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 1-3 of 3 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

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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.

Crossrefs

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

A014739 Expansion of (1+x^2)/(1-2*x+x^3).

Original entry on oeis.org

1, 2, 5, 9, 16, 27, 45, 74, 121, 197, 320, 519, 841, 1362, 2205, 3569, 5776, 9347, 15125, 24474, 39601, 64077, 103680, 167759, 271441, 439202, 710645, 1149849, 1860496, 3010347, 4870845, 7881194, 12752041, 20633237, 33385280, 54018519, 87403801
Offset: 0

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Author

Keywords

Comments

Number of wedged n-spheres in the homotopy type of the Boolean complex of the affine Coxeter group A~ n. - _Bridget Tenner, Jun 04 2008
In an infinite set of sequences such that a(n) = a(n-1) + a(n-2) + k; with a(0) = 1, a(1) = 2, and in A014739, k = 2. Cf. A171516 for a(0) = 1, a(1) = 2, k = 3. - Gary W. Adamson, Dec 10 2009

Examples

			The Boolean complex of the affine Coxeter group \widetilde{A}_3 is homotopy equivalent to the wedge of 5 3-spheres.
		

Crossrefs

Programs

  • GAP
    List([0..40], n-> Lucas(1,-1,n+2)[2] -2); # G. C. Greubel, Jul 22 2019
  • Magma
    [Lucas(n+2)-2: n in [0..40]]; // G. C. Greubel, Jul 22 2019
    
  • Maple
    with(combinat): seq(fibonacci(n)+fibonacci(n+2)-2, n=1..37); # Zerinvary Lajos, Jan 31 2008
    g:=(1+z^2)/(1-z-z^2): gser:=series(g, z=0, 43): seq(coeff(gser, z, n)-2, n=2..38); # Zerinvary Lajos, Jan 09 2009
  • Mathematica
    CoefficientList[Series[(1+x^2)/(1-2*x+x^3), {x,0,40}], x] (* Robert G. Wilson v, Feb 25 2005 *)
    a[0]=1; a[1]=2; a[2]=5; a[n_]:= a[n] = 2a[n-1]-a[n-3]; Array[a, 40, 0]
    LinearRecurrence[{2,0,-1},{1,2,5},40] (* Harvey P. Dale, Jun 26 2011 *)
    LucasL[Range[0,40]+2]-2 (* G. C. Greubel, Jul 22 2019 *)
  • PARI
    Vec((1+x^2)/(1-2*x+x^3)+O(x^40)) \\ Charles R Greathouse IV, Sep 23 2012
    
  • PARI
    vector(40, n, n--; f=fibonacci; f(n+3)+f(n+1)-2) \\ G. C. Greubel, Jul 22 2019
    
  • Sage
    [lucas_number2(n+2,1,-1)-2 for n in (0..40)] # G. C. Greubel, Jul 22 2019
    

Formula

Partial sums of Lucas numbers A000032 less 1.
From Paul Barry, Sep 03 2003: (Start)
G.f.: (1+x^2)/((1-x)*(1-x-x^2)).
a(n) = ((3+sqrt(5))((1+sqrt(5))/2)^n+(3-sqrt(5))((1-sqrt(5))/2)^n)/2-2. (End)
From Zerinvary Lajos, Jan 31 2008: (Start)
a(n) = A001610(n+1)-1.
a(n) = F(n+1) + F(n+3) - 2 = A000071(n+1) + A000071(n+3), where F(n) is the n-th Fibonacci number. [corrected by R. J. Mathar, Mar 14 2011] (End)
a(n) = A000032(n+2) - 2. - Matthew Vandermast, Nov 05 2009
a(n) = 2*a(n-1) - a(n-3). - Vincenzo Librandi, Dec 31 2010

Extensions

More terms from Robert G. Wilson v, Feb 25 2005

A182415 a(0) = 1, a(1) = 2; for n>1, a(n) = a(n-1) + a(n-2) + 4.

Original entry on oeis.org

1, 2, 7, 13, 24, 41, 69, 114, 187, 305, 496, 805, 1305, 2114, 3423, 5541, 8968, 14513, 23485, 38002, 61491, 99497, 160992, 260493, 421489, 681986, 1103479, 1785469, 2888952, 4674425, 7563381, 12237810, 19801195, 32039009, 51840208, 83879221, 135719433, 219598658
Offset: 0

Views

Author

Alex Ratushnyak, Apr 28 2012

Keywords

Examples

			a(3) = 7 + 2 + 4 = 13.
		

Crossrefs

Cf. A000045, A014739, A022095 (first differences), A171516.

Programs

  • GAP
    F:=Fibonacci;; List([0..40], n-> F(n+3)+3*F(n+1)-4); # G. C. Greubel, Jul 22 2019
  • Magma
    F:=Fibonacci; [F(n+3)+3*F(n+1)-4: n in [0..40]]; // G. C. Greubel, Jul 22 2019
    
  • Mathematica
    With[{f=Fibonacci}, Table[F[n+3]+3*F[n+1]-4, {n,0,40}]] (* G. C. Greubel, Jul 22 2019 *)
    RecurrenceTable[{a[0]==1,a[1]==2,a[n]==a[n-1]+a[n-2]+4},a,{n,40}] (* or *) LinearRecurrence[{2,0,-1},{1,2,7},40] (* Harvey P. Dale, Nov 24 2020 *)
  • PARI
    vector(40, n, n--; f=fibonacci; f(n+3) +3*f(n+1) -4 ) \\ G. C. Greubel, Jul 22 2019
    
  • Sage
    f=fibonacci; [f(n+3)+3*f(n+1)-4 for n in (0..40)] # G. C. Greubel, Jul 22 2019
    

Formula

From Colin Barker, May 07 2012: (Start)
a(n) = 2*a(n-1) - a(n-3).
G.f.: (1+3*x^2)/((1-x)*(1-x-x^2)). (End)
a(n) = Fibonacci(n+3) + 3*Fibonacci(n+1) - 4. - G. C. Greubel, Jul 22 2019
Showing 1-3 of 3 results.