0,1

Row sums are:

{2, 10, 170, 2730, 43658, 698670, 11180762, 178915964, 2862907786,

45809074626, 732970281870}

The idea was that the Stirling first kind has an {n,Floor[n/2]}

type boundary, which is like {2,1} as a triangle, so I tried a {3,4,5}

triangle in a binomial.

t0(n,m)=Binomial[4*n, 3*m];

Coefficient reversal applied:

t(n,m)=t0(n,m)+reverse(t0(n,m)).

{2},

{5, 5},

{29, 112, 29},

{221, 1144, 1144, 221},

{1821, 12000, 16016, 12000, 1821},

{15505, 127110, 206720, 206720, 127110, 15505},

{134597, 1309528, 2838752, 2615008, 2838752, 1309528, 134597},

{1184041, 13126386, 37818900, 37328655, 37328655, 37818900, 13126386, 1184041},

{10518301, 129029440, 472341792, 593771520, 451585680, 593771520, 472341792, 129029440, 10518301}, {94143281, 1251684840, 5569850352, 9169278580, 6819580260, 6819580260, 9169278580, 5569850352, 1251684840, 94143281},

{847660529, 12033232760, 62855940030, 131555847280, 118967115280, 80450690112, 118967115280, 131555847280, 62855940030, 12033232760, 847660529}

Clear[t, p, q, n, m, a];

t[n_, m_] = Binomial[4*n, 3*m];

Table[Table[t[n, m], {m, 0, n}] + Reverse[Table[t[n, m], {m, 0, n}]], {n, 0, 10}]

Flatten[%]

Sequence in context: A056396 A085043 A143818 this_sequence A073339 A074636 A127598

Adjacent sequences: A154915 A154916 A154917 this_sequence A154919 A154920 A154921

nonn,tabl,uned,tabl

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 17 2009