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URL: https://oeis.org/A155856

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A155856
Triangle T(n,k) = binomial(2*n-k, k)*(n-k)!, read by rows.
6
1, 1, 1, 2, 3, 1, 6, 10, 6, 1, 24, 42, 30, 10, 1, 120, 216, 168, 70, 15, 1, 720, 1320, 1080, 504, 140, 21, 1, 5040, 9360, 7920, 3960, 1260, 252, 28, 1, 40320, 75600, 65520, 34320, 11880, 2772, 420, 36, 1, 362880, 685440, 604800, 327600, 120120, 30888, 5544, 660, 45, 1
OFFSET
0,4
COMMENTS
Row sums of B^{-1}*A155856*B^{-1} are A000166 with B=A007318.
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
LINKS
Veronica Bitonti, Bishal Deb, and Alan D. Sokal, Thron-type continued fractions (T-fractions) for some classes of increasing trees, arXiv:2412.10214 [math.CO], 2024. See p. 58.
FORMULA
T(n,k) = binomial(2*n-k, k)*(n-k)!.
Sum_{k=0..n} T(n, k) = A155857(n)
Sum_{k=0..floor(n/2)} T(n-k, k) = A155858(n) (diagonal sums).
G.f.: 1/(1-xy-x/(1-xy-x/(1-xy-2x/(1-xy-2x/(1-xy-3x/(1-.... (continued fraction).
From G. C. Greubel, Jun 04 2021: (Start)
T(n, 0) = A000142(n). T(n+1, n) = A000217(n+1).
T(n+1, 1) = A007680(n). T(n+2, n) = A034827(n+4).
T(n+2, 2) = A175925(n). T(n+3, n) = A253946(n).
T(2*n, n) = A064352(n) T(n+4, n) = 4!*A000581(n).
T(n+1, n) = A000217(n+1). T(n+5, n) = 5!*A001287(n). (End)
EXAMPLE
Triangle begins:
1;
1, 1;
2, 3, 1;
6, 10, 6, 1;
24, 42, 30, 10, 1;
120, 216, 168, 70, 15, 1;
720, 1320, 1080, 504, 140, 21, 1;
5040, 9360, 7920, 3960, 1260, 252, 28, 1;
MATHEMATICA
Table[Binomial[2n-k, k](n-k)!, {n, 0, 10}, {k, 0, n}]//Flatten (* Harvey P. Dale, Mar 24 2017 *)
PROG
(SageMath) 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
CROSSREFS
Cf. A155857 (row sums), A155858 (diagonal sums).
Sequence in context: A321625 A132372 A103136 * A086960 A165675 A138771
KEYWORD
easy,nonn,tabl
AUTHOR
Paul Barry, Jan 29 2009
STATUS
approved