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A095666
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Pascal (1,4) triangle.
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16
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4, 1, 4, 1, 5, 4, 1, 6, 9, 4, 1, 7, 15, 13, 4, 1, 8, 22, 28, 17, 4, 1, 9, 30, 50, 45, 21, 4, 1, 10, 39, 80, 95, 66, 25, 4, 1, 11, 49, 119, 175, 161, 91, 29, 4, 1, 12, 60, 168, 294, 336, 252, 120, 33, 4, 1, 13, 72, 228, 462, 630, 588, 372, 153, 37, 4, 1, 14, 85, 300, 690, 1092
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OFFSET
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0,1
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COMMENTS
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This is the fourth member, q=4, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1), A029635 (q=2) (but with a(0,0)=2, not 1), A095660.
This is an example of a Riordan triangle (see A053121 for a comment and the 1991 Shapiro et al. reference on the Riordan group) with o.g.f. of column nr. m of the type g(x)*(x*f(x))^m with f(0)=1. Therefore the o.g.f. for the row polynomials p(n,x) := Sum_{m=0..n} (a(n,m)*x^m is G(z,x) = g(z)/(1 - x*z*f(z)). Here: g(x) = (4-3*x)/(1-x), f(x) = 1/(1-x), hence G(z,x) = (4-3*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022095(n-2), n >= 2, with n=1 value 4. [Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.]
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LINKS
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FORMULA
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Recursion: a(n, m) = 0 if m > n, a(0, 0) = 4; a(n, 0) = 1 if n>=1; a(n, m) = a(n-1, m) + a(n-1, m-1).
G.f. column m (without leading zeros): (4-3*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1 + 3*k/n)*binomial(n,k). [Mircea Merca, Apr 08 2012]
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EXAMPLE
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[4];
[1,4];
[1,5,4];
[1,6,9,4];
[1,7,15,13,4];
...
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MAPLE
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a(n, k):=(1+3*k/n)*binomial(n, k) # Mircea Merca, Apr 08 2012
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PROG
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(Haskell)
a095666 n k = a095666_tabl !! n !! k
a095666_row n = a095666_tabl !! n
a095666_tabl = [4] : iterate
(\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1, 4]
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CROSSREFS
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Row sums: A020714(n-1), n >= 1, 4 if n=0.
Alternating row sums are [4, -3, followed by 0's].
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KEYWORD
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AUTHOR
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STATUS
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approved
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