A122774 Triangle of bifactorial numbers, n B m = (2(n-m)-1)!! (2(n-1))!! / (2(n-m))!!, read by rows.
1, 1, 2, 3, 4, 8, 15, 18, 24, 48, 105, 120, 144, 192, 384, 945, 1050, 1200, 1440, 1920, 3840, 10395, 11340, 12600, 14400, 17280, 23040, 46080, 135135, 145530, 158760, 176400, 201600, 241920, 322560, 645120
Offset: 1
Examples
Examples obtained from the expressions in J 4 B 3 NB. bifactorial 4 B 3, n=4, m=3 24 (B"0 >:(AT)i.)"0 >:i.4 NB. for 1 <= m <= n=4 1 0 0 0 1 2 0 0 3 4 8 0 15 18 24 48
Links
- Oleg Kobchenko, Bifactorial, J Wiki at jsoftware.com
- Oleg Kobchenko, Generalized Monte Hall problem at J Wiki
- B. E. Meserve, Double Factorials, American Mathematical Monthly, 55 (1948), 425-426.
- R. Ondrejka, Tables of double factorials, Math. Comp., 24 (1970), 231.
- Eric Weisstein's World of Mathematics, Double Factorial.
- Index of sequences related to factorial numbers
Crossrefs
Programs
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J
NB. (www.jsoftware.com): Fe=: 2&^ * ! NB. even factorial, 2^n * n! Fo=: !@+: % Fe NB. odd factorial, (2n)! / (2n)!! B =: Fo@- * <:@[ %&Fe - NB. bifactorial, Fo(n-m) Fe(n-1) / Fe(n-m)
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Mathematica
Table[(2 (n - m) - 1)!! (2 (n - 1))!!/(2 (n - m))!!, {n, 8}, {m, n}] // Flatten (* Michael De Vlieger, Jan 25 2017 *)
Formula
(n B m) = (2(n-m)-1)!! (2(n-1))!! / (2(n-m))!!, 1<=m<=n
(n B 1) = (2(n-1)-1)!! = (2n-3)!!, 1<=n
(n B n) = (2(n-1))!!, 1<=n
(n B m+1) = (n B m) 2(n-m) / (2(n-m)-1), 1<=m
(n+1 B m+1) = (n B m) 2n, 1<=m<=n
(n+1 B m+1) = C(n,m) (2(n-m)-1)!!(2m)!!, 1<=m<=n [Corrected by Werner Schulte, Jan 23 2017]
(n+1 B 1) = Sum_{i=1..n} (n B i).
(n B m) = binomial(2*n-2*m,n-m)*((n-1)!)/2^(n+1-2*m) for 1<=m<=n. - Werner Schulte, Jan 23 2017
Comments