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A154918
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A triangular sequence based on: t0(n,m)=Binomial[4*n, 3*m]; by coefficient reversal. t(n,m)=t0(n,m)+reverse(t0(n,m)).
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0
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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
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| 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.
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FORMULA
| t0(n,m)=Binomial[4*n, 3*m];
Coefficient reversal applied:
t(n,m)=t0(n,m)+reverse(t0(n,m)).
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EXAMPLE
| {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}
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MATHEMATICA
| 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[%]
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CROSSREFS
| Sequence in context: A056396 A085043 A143818 * A176862 A176081 A073339
Adjacent sequences: A154915 A154916 A154917 * A154919 A154920 A154921
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KEYWORD
| nonn,tabl,uned,tabl
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AUTHOR
| Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 17 2009
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