OFFSET
1,2
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.
FORMULA
b(n,m)=b(n-1,m]+m; Delta_diagonal=m; m={0,1,2,3,...k}.
EXAMPLE
{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}
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}]
CROSSREFS
KEYWORD
nonn,uned
AUTHOR
Roger L. Bagula and Gary W. Adamson, Sep 17 2008
STATUS
approved