A156050 Triangle T(n,m) = binomial(n,m)+2*P(n,m) read by rows, where P(n,m) = 1+A000041(n)-A000041(m)-A000041(n-m).

1, 1, 1, 1, 4, 1, 1, 5, 5, 1, 1, 8, 10, 8, 1, 1, 9, 16, 16, 9, 1, 1, 14, 25, 32, 25, 14, 1, 1, 15, 35, 51, 51, 35, 15, 1, 1, 22, 48, 82, 96, 82, 48, 22, 1, 1, 25, 64, 118, 164, 164, 118, 64, 25, 1, 1, 34, 83, 170, 264, 310, 264, 170, 83, 34, 1

Offset: 0

Author

Roger L. Bagula, Feb 02 2009

Keywords

• nonn
• tabl
• easy

Comments

The element-by-element sum of: the Pascal triangle A007318 plus two times the elements of P(n,m).

Row sums are 2^n+2*( (n+1)*(1+A000041(n)) -2*A000070(n) ), starting 1, 2, 6, 12, 28, 52, 112, 204, 402, 744, 1414,..., always below factorial(n+1).

The remarkable thing about this sub-Eulerian numbers triangle is that it a Sierpinski gasket modulo 2.

Examples

			P(n,m) starts in row n= 0 as
0
0, 0
0, 1, 0
0, 1, 1, 0
0, 2, 2, 2, 0
0, 2, 3, 3, 2, 0
0, 4, 5, 6, 5, 4, 0
0, 4, 7, 8, 8, 7, 4, 0
0, 7, 10, 13, 13, 13, 10, 7, 0
0, 8, 14, 17, 19, 19, 17, 14, 8, 0
0, 12, 19, 25, 27, 29, 27, 25, 19, 12, 0
to yield T(n,m) from row n=0 on:
1,
1, 1,
1, 4, 1,
1, 5, 5, 1,
1, 8, 10, 8, 1,
1, 9, 16, 16, 9, 1,
1, 14, 25, 32, 25, 14, 1,
1, 15, 35, 51, 51, 35, 15, 1,
1, 22, 48, 82, 96, 82, 48, 22, 1,
1, 25, 64, 118, 164, 164, 118, 64, 25, 1,
1, 34, 83, 170, 264, 310, 264, 170, 83, 34, 1
		

Programs

Mathematica

Clear[f];
t[n_, m_] = 1 + PartitionsP[n] - PartitionsP[m] - PartitionsP[n - m]; \! Table[(Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]])/2, {n, 0, 10}];
Table[Table[Binomial[n, m], {m, 0, n}] + (Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]]), {n, 0, 10}];
Flatten[%]

Extensions

Row sum formula and P(n,m) examples added - The Assoc. Eds. of the OEIS, Aug 29 2010