OFFSET
0,3
COMMENTS
Row sums are: {1, 1, 6, 138, 18768, 3878352, 2906541648, 5191071615792, 30559282412221440, 552452422047369288192, 31282693282240728431406336,...}.
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.
FORMULA
t(n,k)=A060187[n,m] from polynomials; f(n, m) = If[m <= Floor[n/2], f(m, 1)*f(n - m, 1)*t(n + 1, m)];
EXAMPLE
Half MacMahon numbers: Table[Table[f[n, m]/(f[m, 1]*f[n - m, 1]), {m, 0, Floor[n/2]}], {n, 0, 10}];
{1},
{1},
{1, 6},
{1, 23},
{1, 76, 230},
{1, 237, 1682},
{1, 722, 10543, 23548},
{1, 2179, 60657, 259723},
{1, 6552, 331612, 2485288, 4675014},
{1, 19673, 1756340, 21707972, 69413294},
{1, 59038, 9116141, 178300904, 906923282, 1527092468}
Factorial type triangle is:
{1},
{1},
{6},
{138},
{10488, 8280},
{2485656, 1392696},
{1794643632, 663449904, 448448112},
{3910528474128, 904634615952, 375908525712},
{25621782562486656, 3570752184568704, 852504801992064, 514242863174016},
...
MATHEMATICA
Clear[t, n, m, f, x, p];
p[x_, n_] = (-1)^(n + 1)*(x - 1)^(n + 1)*Sum[(2* m + 1)^n*x^m, {m, 0, Infinity}];
t[n_, m_] := Table[CoefficientList[FullSimplify[ExpandAll[p[x, k]]], x], { k, 0, 10}][[n + 1, m + 1]];
f[0, 1] = 1; f[1, 1] = 1; f[2, 1] = 6;
f[n_, m_] := f[n, m] = If[m <= Floor[n/2], f[m, 1]*f[n - m, 1]*t[n, m]];
a = Join[{{1}}, {{1}}, Table[Table[f[n, m], {m, 1, Floor[n/2]}], {n, 2, 10}]];
Flatten[%]
CROSSREFS
KEYWORD
nonn,tabf,uned
AUTHOR
Roger L. Bagula, Jan 24 2009
STATUS
approved