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A096956
Pascal (1,6) triangle.
11
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
OFFSET
0,1
COMMENTS
Except for the first row this is the row reversed (6,1)-Pascal triangle A093563.
This is the sixth member, q=6, in the family of (1,q) Pascal triangles: A007318 (Pascal (q=1), A029635 (q=2) (but with a(0,0)=2, not 1), A095660, A095666, A096940.
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.
FORMULA
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
EXAMPLE
[6]; [1,6]; [1,7,6]; [1,8,13,6]; [1,9,21,16,6]; ...
MAPLE
a(n, k):=piecewise(n=0, 6, 0<n, (1+5*k/n)*binomial(n, k)) # Mircea Merca, Apr 08 2012
CROSSREFS
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].
Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+6), A056115, A096957-9, A097297-A097300.
Sequence in context: A109918 A339433 A263494 * A326126 A078300 A176398
KEYWORD
nonn,easy,tabl
AUTHOR
Wolfdieter Lang, Aug 13 2004
STATUS
approved