|
| |
|
|
A096940
|
|
Pascal (1,5) triangle.
|
|
13
|
|
|
|
5, 1, 5, 1, 6, 5, 1, 7, 11, 5, 1, 8, 18, 16, 5, 1, 9, 26, 34, 21, 5, 1, 10, 35, 60, 55, 26, 5, 1, 11, 45, 95, 115, 81, 31, 5, 1, 12, 56, 140, 210, 196, 112, 36, 5, 1, 13, 68, 196, 350, 406, 308, 148, 41, 5, 1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5, 1, 15, 95, 345, 810, 1302
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,1
|
|
|
COMMENTS
|
This is the fifth member, q=5, 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.
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)=(5-4*x)/(1-x), f(x)=1/(1-x), hence G(z,x)=(5-4*z)/(1-(1+x)*z).
The SW-NE diagonals give Sum_{k=0..ceiling((n-1)/2)} a(n-1-k, k) = A022096(n-2), n>=2, with n=1 value 5. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs.
|
|
|
LINKS
|
|
|
|
FORMULA
|
Recursion: a(n, m)=0 if m>n, a(0, 0)= 5; 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): (5-4*x)/(1-x)^(m+1), m>=0.
a(n,k) = (1+4*k/n)*binomial(n,k), for n>0. - Mircea Merca, Apr 08 2012
|
|
|
EXAMPLE
|
Triangle begins:
5;
1, 5;
1, 6, 5;
1, 7, 11, 5;
1, 8, 18, 16, 5;
1, 9, 26, 34, 21, 5;
1, 10, 35, 60, 55, 26, 5;
1, 11, 45, 95, 115, 81, 31, 5;
1, 12, 56, 140, 210, 196, 112, 36, 5;
1, 13, 68, 196, 350, 406, 308, 148, 41, 5;
1, 14, 81, 264, 546, 756, 714, 456, 189, 46, 5; etc.
|
|
|
MAPLE
|
a(n, k):=piecewise(n=0, 5, 0<n, (1+4*k/n)*binomial(n, k)) # Mircea Merca, Apr 08 2012
|
|
|
PROG
|
(PARI) a(n) = {if(n <= 1, return(5 - 4*(n==1))); my(m = (sqrtint(8*n + 1) - 1)\2, t = n - binomial(m + 1, 2)); (1+4*t/m)*binomial(m, t)} \\ David A. Corneth, Aug 28 2019
|
|
|
CROSSREFS
|
Row sums: A007283(n-1), n>=1, 5 if n=0; g.f.: (5-4*x)/(1-2*x). Alternating row sums are [5, -4, followed by 0's].
Column sequences (without leading zeros) give for m=1..9, with n>=0: A000027(n+5), A056000(n-1), A096941-7.
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
