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A119301
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Triangle read by rows: T(n,k) = binomial(3*n-k,n-k).
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7
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1, 3, 1, 15, 5, 1, 84, 28, 7, 1, 495, 165, 45, 9, 1, 3003, 1001, 286, 66, 11, 1, 18564, 6188, 1820, 455, 91, 13, 1, 116280, 38760, 11628, 3060, 680, 120, 15, 1, 735471, 245157, 74613, 20349, 4845, 969, 153, 17, 1, 4686825, 1562275, 480700, 134596, 33649, 7315
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OFFSET
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0,2
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COMMENTS
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Row sums are A045721. Inverse is Riordan array (1-3x,x(1-x)^2), A119302.
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LINKS
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FORMULA
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G.f. g(x) = 2*sin(arcsin(3*sqrt(3*x)/2)/3)/sqrt(3*x) satisfies g(x) = 1/(1-x*g(x)^2).
Riordan array (1/(1-3*x*g(x)^2),x*g(x)^2) where g(x)=1+x*g(x)^3.
'Horizontal' recurrence equation: T(n,0) = binomial(3*n,n) and for k >= 1, T(n,k) = Sum_{i = 1..n+1-k} i*T(n-1,k-2+i). - Peter Bala, Dec 28 2014
T(n, k) = Sum_{j = 0..n} binomial(n+j-1, j)*binomial(2*n-k-j, n). - Peter Bala, Jun 04 2024
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EXAMPLE
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Triangle begins
1,
3, 1,
15, 5, 1,
84, 28, 7, 1,
495, 165, 45, 9, 1,
3003, 1001, 286, 66, 11, 1,
18564, 6188, 1820, 455, 91, 13, 1,
116280, 38760, 11628, 3060, 680, 120, 15, 1
...
Horizontal recurrence: T(4,1) = 1*84 + 2*28 + 3*7 + 4*1 = 165. - Peter Bala, Dec 29 2014
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MAPLE
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T := proc(n, k) option remember;
`if`(n = 0, 1, add(i*T(n-1, k-2+i), i=1..n+1-k)) end:
for n from 0 to 9 do print(seq(T(n, k), k=0..n)) od; # Peter Luschny, Dec 30 2014
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MATHEMATICA
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Flatten[Table[Binomial[3n-k, n-k], {n, 0, 10}, {k, 0, n}]] (* Harvey P. Dale, Jul 28 2012 *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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