cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A192951 Coefficient of x in the reduction by x^2 -> x+1 of the polynomial p(n,x) defined at Comments.

Original entry on oeis.org

0, 1, 3, 9, 20, 40, 74, 131, 225, 379, 630, 1038, 1700, 2773, 4511, 7325, 11880, 19252, 31182, 50487, 81725, 132271, 214058, 346394, 560520, 906985, 1467579, 2374641, 3842300, 6217024, 10059410, 16276523, 26336025, 42612643, 68948766
Offset: 0

Views

Author

Clark Kimberling, Jul 13 2011

Keywords

Comments

The titular polynomials are defined recursively: p(n,x) = x*p(n-1,x) + 3n - 1, with p(0,x)=1. For an introduction to reductions of polynomials by substitutions such as x^2 -> x+1, see A192232 and A192744.
...
The list of examples at A192744 is extended here; the recurrence is given by p(n,x) = x*p(n-1,x) + v(n), with p(0,x)=1, and the reduction of p(n,x) by x^2 -> x+1 is represented by u1 + u2*x:
...
If v(n)= n, then u1=A001595, u2=A104161.
If v(n)= n-1, then u1=A001610, u2=A066982.
If v(n)= 3n-1, then u1=A171516, u2=A192951.
If v(n)= 3n-2, then u1=A192746, u2=A192952.
If v(n)= 2n-1, then u1=A111314, u2=A192953.
If v(n)= n^2, then u1=A192954, u2=A192955.
If v(n)= -1+n^2, then u1=A192956, u2=A192957.
If v(n)= 1+n^2, then u1=A192953, u2=A192389.
If v(n)= -2+n^2, then u1=A192958, u2=A192959.
If v(n)= 2+n^2, then u1=A192960, u2=A192961.
If v(n)= n+n^2, then u1=A192962, u2=A192963.
If v(n)= -n+n^2, then u1=A192964, u2=A192965.
If v(n)= n(n+1)/2, then u1=A030119, u2=A192966.
If v(n)= n(n-1)/2, then u1=A192967, u2=A192968.
If v(n)= n(n+3)/2, then u1=A192969, u2=A192970.
If v(n)= 2n^2, then u1=A192971, u2=A192972.
If v(n)= 1+2n^2, then u1=A192973, u2=A192974.
If v(n)= -1+2n^2, then u1=A192975, u2=A192976.
If v(n)= 1+n+n^2, then u1=A027181, u2=A192978.
If v(n)= 1-n+n^2, then u1=A192979, u2=A192980.
If v(n)= (n+1)^2, then u1=A001891, u2=A053808.
If v(n)= (n-1)^2, then u1=A192981, u2=A192982.

Links

Crossrefs

Cf. A000045, A192232, A192744.

Programs

  • GAP
    F:=Fibonacci;; List([0..40], n-> F(n+4)+2*F(n+2)-(3*n+5)); # G. C. Greubel, Jul 12 2019
  • Magma
    I:=[0, 1, 3, 9]; [n le 4 select I[n] else 3*Self(n-1)-2*Self(n-2)-1*Self(n-3)+Self(n-4): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011
  • Magma
    F:=Fibonacci; [F(n+4)+2*F(n+2)-(3*n+5): n in [0..40]]; // G. C. Greubel, Jul 12 2019
  • Mathematica
    (* First program *)
q = x^2; s = x + 1; z = 40;
p[0, x]:= 1;
p[n_, x_]:= x*p[n-1, x] + 3n - 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}] (* A171516 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192951 *)
(* Additional programs *)
LinearRecurrence[{3,-2,-1,1},{0,1,3,9},40] (* Vincenzo Librandi, Nov 16 2011 *)
With[{F=Fibonacci}, Table[F[n+4]+2*F[n+2]-(3*n+5), {n,0,40}]] (* G. C. Greubel, Jul 12 2019 *)
  • PARI
    a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 1,-1,-2,3]^n*[0;1;3;9])[1,1] \\ Charles R Greathouse IV, Mar 22 2016
  • PARI
    vector(40, n, n--; f=fibonacci; f(n+4)+2*f(n+2)-(3*n+5)) \\ G. C. Greubel, Jul 12 2019
  • Sage
    f=fibonacci; [f(n+4)+2*f(n+2)-(3*n+5) for n in (0..40)] # G. C. Greubel, Jul 12 2019

Formula

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4).
From Bruno Berselli, Nov 16 2011: (Start)
G.f.: x*(1+2*x^2)/((1-x)^2*(1 - x - x^2)).
a(n) = ((25+13*t)*(1+t)^n + (25-13*t)*(1-t)^n)/(10*2^n) - 3*n - 5 = A000285(n+2) - 3*n - 5 where t=sqrt(5). (End)
a(n) = Fibonacci(n+4) + 2*Fibonacci(n+2) - (3*n+5). - G. C. Greubel, Jul 12 2019

A027181 a(n) = Lucas(n+4) - 2*(n+3).

Original entry on oeis.org

1, 3, 8, 17, 33, 60, 105, 179, 300, 497, 817, 1336, 2177, 3539, 5744, 9313, 15089, 24436, 39561, 64035, 103636, 167713, 271393, 439152, 710593, 1149795, 1860440, 3010289, 4870785, 7881132, 12751977, 20633171, 33385212, 54018449, 87403729, 141422248
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

Let F be a homogeneous polynomial in n + 4 variables f0, f1, f2, g0, g1, ..., gn, defined as the determinant of a Sylvester matrix of polynomials f2*x^2 + f1*x + f0 and gn*x^n + ... + g1*x + g0. It appears that a(n) is equal to the l1-norm of F, i.e., the sum of absolute values of coefficients of F. - Anton Mosunov, Apr 13 2019

Links

Crossrefs

Cf. A000032, A000045, A027170, A192978.

Programs

  • GAP
    List([0..40], n-> Lucas(1,-1,n+4)[2] -2*(n+3)); # G. C. Greubel, Jul 24 2019
  • Magma
    [Lucas(n+4) - (2*n+6): n in [0..40]]; // Vincenzo Librandi, Apr 16 2011
  • Mathematica
    LinearRecurrence[{3,-2,-1,1},{1,3,8,17},40] (* Vladimir Joseph Stephan Orlovsky, Jan 25 2012 *)
Table[LucasL[n+4]-2*(n+3), {n,0,40}] (* G. C. Greubel, Jul 24 2019 *)
  • PARI
    Vec((1 + x^2)/((1 - x)^2*(1 - x - x^2)) + O(x^40)) \\ Colin Barker, Mar 10 2017
  • PARI
    vector(40, n, n--; f=fibonacci; f(n+5)+f(n+3)-2*(n+3)) \\ G. C. Greubel, Jul 24 2019
  • Sage
    [lucas_number2(n+4,1,-1) -2*(n+3) for n in range(40)] # G. C. Greubel, Apr 14 2019

Formula

a(n) = Sum_{k=0..floor(n/2)} A027170(n-k, k).
G.f.: (1 + x^2)/((1 - x)^2*(1 - x - x^2)).
From Colin Barker, Mar 10 2017: (Start)
a(n) = -4 + (2^(-1-n)*((1-sqrt(5))^n*(-15+7*sqrt(5)) + (1+sqrt(5))^n*(15+7*sqrt(5))))/sqrt(5) - 2*(1+n).
a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>3.
(End)
Showing 1-2 of 2 results.

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