A142240 A triangular sequence from the pattern in row sums of Pascal's triangle A007318, Eulerian numbers A008292 and A060187: Delta_diagonal=m; m={0,1,2,3,...k}.

1, 2, 2, 2, 3, 2, 2, 4, 4, 2, 2, 5, 6, 5, 2, 2, 6, 8, 8, 6, 2, 2, 7, 10, 11, 10, 7, 2, 2, 8, 12, 14, 14, 12, 8, 2, 2, 9, 14, 17, 18, 17, 14, 9, 2, 2, 10, 16, 20, 22, 22, 20, 16, 10, 2

Offset: 1

Author

Roger L. Bagula and Gary W. Adamson, Sep 17 2008

Keywords

• nonn
• uned

Comments

Row sums are:

{1, 4, 7, 12, 20, 32, 49, 72, 102, 140}.

The triangle is calculated by hand.

Row sums are:

1) Pascal A007318:ratio =2: delta row zero:a(n)=2*a(n-1);a(1)=1;

b(n)->0

1,2,4,8,16,32,64,128,256,512

2) Eulerian numbers A008292: ratio =n: delta=1:a(n)=n*a(n-1)

1,2,6,24,120,720,...n!

b(n)->1,2,3,4,...

3) A060187:ratio=2*n: delta=2:a(n)=2*n*a(n-1)

b(n)->2,4,6,8,...

{1, 2, 8, 48, 384, 3840, 46080, 645120, 10321920, 185794560, 3715891200}.

4) hypothetical next level sums:delta=3:b(n)=b(n-1)+3;a(n)=a(n-1)*b(n);

b(n)->{2, 5, 8, 11, 14, 17, 20, 23, 26, 29} diagonal

{1, 2, 10, 80, 880, 12320, 209440, 4188800, 96342400, 2504902400, 72642169600}

5)hypothetical next level sums:delta=3:b(n)=b(n-1)+4;a(n)=a(n-1)*b(n);

b(n)->{2, 6, 10, 14, 18, 22, 26, 30, 34, 38} diagonal

{1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 670442572800}

The conjecture that goes with this triangular sequence is that there on n levels like these for Pascal combinatorial quantum levels.

Examples

			{1},
{2, 2},
{2, 3, 2},
{2, 4, 4, 2},
{2, 5, 6, 5, 2},
{2, 6, 8, 8, 6, 2},
{2, 7, 10, 11, 10, 7, 2},
{2, 8, 12, 14, 14, 12, 8, 2},
{2, 9, 14, 17, 18, 17, 14, 9, 2},
{2, 10, 16, 20, 22, 22, 20, 16, 10, 2}
		

Crossrefs

Cf. A060187, A007318, A008292.

Programs

Mathematica

a={{1},{2,2},{2,3,2},{2,4,4,2}, {2,5,6,5,2},{2,6,8,8,6,2},{2,7,10,11,10,7,2},{2,8,12,14,14, 12,8,2},{2,9,14,17,18,17,14,9,2},{2,10,16,20,22,22,20,16,10,2}} Flatten[a] Table[Apply[Plus,a[[n]]],{n,1,10}]

Formula

b(n,m)=b(n-1,m]+m; Delta_diagonal=m; m={0,1,2,3,...k}.