A211956 - OEIS

%I #7 Apr 30 2012 14:28:12

%S 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,

%T 16,1,25,100,140,80,16,1,30,140,224,144,32,1,36,210,448,432,192,32,1,

%U 42,280,672,720,352,64,1,49,392,1176,1680,1232,448,64

%N Coefficients of a sequence of polynomials related to the Morgan-Voyce polynomials.

%C 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).

%C 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.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Morgan-VoycePolynomials.html">Morgan-Voyce polynomials</a>

%F 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).

%F 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 + ....

%F 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.

%F The product P(n,x)*P(n+1,x) is the n-th row polynomial of A211955.

%F 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).

%F Other representations for the row polynomials include

%F P(2*n,x) = 1/2*(1+x+sqrt(x^2+2*x))^n + 1/2*(1+x-sqrt(x^2+2*x))^n;

%F 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.

%F Recurrence equation: P(n+1,x)*P(n-2,x) - P(n,x)*P(n-1,x) = x.

%F Row sums A005246(n+2).

%e Triangle begins

%e .n\k.|..0....1....2....3....4

%e = = = = = = = = = = = = = = =

%e ..0..|..1

%e ..1..|..1

%e ..2..|..1....1

%e ..3..|..1....2

%e ..4..|..1....4....2

%e ..5..|..1....6....4

%e ..6..|..1....9...12....4

%e ..7..|..1...12...20....8

%e ..8..|..1...16...40...32....8

%e ..9..|..1...20...60...56...16

%e ...

%Y Cf. A005246 (row sums), A085478, A123519, A204021, A208513, A211955, A211957.

%K nonn,easy,tabf

%O 0,6

%A _Peter Bala_, Apr 30 2012