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A096956
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Pascal (1,6) triangle.
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11
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6, 1, 6, 1, 7, 6, 1, 8, 13, 6, 1, 9, 21, 19, 6, 1, 10, 30, 40, 25, 6, 1, 11, 40, 70, 65, 31, 6, 1, 12, 51, 110, 135, 96, 37, 6, 1, 13, 63, 161, 245, 231, 133, 43, 6, 1, 14, 76, 224, 406, 476, 364, 176, 49, 6, 1, 15, 90, 300, 630, 882, 840, 540, 225, 55, 6, 1, 16, 105, 390, 930
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OFFSET
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0,1
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COMMENTS
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Except for the first row this is the row reversed (6,1)-Pascal triangle A093563.
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)=(6-5*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(6-5*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k,k) = A022097(n-2), n >= 2, with n=1 value 6. 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) = 6; 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): (6-5*x)/(1-x)^(m+1), m >= 0.
a(n,k) = (1+5*k/n)*binomial(n,k)), for n > 0. - Mircea Merca, Apr 08 2012
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EXAMPLE
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[6]; [1,6]; [1,7,6]; [1,8,13,6]; [1,9,21,16,6]; ...
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MAPLE
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a(n, k):=piecewise(n=0, 6, 0<n, (1+5*k/n)*binomial(n, k)) # Mircea Merca, Apr 08 2012
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CROSSREFS
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Row sums: A005009(n-1), n>=1, 6 if n=0; g.f.: (6-5*x)/(1-2*x). Alternating row sums are [6, -5, 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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