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A192744 Constant term of the reduction by x^2->x+1 of the polynomial p(n,x) defined below in Comments. 124
1, 1, 3, 8, 29, 133, 762, 5215, 41257, 369032, 3676209, 40333241, 483094250, 6271446691, 87705811341, 1314473334832, 21017294666173, 357096406209005, 6424799978507178, 122024623087820183, 2439706330834135361, 51219771117454755544 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

The titular polynomial is defined recursively by p(n,x)=x*p(n-1,x)+n! for n>0, where p(0,x)=1; see the Example.  For an introduction to polynomial reduction, see A192232.  The discussion at A192232 Comments continues here:

...

Let R(p,q,s) denote the "reduction of polynomial p by q->s" as defined at A192232.  Suppose that q(x)=x^k for some k>0 and that s(x)=s(k,0)*x^(k-1)+s(k,1)*x^(k-2)+...+s(k,k-2)x+s(k,k-1).

...

First, we shall take p(x)=x^n, where n>=0; the results will be used to formulate R(p,q,s) for general p.  Represent R(x^n,q,s) by

...

R(x^n)=s(n,0)*x^(k-1)+s(n,1)*x^(k-2)+...+s(n,k-2)*x+s(n,k-1).

...

Then each of the sequences u(n)=s(n,h), for h=0,1,...,k-1, satisfies this linear recurrence relation:

...

u(n)=s(k,0)*u(n-1)+s(k,1)*u(n-2)+...+s(k,k-2)*u(n-k-1)+s(k,k-1)*u(n-k), with initial values tabulated here:

...

n: ..s(n,0)...s(n,1)..s(n,2).......s(n,k-2)..s(n,k-1)

0: ....0........0.......0..............0.......1

1: ....0........0.......0..............1.......0

...

k-2: ..0........1.......0..............0.......0

k-1: ..0........0.......0..............0.......0

k: ..s(k,0)...s(k,1)..s(k,2).......s(k,k-2)..s(k,k-1)

...

That completes the formulation for p(x)=x^n.  Turning to the general case, suppose that

...

p(n,x)=p(n,0)*x^n+p(n,1)*x^(n-1)+...+p(n,n-1)*x+p(n,n)

...

is a polynomial of degree n>=0.  Then the reduction denoted by (R(p(n,x) by x^k -> s(x)) is the polynomial of degree k-1 given by the matrix product P*S*X, where P=(p(n,0)...p(n,1).........p(n-k)...p(n,n-k+1); X has all 0's except for main diagonal (x^(k-1), x^(k-2)...x,1); and S has

...

row 1: ... s(n,0) ... s(n,1) ...... s(n,k-2) . s(n,k-1)

row 2: ... s(n-1,0) . s(n-1,1) .... s(n-1,k-2) s(n-1,k-1)

...

row n-k+1: s(k,0).... s(k,1) ...... s(k,k-2) ..s(k,k-1)

row n-k+2: p(n,n-k+1) p(n,n-k+2) .. p(n,n-1) ..p(n,n)

*****

As a class of examples, suppose that (v(n)), for n>=0, is a sequence, that p(0,x)=1, and p(n,x)=v(n)+p(n-1,x) for n>0.  If q(x)=x^2 and s(x)=x+1, and we write the reduction R(p(n,x)) as u1(n)*x+u2(n), then the sequences u1 and u2 are convolutions with the Fibonacci sequence, viz., let F=(0,1,1,2,3,5,8,...)=A000045 and let G=(1,0,1,1,2,3,5,8...); then u1=G**v and u2=F**v, where ** denotes convolution.  Examples (with a few exceptions for initial terms):

...

If v(n)=n! then u1=A192744, u2=A192745.

If v(n)=n+1 then u1=A000071, u2=A001924.

If v(n)=2n then u1=A014739, u2=A027181.

If v(n)=2n+1 then u1=A001911, u2=A001891.

If v(n)=3n+1 then u1=A027961, u2=A023537.

If v(n)=3n+2 then u1=A192746, u2=A192747.

If v(n)=3n then u1=A154691, u2=A192748.

If v(n)=4n+1 then u1=A053311, u2=A192749.

If v(n)=4n+2 then u1=A192750, u2=A192751.

If v(n)=4n+3 then u1=A192752, u2=A192753.

If v(n)=4n then u1=A147728, u2=A023654.

If v(n)=5n+1 then u1=A192754, u2=A192755.

If v(n)=5n then u1=A166863, u2=A192756.

If v(n)=floor((n+1)tau) then u1=A192457, u2=A023611.

If v(n)=floor((n+2)/2) then u1=A052952, u2=A129696.

If v(n)=floor((n+3)/3) then u1=A004695, u2=A178982.

If v(n)=floor((n+4)/4) then u1=A080239, u2=A192758.

If v(n)=floor((n+5)/5) then u1=A124502, u2=A192759.

If v(n)=n+2 then u1=A001594, u2=A192760.

If v(n)=n+3 then u1=A022318, u2=A192761.

If v(n)=n+4 then u1=A022319, u2=A192762.

If v(n)=2^n then u1=A027934, u2=A008766.

If v(n)=3^n then u1=A106517, u2=A094688.

LINKS

Table of n, a(n) for n=0..21.

FORMULA

G.f.: (1-x)/(1-x-x^2)/Q(0), where Q(k)= 1 - x*(k+1)/(1 - x*(k+1)/Q(k+1)); (continued fraction). - Sergei N. Gladkovskii, May 20 2013

Conjecture: a(n) +(-n-2)*a(n-1) +2*(n-1)*a(n-2) +3*a(n-3) +(-n+2)*a(n-4)=0. - R. J. Mathar, May 04 2014

Conjecture: (-n+2)*a(n) +(n^2-n-1)*a(n-1) +(-n^2+3*n-3)*a(n-2) -(n-1)^2*a(n-3)

=0. - R. J. Mathar, Dec 16 2015

EXAMPLE

The first five polynomials and their reductions:

1 -> 1

1+x -> 1+x

2+x+x^2 -> 3+2x

6+2x+x^2+x^3 -> 8+5x

24+6x+2x^2+x^3+x^4 -> 29+13x, so that

A192744=(1,1,3,8,29,...) and A192745=(0,1,2,5,13,...).

MAPLE

A192744p := proc(n, x)

    option remember;

    if n = 0 then

        1;

    else

        x*procname(n-1, x)+n! ;

        expand(%) ;

    end if;

end proc:

A192744 := proc(n)

    local p;

    p := A192744p(n, x) ;

    while degree(p, x) > 1 do

        p := algsubs(x^2=x+1, p) ;

        p := expand(p) ;

    end do:

    coeftayl(p, x=0, 0) ;

end proc: # R. J. Mathar, Dec 16 2015

MATHEMATICA

q = x^2; s = x + 1; z = 40;

p[0, n_] := 1; p[n_, x_] := x*p[n - 1, x] + n!;

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}]

  (* A192744 *)

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

  (* A192745 *)

CROSSREFS

Cf. A192232.

Sequence in context: A013309 A058378 A063839 * A130470 A275166 A182117

Adjacent sequences:  A192741 A192742 A192743 * A192745 A192746 A192747

KEYWORD

nonn

AUTHOR

Clark Kimberling, Jul 09 2011

STATUS

approved

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Last modified August 17 06:03 EDT 2017. Contains 290635 sequences.