0,3

Row sums are:

{1, 1, 4, 44, 2200, 118360, 17952352, 3838714880, 1885044386368,

1607186778033408, 2934910973174349312,...}.

The Eulerian numbers factored as factorial like to middle Floor[n/2]:

t(n,m)=f(n,m)/(f[m,1]*f[n-m,1]).

The idea is to factor the Eulerian numbers as

if the coefficients were made up of equivalents to factorials.

The result is a multi-bifurcating recursion thast fits the Eulerian numbers.

Half Eulerian numbers: Table[Table[f[n, m]/(f[m, 1]*f[n - m, 1]), {m, 0, Floor[n/2]}], {n, 0, 10}];

{1},

{1},

{1, 4},

{1, 11},

{1, 26, 66},

{1, 57, 302},

{1, 120, 1191, 2416},

{1, 247, 4293, 15619},

{1, 502, 14608, 88234, 156190},

{1, 1013, 47840, 455192, 1310354},

{1, 2036, 152637, 2203488, 9738114, 15724248}...

Factorial type triangle is:

{1},

{1},

{4},

{44},

{1144, 1056},

{65208, 53152},

{7824960, 5450016, 4677376},

{1932765120, 1119751776, 786197984},

{970248090240, 457228062720, 253156757568, 204411475840},

{982861315413120, 369853933363200, 156721804462080, 97749724795008},

{2001105638181112320, 592383030999851520, 187388288944496640, 87173203289103360, 66860811759785472}

Clear[t, n, m, f, x];

t[n_, k_] = Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}];

f[0, 1] = 1; f[1, 1] = 1; f[2, 1] = 4;

f[n_, m_] := f[n, m] = If[m <= Floor[n/2], f[m, 1]*f[n - m, 1]*t[n + 1, m]];

a = Join[{{1}}, {{1}}, Table[Table[f[n, m], {m, 1, Floor[n/2]}], {n, 2, 10}]];

Flatten[%]

A008292

Sequence in context: A053333 A137783 A136552 this_sequence A127635 A134174 A157193

Adjacent sequences: A155553 A155554 A155555 this_sequence A155557 A155558 A155559

nonn,tabl,uned

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 24 2009