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Triangle T(n,k) = binomial(2*n-k, k)*(n-k)!, read by rows.
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%I #8 Jun 04 2021 09:48:13

%S 1,1,1,2,3,1,6,10,6,1,24,42,30,10,1,120,216,168,70,15,1,720,1320,1080,

%T 504,140,21,1,5040,9360,7920,3960,1260,252,28,1,40320,75600,65520,

%U 34320,11880,2772,420,36,1,362880,685440,604800,327600,120120,30888,5544,660,45,1

%N Triangle T(n,k) = binomial(2*n-k, k)*(n-k)!, read by rows.

%C Row sums of B^{-1}*A155856*B^{-1} are A000166 with B=A007318.

%C Downward diagonals T(n+j, n) = j!*binomial(n+j, n) = j!*seq(j), where seq(j) are sequences A010965, A010967, ..., A011101, A017714, A017716, ..., A017764, for 6 <= j <= 50, respectively. - _G. C. Greubel_, Jun 04 2021

%H G. C. Greubel, <a href="/A155856/b155856.txt">Rows n = 0..50 of the triangle, flattened</a>

%F T(n,k) = binomial(2*n-k, k)*(n-k)!.

%F Sum_{k=0..n} T(n, k) = A155857(n)

%F Sum_{k=0..floor(n/2)} T(n-k, k) = A155858(n) (diagonal sums).

%F G.f.: 1/(1-xy-x/(1-xy-x/(1-xy-2x/(1-xy-2x/(1-xy-3x/(1-.... (continued fraction).

%F From _G. C. Greubel_, Jun 04 2021:

%F T(n, 0) = A000142(n). T(n+1, n) = A000217(n+1).

%F T(n+1, 1) = A007680(n). T(n+2, n) = A034827(n+4).

%F T(n+2, 2) = A175925(n). T(n+3, n) = A253946(n).

%F T(2*n, n) = A064352(n) T(n+4, n) = 4!*A000581(n).

%F T(n+1, n) = A000217(n+1). T(n+5, n) = 5!*A001287(n). (End)

%e Triangle begins:

%e 1;

%e 1, 1;

%e 2, 3, 1;

%e 6, 10, 6, 1;

%e 24, 42, 30, 10, 1;

%e 120, 216, 168, 70, 15, 1;

%e 720, 1320, 1080, 504, 140, 21, 1;

%e 5040, 9360, 7920, 3960, 1260, 252, 28, 1;

%t Table[Binomial[2n-k,k](n-k)!,{n,0,10},{k,0,n}]//Flatten (* _Harvey P. Dale_, Mar 24 2017 *)

%o (Sage) flatten([[factorial(n-k)*binomial(2*n-k, k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jun 04 2021

%Y Cf. A155857 (row sums), A155858 (diagonal sums).

%K easy,nonn,tabl

%O 0,4

%A _Paul Barry_, Jan 29 2009