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%I #17 Aug 28 2019 16:39:01
%S 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,
%T 1,12,51,110,135,96,37,6,1,13,63,161,245,231,133,43,6,1,14,76,224,406,
%U 476,364,176,49,6,1,15,90,300,630,882,840,540,225,55,6,1,16,105,390,930
%N Pascal (1,6) triangle.
%C Except for the first row this is the row reversed (6,1)-Pascal triangle A093563.
%C 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.
%C 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).
%C 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.
%H W. Lang, <a href="/A096956/a096956.txt">First 10 rows</a>.
%F 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).
%F G.f. column m (without leading zeros): (6-5*x)/(1-x)^(m+1), m >= 0.
%F a(n,k) = (1+5*k/n)*binomial(n,k)), for n > 0. - _Mircea Merca_, Apr 08 2012
%e [6]; [1,6]; [1,7,6]; [1,8,13,6]; [1,9,21,16,6]; ...
%p a(n,k):=piecewise(n=0,6,0<n,(1+5*k/n)*binomial(n,k)) # _Mircea Merca_, Apr 08 2012
%Y 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].
%Y Column sequences (without leading zeros) give for m=1..9, with n >= 0: A000027(n+6), A056115, A096957-9, A097297-A097300.
%K nonn,easy,tabl
%O 0,1
%A _Wolfdieter Lang_, Aug 13 2004
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