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 A192872 Constant term in the reduction by (x^2 -> x+1) of the polynomial p(n,x) given in Comments. 32
 1, 0, 3, 4, 13, 30, 81, 208, 547, 1428, 3741, 9790, 25633, 67104, 175683, 459940, 1204141, 3152478, 8253297, 21607408, 56568931, 148099380, 387729213, 1015088254, 2657535553, 6957518400 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS The polynomial p(n,x) is defined by p(0,x)=1, p(1,x)=x, and p(n,x) = x*p(n-1,x) + (x^2)*p(n-1,x) + 1. The resulting sequence typifies a general class which we shall describe here. Suppose that u,v,a,b,c,d,e,f are numbers used to define these polynomials: ... q(x) = x^2 s(x) = u*x + v p(0,x) = a, p(1,x) = b*x + c p(n,x) = d*x*p(n-1,x) + e*(x^2)*p(n-2,x) + f. ... We shall assume that u is not 0 and that {d,e} is not {0}. The reduction of p(n,x) by the repeated substitution q(x)->s(x), as defined and described at A192232 and A192744, has the form h(n)+k(n)*x. The numerical sequences h and k are, formally, linear recurrence sequences of order 5. The second Mathematica program below shows initial terms and the recurrence coefficients, which are too long to be included here, which imply these properties: (1) The numbers a,b,c,f affect initial terms but not the recurrence coefficients, which depend only on u,v,d,e. (2) If v=0 or e=0, the order of recurrence is <= 3. (3) If v=0 and e=0, the order of recurrence is 2, and the coefficients are 1+d*u and d*u. (See A192904 for similar results for other p(n,x).) ... Examples: u v a b c d e f seq h.....seq k 1 1 1 2 0 1 1 0 -A121646..A059929 1 1 1 3 0 1 1 0 A128533...A081714 1 1 2 1 0 1 1 0 A081714...A001906 1 1 1 1 1 1 1 0 A000045...A001906 1 1 2 1 1 1 1 0 A129905...A192879 1 1 1 2 1 1 1 0 A061646...A079472 1 1 1 1 0 1 1 1 A192872...A192873 1 1 1 1 1 2 1 1 A192874...A192875 1 1 1 1 1 2 1 1 A192876...A192877 1 1 1 1 1 1 2 1 A192880...A192882 1 1 1 1 1 1 1 1 A166536...A064831 The terms of several of these sequences are products of Fibonacci numbers (A000045), or Fibonacci numbers and Lucas numbers (A000032). LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (3,0,-3,1). FORMULA a(n) = 3*a(n-1) - 3*a(n-3) + a(n-4). G.f.: (2*x-1)*(x^2-x+1) / ( (x-1)*(1+x)*(x^2-3*x+1) ). - R. J. Mathar, Oct 26 2011 EXAMPLE The coefficients in all the polynomials p(n,x) are Fibonacci numbers (A000045). The first six and their reductions: p(0,x) = 1 -> 1 p(1,x) = x -> x p(2,x) = 1 + 2*x^2 -> 3 + 2*x p(3,x) = 1 + x + 3*x^3 -> 4 + 7*x p(4,x) = 1 + x + 2*x^2 + 5*x^4 -> 13 + 18*x p(5,x) = 1 + x + 2*x^2 + 3*x^3 + 8*x^5 -> 30 + 49*x MATHEMATICA (* First program *) q = x^2; s = x + 1; z = 26; p[0, x_] := 1; p[1, x_] := x; p[n_, x_] := p[n - 1, x]*x + p[n - 2, x]*x^2 + 1; 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}] (* A192872 *) u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192873 *) (* End of 1st program *) (* ******************************************** *) (* Second program: much more general *) (* u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 1; Nine degrees of freedom for user; shown values generate A192872. *) q = x^2; s = u*x + v; z = 11; (* will apply reduction (x^2 -> u*x+v) to p(n, x) *) p[0, x_] := a; p[1, x_] := b*x + c; (* initial values of polynomial sequence p(n, x) *) p[n_, x_] := d*x*p[n - 1, x] + e*(x^2)*p[n - 2, x] + f; (* recurrence for p(n, 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}]; u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}]; Simplify[FindLinearRecurrence[u1]] (* for 0-sequence *) Simplify[FindLinearRecurrence[u2]] (* for 1-sequence *) u1 = Table[Coefficient[Part[t, n], x, 0], {n, 1, 4}] (* initial values for 0-sequence *) u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, 4}] (* initial values for 1-sequence *) LinearRecurrence[{3, 0, -3, 1}, {1, 0, 3, 4}, 26] (* Ray Chandler, Aug 02 2015 *) PROG (PARI) my(x='x+O('x^30)); Vec((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))) \\ G. C. Greubel, Jan 06 2019 (Magma) m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1)) )); // G. C. Greubel, Jan 06 2019 (Sage) ((2*x-1)*(x^2-x+1)/((x-1)*(1+x)*(x^2-3*x +1))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jan 06 2019 (GAP) a:=[1, 0, 3, 4];; for n in [5..30] do a[n]:=3*a[n-1]-3*a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 06 2019 CROSSREFS Cf. A192232, A192744, A192873, A192908 (sums of adjacent terms). Sequence in context: A151521 A142860 A111954 * A036672 A174684 A286917 Adjacent sequences: A192869 A192870 A192871 * A192873 A192874 A192875 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jul 11 2011 STATUS approved

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Last modified July 15 16:36 EDT 2024. Contains 374333 sequences. (Running on oeis4.)