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.
LINKS
Wolfdieter Lang, First 10 rows.
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
KEYWORD
AUTHOR
Wolfdieter Lang, Aug 13 2004
STATUS
approved