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A116925
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Triangle read by rows: row n (n >= 0) consists of the elements g(i, n-i) (0 <= i <= n), where g(r,s) = 1 + Sum_{k=1..r} Product_{i=0..k-1} binomial(r+s-1, s+i) / binomial(r+s-1, i).
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8
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1, 1, 2, 1, 2, 3, 1, 2, 4, 4, 1, 2, 5, 8, 5, 1, 2, 6, 14, 16, 6, 1, 2, 7, 22, 42, 32, 7, 1, 2, 8, 32, 92, 132, 64, 8, 1, 2, 9, 44, 177, 422, 429, 128, 9, 1, 2, 10, 58, 310, 1122, 2074, 1430, 256, 10, 1, 2, 11, 74, 506, 2606, 7898, 10754, 4862, 512, 11, 1, 2, 12, 92, 782, 5462, 25202, 60398, 58202, 16796, 1024, 12
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OFFSET
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0,3
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COMMENTS
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A generalized Catalan number triangle.
An alternative construction of this triangle. Begin with the Pascal triangle array, written as:
1 1 1 1 1 1 ...
1 2 3 4 5 6 ...
1 3 6 10 15 21 ...
1 4 10 20 35 56 ...
1 5 15 35 70 126 ...
...
For each row r (r >= 0) in the above array, construct a triangle U(r) by applying the operation H defined below.
Then the r-th diagonal from the right in the new triangle is given by the row sums of U(r).
To define H, let us use row r=2, {1 3 6 10 15 ...}, as an illustration.
To get the 4th entry, take the first 4 terms of the row, reverse them and write them under the first 4 terms:
A: 1 3 6 10
B: 10 6 3 1
and form a new row C by beginning with 1 and iterating the map C' = C*B/A until we reach 1:
C: 1 10 20 10 1
E.g., 20 = (6 *10) / 3.
The sum of the terms {1 10 20 10 1} is 42, which is the 4th entry in the r=2 diagonal of the new triangle.
The full triangle U(2) begins
1
1 1
1 3 1
1 6 6 1
1 10 20 10 1
...
(this is the Narayana triangle A001263)
and the row sums are the Catalan numbers, which give our r=2 diagonal.
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LINKS
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FORMULA
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The n-th entry in the r-th diagonal from the right (r >= 0, n >= 1) is given by the quotient:
Sum_{k=1..n} Product_{i=0..r-1} binomial(n+r-2, k-1+i)
------------------------------------------------------
Product_{i=1..r-1} binomial(n+r-2, i)
(End)
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EXAMPLE
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The first few rows of the triangle are:
1
1 2
1 2 3
1 2 4 4
1 2 5 8 5
1 2 6 14 16 6
1 2 7 22 42 32 7
1 2 8 32 92 132 64 8
1 2 9 44 177 422 429 128 9
1 2 10 58 310 1122 2074 1430 256 10
...
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MAPLE
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g:=proc(n, p) local k, i; 1 + add( mul( binomial(n+p-1, p+i) / binomial(n+p-1, i), i=0..k-1 ), k=1..n); end; (N. J. A. Sloane, based on the formula from Hsueh-Hsing Hung)
f:=proc(n, r) local k, b, i; b:=binomial; add( mul( b(n+r-2, k-1+i), i=0..r-1)/ mul( b(n+r-2, i), i=1..r-1), k=1..n); end; M:=30; for j from 0 to M do lprint(seq(f(i, j+1-i), i=1..j+1)); od; # N. J. A. Sloane
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MATHEMATICA
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rows = 11; t[n_, p_] := 1 + Sum[Product[ Binomial[ n+p-1, p+i] / Binomial[ n+p-1, i], {i, 0, k-1}], {k, 1, n}]; Flatten[ Table[ t[p, n-p], {n, 0, rows}, {p, 0, n}]](* Jean-François Alcover, Nov 18 2011, after Maple *)
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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One entry corrected by Hsueh-Hsing Hung (hhh(AT)mail.nhcue.edu.tw), Sep 06 2006
Simpler formula provided by Hsueh-Hsing Hung (hhh(AT)mail.nhcue.edu.tw), Sep 08 2006, which is now taken as the definition of this triangle
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STATUS
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approved
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