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A112333
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An invertible triangle of ratios of triple factorials.
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5
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1, 2, 1, 10, 5, 1, 80, 40, 8, 1, 880, 440, 88, 11, 1, 12320, 6160, 1232, 154, 14, 1, 209440, 104720, 20944, 2618, 238, 17, 1, 4188800, 2094400, 418880, 52360, 4760, 340, 20, 1, 96342400, 48171200, 9634240, 1204280, 109480, 7820, 460, 23, 1, 2504902400
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| First column is A008544. Second column is A034000. Third column is A051605. As a square array read by anti-diagonals, columns have e.g.f. (1/(1-3x)^(2/3))*(1/(1-3x))^k
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FORMULA
| Number triangle T(n, k)=if(k<=n, Product{k=1..n, 3k-1}/Product{j=1..k, 3j-1}, 0); T(n, k)=if(k<=n, 3^(n-k)*(n-1/3)!/(k-1/3)!, 0).
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EXAMPLE
| Triangle begins
1;
2,1;
10,5,1;
80,40,8,1;
880,440,88,11,1;
12320,6160,1232,154,14,1;
Inverse triangle A112334 begins
1;
-2,1;
0,-5,1;
0,0,-8,1;
0,0,0,-11,1;
0,0,0,0,-14,1;
0,0,0,0,0,-17,1;
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MAPLE
| nmax:=8: for n from 0 to nmax do for k from 0 to n do if k<=n then T(n, k):= product(3*k1-1, k1=1..n)/ product(3*j-1, j=1..k) else T(n, k):= 0: fi: od: od: for n from 0 to nmax do seq(T(n, k), k=0..n) od: Tx:=0: for n from 0 to nmax do for k from 0 to n do a(Tx):=T(n, k): Tx:=Tx+1: od: od: seq(a(n), n=0..Tx-1); [Johannes W. Meijer, Jul 04 2011]
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CROSSREFS
| Sequence in context: A136216 A121334 A126450 * A066868 A193900 A143172
Adjacent sequences: A112330 A112331 A112332 * A112334 A112335 A112336
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KEYWORD
| easy,nonn,tabl
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Sep 04 2005
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