0,1

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(a(n,m)*x^m,m=0..n) 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(a(n-1-k,k),k=0..ceiling((n-1)/2)) = A022097(n-2), n>=2, with n=1 value 6. Observation by Paul Barry (pbarry(AT)wit.ie, Apr 29 2004. Proof via recursion relations and comparison of inputs.

W. Lang, First 10 rows.

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.

[6];[1,6];[1,7,6];[1,8,13,6];[1,9,21,16,6];...

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: A138116 A010687 A109918 this_sequence A010492 A070514 A070472

Adjacent sequences: A096953 A096954 A096955 this_sequence A096957 A096958 A096959

nonn,easy,tabl

Wolfdieter Lang (wolfdieter.lang_AT_physik_DOT_uni-karlsruhe_DOT_de), Aug 13 2004