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 A192904 Constant term in the reduction by (x^2 -> x + 1) of the polynomial p(n,x) defined below at Comments. 7
 1, 0, 1, 5, 16, 49, 153, 480, 1505, 4717, 14784, 46337, 145233, 455200, 1426721, 4471733, 14015632, 43928817, 137684905, 431542080, 1352570689, 4239325789, 13287204352, 41645725825, 130529073953, 409113752000, 1282274186177 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 COMMENTS The titular polynomial is defined by p(n,x) = (x^2)*p(n-1,x) + x*p(n-2,x), with p(0,x) = 1, p(1,x) = x.  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^2)*p(n-1,x) + e*x*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 linear recurrence sequences, formally of order 5.  The Mathematica program below, with first line deleted, shows initial terms and recurrence coefficients, which imply these properties: (1)  the recurrence coefficients depend only on u,v,d,e; the parameters a,b,c,f affect only the initial terms. (2)  if e=0 or v=0, the order of recurrence is <= 3; (3)  if e=0 and v=0, the recurrence coefficients are 1+d*u^2 and -d*u^2 (cf. similar results at A192872). ... Examples: u v a b c d e f... seq h.....seq k 1 1 1 1 1 1 0 0... A001906..A001519 1 1 1 1 0 0 1 0... A103609..A193609 1 1 1 1 0 1 1 0... A192904..A192905 1 1 1 1 1 1 0 0... A001519..A001906 1 1 1 1 1 1 1 0... A192907..A192907 1 1 1 1 1 1 0 1... A192908..A069403 1 1 1 1 1 1 1 1... A192909..A192910 The terms of these sequences involve Fibonacci numbers, F(n)=A000045(n); e.g., A001906: even-indexed F(n) A001519: odd-indexed F(n) A103609: (1,1,1,1,2,2,3,3,5,5,8,8,...) LINKS G. C. Greubel, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (3,0,1,1). FORMULA a(n) = 3*a(n-1) + a(n-3) + a(n-4). G.f.: (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4). - Colin Barker, Aug 31 2012 EXAMPLE The first six polynomials and reductions: 1 -> 1 x -> x x + x^3 -> 1 + 3*x x^2 + x^3 + x^5 -> 5 + 8*x x^2 + 2*x^4 + x^5 + x^7 -> 16 + 25*x x^3 + 2*x^4 + 3*x^6 + x^7 + x^9 -> 49 + 79*x, so that A192904 = (1,0,1,5,16,49,...) and A192905 = (0,1,3,8,25,79,...) MATHEMATICA (* To obtain general results, delete the next line. *) u = 1; v = 1; a = 1; b = 1; c = 0; d = 1; e = 1; f = 0; q = x^2; s = u*x + v; z = 24; p[0, x_] := a; p[1, x_] := b*x + c; p[n_, x_] :=  d*(x^2)*p[n - 1, x] + e*x*p[n - 2, x] + f; Table[Expand[p[n, x]], {n, 0, 8}] 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}]; u0 = Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192904 *) u1 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192905 *) Simplify[FindLinearRecurrence[u0]] (* recurrence for 0-sequence *) Simplify[FindLinearRecurrence[u1]] (* recurrence for 1-sequence *) LinearRecurrence[{3, 0, 1, 1}, {1, 0, 1, 5}, 40] (* G. C. Greubel, Jan 10 2019 *) PROG (PARI) my(x='x+O('x^40)); Vec((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)) \\ G. C. Greubel, Jan 10 2019 (MAGMA) m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4) )); // G. C. Greubel, Jan 10 2019 (Sage) ((1-x)*(1-2*x-x^2)/(1-3*x-x^3-x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Jan 10 2019 (GAP) a:=[1, 0, 1, 5];; for n in [5..40] do a[n]:=3*a[n-1]+a[n-3]+a[n-4]; od; a; # G. C. Greubel, Jan 10 2019 CROSSREFS Cf. A192232, A192744, A192905, A192872. Sequence in context: A244410 A052909 A037536 * A082001 A084356 A007806 Adjacent sequences:  A192901 A192902 A192903 * A192905 A192906 A192907 KEYWORD nonn,easy AUTHOR Clark Kimberling, Jul 12 2011 STATUS approved

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Last modified March 29 15:03 EDT 2020. Contains 333107 sequences. (Running on oeis4.)