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%I #7 May 01 2012 13:02:43
%S 1,1,1,1,3,2,1,6,10,4,1,10,30,28,8,1,15,70,112,72,16,1,21,140,336,360,
%T 176,32,1,28,252,840,1320,1056,416,64,1,36,420,1848,3960,4576,2912,
%U 960,128,1,45,660,3696,10296,16016,14560,7680,2176,256
%N Triangle of coefficients of a polynomial sequence related to the Morgan-Voyce polynomials A085478.
%C Let b(n,x) = sum {k = 0..n} binomial(n+k,2*k)*x^k denote the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients (in ascending powers of x) of the related polynomial sequence R(n,x) := 1/2*b(n,2*x) + 1/2. Several sequences already in the database are of the form (R(n,x))n>=0 for a fixed value of x. These include A101265 (x = 1), A011900 (x = 2), A182432 (x = 3), A054318 (x = 4) as well as signed versions of A133872 (x = -1), A109613(x = -2), A146983 (x = -3) and A084159(x = -4).
%C The polynomials R(n,x) 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). The coefficients of P(n,x) occur in several tables in the database, although without the connection to the Morgan-Voyce polynomials being noted - see A211956 for more details. 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). Hence R(n,x) = 1/u*T(n,u)*T(n+1,u).
%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; T(n,k) = 2^(k-1)*binomial(n+k,2*k) for k > 0.
%F O.g.f. for column k (except column 0): 2^(k-1)*x^k/(1-x)^(2*k+1). O.g.f.: (1-t*(x+2)+t^2)/((1-t)*(1-2*t(x+1)+t^2)) = 1 + (1+x)*t + (1+3*x+2*x^2)*t^2 + ....
%F Removing the first column from the triangle produces the Riordan array [x/(1-x)^3, 2*x/(1-x)^2].
%F The row polynomials R(n,x) := 1/2*b(n,2*x) + 1/2 = 1 + x*sum {k = 1..n} binomial(n+k,2*k)*(2*x)^(k-1).
%F Recurrence equation: R(n,x) = 2*(1+x)*R(n-1,x) - R(n-2,x) - x with initial conditions R(0,x) = 1, R(1,x) = 1+x. Another recurrence is R(n,x)*R(n-2,x) = R(n-1,x)*(R(n-1,x) + x).
%F With P(n,x) as defined in the Comments section we have (x+2)/x - {sum {k = 0..2n} 1/R(k,x)}^2 = 2/(x*P(2*n+1,x)^2); (x+2)/x - {sum {k = 0..2n+1} 1/R(k,x)}^2 = (x+2)/(x*P(2*n+2,x)^2); consequently sum {k = 0..inf} 1/R(k,x) = sqrt((x+2)/x) for either x > 0 or x <= -2.
%F Row sums R(n,1) = A101265(n+1); Alt. row sums R(n,-1) = A133872(n+1);
%F R(n,2) = A011900(n); R(n,-2) = (-1)^n*A109613(n); R(n,3) = A182432;
%F R(n,-3) = (-1)^n*A146983(n); R(n,4) = A054318(n+1); R(n,-4) = (-1)^n*A084159(n).
%e Triangle begins
%e .n\k.|..0....1....2....3....4....5....6
%e = = = = = = = = = = = = = = = = = = = =
%e ..0..|..1
%e ..1..|..1....1
%e ..2..|..1....3....2
%e ..3..|..1....6...10....4
%e ..4..|..1...10...30...28....8
%e ..5..|..1...15...70..112...72...16
%e ..6..|..1...21..140..336..360..176...32
%Y Cf. A011900, A084159, A085478, A101265 (row sums), A109613, A112373, A123519, A133872 (alt row sums), A146983, A182432, A204021, A208513, A211956, A211957.
%K nonn,easy,tabl
%O 0,5
%A _Peter Bala_, Apr 30 2012
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