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A192939 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x)=(x+2)(x+4)...(x+2n). 2
1, 2, 9, 61, 546, 6043, 79475, 1209160, 20873685, 402896615, 8595041400, 200773098515, 5095839723205, 139624739872970, 4107177047046645, 129087781738773385, 4316962772836390050, 153048896045632212175, 5733602882337419294975 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

For an introduction to reductions of polynomials by substitutions such as x^2->x+1, see A192232.

LINKS

Table of n, a(n) for n=0..18.

FORMULA

Conjecture: a(n) +(-4*n+1)*a(n-1) +(4*n^2-6*n+1)*a(n-2)=0. - R. J. Mathar, May 08 2014

EXAMPLE

The first four polynomials p(n,x) and their reductions are as follows:

p(0,x)=1

p(1,x)=x+2 -> x+2

p(2,x)=(x+2)(x+4) -> 9+7x

p(3,x)=(x+2)(x+4)(x+6) -> 61+58x

From these, read

A192939=(1,2,9,61,...) and A192940=(0,1,7,58,...)

MATHEMATICA

q = x^2; s = x + 1; z = 26;

p[0, x] := 1;

p[n_, x_] := (x + 2 n)*p[n - 1, x];

Table[Expand[p[n, x]], {n, 0, 7}]

reduce[{p1_, q_, s_, x_}] :=

FixedPoint[(s PolynomialQuotient @@ #1 +

       PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1]

t = Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];

u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}]

  (* A192939 *)

u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]

  (* A192940 *)

CROSSREFS

Cf. A192232, A192744, A192940.

Sequence in context: A207649 A289713 A053983 * A107883 A088182 A006155

Adjacent sequences:  A192936 A192937 A192938 * A192940 A192941 A192942

KEYWORD

nonn

AUTHOR

Clark Kimberling, Jul 13 2011

STATUS

approved

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Last modified February 20 00:51 EST 2018. Contains 299357 sequences. (Running on oeis4.)