/(m + 1)) + k*Eulerian[n + 1, m];
Table[Flatten[Table[Table[t[n, m, k], {m, 0, n}], {n, 0, 10}]], {k, 0, 10}]

Crossrefs

Cf. A007318, A008292, A001263, A060187

Sequence in context: A347974 A029847 A154334 * A168551 A262307 A144397

Adjacent sequences: A178343 A178344 A178345 * A178347 A178348 A178349

Keyword

nonn,tabl,uned

Author

Roger L. Bagula, May 25 2010

Status

approved