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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 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 November 17 22:28 EST 2018. Contains 317279 sequences. (Running on oeis4.)