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Triangle of coefficients of polynomials v(n,x) jointly generated with A208751; see the Formula section.
3

%I #16 Feb 07 2024 09:02:23

%S 1,2,1,3,5,1,4,14,8,1,5,30,34,11,1,6,55,104,63,14,1,7,91,259,253,101,

%T 17,1,8,140,560,806,504,148,20,1,9,204,1092,2178,1966,884,204,23,1,10,

%U 285,1968,5202,6412,4090,1420,269,26,1,11,385,3333,11286,18238

%N Triangle of coefficients of polynomials v(n,x) jointly generated with A208751; see the Formula section.

%C For a discussion and guide to related arrays, see A208510.

%C Subtriangle of the triangle given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 17 2012

%C Setting v(0,x) = 0, the sequence of polynomials {v(n,x) : n >= 0} satisfies the second-order recurrence v(n,x) = (x + 2)*v(n-1,x) + (x - 1)*v(n-2,x) with v(0,x) = 0 and v(1,x) = 1. Then by Norfleet, this sequence of polynomials is a strong divisibility sequence of polynomials in the ring Z[x], that is gcd(v(n,x), v(m,x)) = v(gcd(n,m),x). In particular, if n divides m then v(n,x) divides v(m,x) in Z[x]. - _Peter Bala_, Feb 07 2024

%H M. Norfleet, <a href="http://www.fq.math.ca/43-2.html">Characterization of second-order strong divisibility sequences of polynomials</a>, The Fibonacci Quarterly, 43(2) (2005), 166-169.

%F u(n,x)=u(n-1,x)+2x*v(n-1,x),

%F v(n,x)=u(n-1,x)+(x+1)*v(n-1,x),

%F where u(1,x)=1, v(1,x)=1.

%F T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) + T(n-2,k-1), T(1,0) = 1, T(2,0) = 2, T(2,1) = 1, T(n,k) = 0 if k<0 or if k>=n. - _Philippe Deléham_, Mar 17 2012

%F G.f.: -x*y/(-1+2*x-x^2+x^2*y+x*y). - _R. J. Mathar_, Aug 12 201

%e First five rows:

%e 1

%e 2...1

%e 3...5....1

%e 4...14...8....1

%e 5...30...34...11...1

%e First five polynomials u(n,x) - see A208751:

%e 1

%e 1 + 2x

%e 1 + 6x + 2x^2

%e 1 + 12x + 12x^2 + 2x^3

%e 1 + 20x + 40x^2 + 18x^3 + 2x^4

%e (0, 2, -1/2, 1/2, 0, 0, ...) DELTA (1, 0, 1/2, -1/2, 0, 0, ...) begins :

%e 1

%e 0, 1

%e 0, 2, 1

%e 0, 3, 5, 1

%e 0, 4, 14, 8, 1

%e 0, 5, 30, 34, 11, 1. - _Philippe Deléham_, Mar 17 2012

%t u[1, x_] := 1; v[1, x_] := 1; z = 16;

%t u[n_, x_] := u[n - 1, x] + 2 x*v[n - 1, x];

%t v[n_, x_] := u[n - 1, x] + (x + 1) v[n - 1, x];

%t Table[Expand[u[n, x]], {n, 1, z/2}]

%t Table[Expand[v[n, x]], {n, 1, z/2}]

%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];

%t TableForm[cu]

%t Flatten[%] (* A208751 *)

%t Table[Expand[v[n, x]], {n, 1, z}]

%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];

%t TableForm[cv]

%t Flatten[%] (* A208752 *)

%Y Cf. A208751, A208510.

%K nonn,tabl

%O 1,2

%A _Clark Kimberling_, Mar 01 2012