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A112292
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An invertible triangle of ratios of double factorials.
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6
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1, 1, 1, 3, 3, 1, 15, 15, 5, 1, 105, 105, 35, 7, 1, 945, 945, 315, 63, 9, 1, 10395, 10395, 3465, 693, 99, 11, 1, 135135, 135135, 45045, 9009, 1287, 143, 13, 1, 2027025, 2027025, 675675, 135135, 19305, 2145, 195, 15, 1, 34459425, 34459425, 11486475
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| Columns include A001147,A051577,A051579. Row sums are A112293. Diagonal sums are A112294.
As a square array read by anti-diagonals, column k has e.g.f. (1/(1-2x)^(1/2))*(1/(1-2x))^k - Paul Barry (pbarry(AT)wit.ie), Sep 04 2005
Contribution from Peter Luschny (peter(AT)luschny.de), Jun 01 2009: (Start)
If the first column is deleted and the triangle read from right to left resulting in
1|1,3|1,5,15|1,7,35,105|1,9,63,315,945|...,
then this triangle T'(m,k) (m>=0,m>=k>=0) has the definition
T'(m,k) = (-2)^k prod_{j=0..k-1} (j-m-1/2)
for n from -1 to 7 do print(seq(T'(n,n-k),k=-1..n)) od: (End)
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FORMULA
| Number triangle T(n, k)=if(k<=n, (2n-1)!!/(2k-1)!!, 0); T(n, k)=if(k<=n, n!*C(2n, n)2^(k-n)/(k!*C(2k, k)), 0); T(n, k)=if(k<=n, 2^(n-k)(n-1/2)!/(k-1/2)!, 0); T(n, k)=if(k<=n, (n+1)!*C(n)2^(k-n)/((k+1)!*C(k)), 0).
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EXAMPLE
| Triangle begins
1;
1,1;
3,3,1;
15,15,5,1;
105,105,35,7,1;
945,945,315,63,9,1;
10395,10395,3465,693,99,11,1;
Inverse is A112295, which begins
1;
-1,1;
0,-3,1;
0,0,-5,1;
0,0,0,-7,1;
0,0,0,0,-9,1;
Similar results arise for higher factorials.
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MAPLE
| The triangle A112292(n, k) can be obtained from T' by (Maple-parlance)
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CROSSREFS
| Sequence in context: A115193 A039797 A143171 * A001497 A123244 A105599
Adjacent sequences: A112289 A112290 A112291 * A112293 A112294 A112295
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KEYWORD
| easy,nonn,tabl
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Sep 01 2005
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