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%I #13 Oct 19 2021 17:54:40
%S 1,0,0,1,0,2,0,6,1,0,24,6,0,120,36,1,0,720,240,12,0,5040,1800,120,1,0,
%T 40320,15120,1200,20,0,362880,141120,12600,300,1,0,3628800,1451520,
%U 141120,4200,30,0,39916800,16329600,1693440,58800,630,1
%N T(n, k) = binomial(n-k-1, k-1)*(n-k)!/k! for n >= 0 and 0 <= k <= floor(n/2). Irregular triangle read by rows.
%C Also the antidiagonals of the Lah triangle A271703.
%H Harvey P. Dale, <a href="/A330609/b330609.txt">Table of n, a(n) for n = 0..1000</a>
%F T(0,0) = T(2,1) = 1. If k < 1 or k > ceiling(n/2) then T(n,k) = 0. Otherwise:
%F T(n, k) = (n-1)*T(n-1, k) + T(n-2, k-1)
%e Triangle begins:
%e [0] 1
%e [1] 0
%e [2] 0, 1
%e [3] 0, 2
%e [4] 0, 6, 1
%e [5] 0, 24, 6
%e [6] 0, 120, 36, 1
%e [7] 0, 720, 240, 12
%e [8] 0, 5040, 1800, 120, 1
%e [9] 0, 40320, 15120, 1200, 20
%p T := (n, k) -> binomial(n-k-1, k-1)*(n-k)!/k!:
%p seq(seq(T(n, k), k=0..floor(n/2)), n=0..12);
%p # Alternative:
%p T := proc(n, k) option remember;
%p if (n=0 and k=0) or (n=2 and k=1) then 1 elif (k < 1) or (k > ceil(n/2)) then 0
%p else (n-1)*T(n-1, k) + T(n-2, k-1) fi end: seq(seq(T(n, k), k=0..n/2), n=0..12);
%t Table[Binomial[n-k-1,k-1] (n-k)!/k!,{n,0,20},{k,0,Floor[n/2]}]//Flatten (* _Harvey P. Dale_, Oct 19 2021 *)
%Y Variants: A180047, A221913. Row sums: A001053.
%Y Cf. A271703.
%K nonn,tabf
%O 0,6
%A _Peter Luschny_, Dec 27 2019
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