

A127893


Riordan array (1/(1x)^3, x/(1x)^3).


7



1, 3, 1, 6, 6, 1, 10, 21, 9, 1, 15, 56, 45, 12, 1, 21, 126, 165, 78, 15, 1, 28, 252, 495, 364, 120, 18, 1, 36, 462, 1287, 1365, 680, 171, 21, 1, 45, 792, 3003, 4368, 3060, 1140, 231, 24, 1, 55, 1287, 6435, 12376, 11628, 5985, 1771, 300, 27, 1
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OFFSET

0,2


COMMENTS

Inverse is A127894.
From Peter Bala, Jul 22 2014: (Start)
Let M denote the unsigned version of the lower unit triangular array A122432 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k x k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly welldefined). See the Example section. (End)


LINKS

G. C. Greubel, Rows n = 0..99, flattened
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.


FORMULA

T(n,k) = binomial(n+2*k+2, nk).
Sum_{k=0..n} T(n, k) = A052529(n+1) (row sums).
Sum_{k=0..floor(n/2)} T(nk, k) = A095263(n+1) (diagonal sums).
Recurrence: T(n+1, k+1) = Sum_{i = 0..nk} binomial(i+2, 2)*T(ni,k).  Peter Bala, Jul 22 2014
G.f.: 1/((1x)^3x*y).  Vladimir Kruchinin, Apr 27 2015


EXAMPLE

Triangle begins
1;
3, 1;
6, 6, 1;
10, 21, 9, 1;
15, 56, 45, 12, 1;
21, 126, 165, 78, 15, 1;
28, 252, 495, 364, 120, 18, 1;
36, 462, 1287, 1365, 680, 171, 21, 1;
45, 792, 3003, 4368, 3060, 1140, 231, 24, 1;
55, 1287, 6435, 12376, 11628, 5985, 1771, 300, 27, 1;
66, 2002, 12870, 31824, 38760, 26334, 10626, 2600, 378, 30, 1;
...
From Peter Bala, Jul 22 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
/ 1 \/1 \/1 \ / 1 \
 3 1 0 1 0 1   3 1 
 6 3 1 0 3 1 0 0 1 ... =  6 6 1 
10 6 3 1 0 6 3 1 0 0 3 1  10 21 9 1
15 10 6 3 10 10 6 3 10 0 6 3 1 ... 
... ... ...  ... 
(End)


MAPLE

seq(seq(binomial(n+2*k+2, nk), k=0..n), n=0..10); # Robert Israel, Apr 28 2015


MATHEMATICA

Flatten@ Table[Binomial[n+2k1, nk], {n, 10}, {k, n}] (* Michael De Vlieger, Apr 27 2015 *)


PROG

(PARI) for(n=0, 10, for(k=0, n, print1(binomial(n+2*k+2, nk), ", "))) \\ G. C. Greubel, Apr 29 2018
(Magma) [Binomial(n+2*k+2, nk): k in [0..n], n in [0..10]]; // G. C. Greubel, Apr 29 2018
(GAP) Flat(List([0..10], n>List([0..n], k>Binomial(n+2*k+2, nk)))); # Muniru A Asiru, Apr 30 2018
(Sage) flatten([[binomial(n+2*k+2, nk) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 16 2021


CROSSREFS

Cf. A052529, A095263, A122432, A127894.
Sequence in context: A235706 A124847 A249251 * A127895 A325005 A325013
Adjacent sequences: A127890 A127891 A127892 * A127894 A127895 A127896


KEYWORD

easy,nonn,tabl


AUTHOR

Paul Barry, Feb 04 2007


STATUS

approved



