|
| |
|
|
A211956
|
|
Coefficients of a sequence of polynomials related to the Morgan-Voyce polynomials.
|
|
5
|
|
|
|
1, 1, 1, 1, 1, 2, 1, 4, 2, 1, 6, 4, 1, 9, 12, 4, 1, 12, 20, 8, 1, 16, 40, 32, 8, 1, 20, 60, 56, 16, 1, 25, 100, 140, 80, 16, 1, 30, 140, 224, 144, 32, 1, 36, 210, 448, 432, 192, 32, 1, 42, 280, 672, 720, 352, 64, 1, 49, 392, 1176, 1680, 1232, 448, 64
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,6
|
|
|
COMMENTS
|
The row generating polynomials R(n,x) of A211955 factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x), where b(n,x) := sum {k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients in ascending powers of x of the polynomials P(n,x).
The odd numbered rows of the present triangle produce triangle A123519; the even numbered row entries are recorded separately in A211957 and appear to equal the unsigned and row reversed form of A204021. The even numbered rows with a factor of 2^(k-1) removed from the k-th column entries produce triangle A208513.
|
|
|
LINKS
|
|
|
|
FORMULA
|
T(n,0) = 1; for k > 0, T(2*n,k) = 2^k*binomial(n+k,2*k) = A123519(n,k); for k > 0, T(2*n-1,k) = n/(n+k)*2^k*binomial(n+k,2*k) = 2^(k-1)*A208513(n,k).
O.g.f.: ((1+t)*(1-t^2)-t^2*x)/((1-t^2)^2-2*t^2*x) = 1 + t + (1+x)*t^2 + (1+2*x)*t^3 + (1+4*x+2*x^2)*t^4 + ....
Row generating polynomials: P(2*n,x) := 1/2*(b(2*n,2*x)+1)/b(n,2*x) and P(2*n+1,x) := b(n,2*x), where b(n,x) := sum {k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478.
The product P(n,x)*P(n+1,x) is the n-th row polynomial of A211955.
In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u*T(2*n+1,u), where u = sqrt((x+2)/2).
Other representations for the row polynomials include
P(2*n,x) = 1/2*(1+x+sqrt(x^2+2*x))^n + 1/2*(1+x-sqrt(x^2+2*x))^n;
P(2*n,x) = n*sum {k = 0..n}(-1)^(n-k)/(n+k)*binomial(n+k,2*k) *(2*x+4)^k for n >= 1; and P(2*n+1,x) = (2*n+1)*sum {k=0..n} (-1)^(n-k)/(n+k+1)* binomial(n+k+1,2*k+1)*(2*x+4)^k.
Recurrence equation: P(n+1,x)*P(n-2,x) - P(n,x)*P(n-1,x) = x.
|
|
|
EXAMPLE
|
Triangle begins
.n\k.|..0....1....2....3....4
= = = = = = = = = = = = = = =
..0..|..1
..1..|..1
..2..|..1....1
..3..|..1....2
..4..|..1....4....2
..5..|..1....6....4
..6..|..1....9...12....4
..7..|..1...12...20....8
..8..|..1...16...40...32....8
..9..|..1...20...60...56...16
...
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy,tabf
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
