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 1-4 of 4 results.

A292612 a(n) = F(n)^2 + 4*(-1)^n = F(n+3)*F(n-3), where F = A000045.

Original entry on oeis.org

4, -3, 5, 0, 13, 21, 68, 165, 445, 1152, 3029, 7917, 20740, 54285, 142133, 372096, 974173, 2550405, 6677060, 17480757, 45765229, 119814912, 313679525, 821223645, 2149991428, 5628750621, 14736260453, 38580030720, 101003831725, 264431464437, 692290561604, 1812440220357
Offset: 0

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Author

Bruno Berselli, Sep 20 2017

Keywords

Comments

This is the case k=3 of the identity F(n)^2 - F(k)^2*(-1)^(n+k) = F(n+k)*F(n-k), known also as Catalan's identity.

Links

Crossrefs

Cf. A000045, A001622, A005248: Lucas(2*n), A001654: F(n)*F(n+1).
Cf. A007598 (k=0), A059929 (k=1, without initial 1), A192883 (k=2, without initial -1), this sequence (k=3).

Programs

  • GAP
    List([0..10^2],n ->Fibonacci(n)^2+4*(-1)^n); # Muniru A Asiru, Sep 26 2017
  • Magma
    [Fibonacci(n)^2+4*(-1)^n: n in [0..40]];
  • Maple
    with(combinat,fibonacci):  A292612:=seq(fibonacci(n)^2+4*(-1)^n, n=0..10^2); # Muniru A Asiru, Sep 26 2017
  • Mathematica
    Table[Fibonacci[n]^2 + 4 (-1)^n, {n, 0, 40}]
  • PARI
    for(n=0, 40, print1(fibonacci(n)^2+4*(-1)^n", "));
  • PARI
    Vec((4-11*x+3*x^2)/((1+x)*(1-3*x+x^2))+O(x^30)) \\ Colin Barker, Sep 20 2017
  • Sage
    [fibonacci(n)^2+4*(-1)^n for n in range(40)]

Formula

G.f.: (4 - 11*x + 3*x^2)/((1 + x)*(1 - 3*x + x^2)).
a(n) = a(-n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = 4*A001654(n+1) - 11*A001654(n) + 3*A001654(n-1) with A001654(-1)=0.
5*a(n) = Lucas(2*n) + 18*(-1)^n. Note that Lucas(2*n) + r*(-1)^n is divisible by 5 for r = -2, 3, -7, 8, -12, 13, -17, 18, -22, 23, -27, ... = (-1/4)*(3 + 5*(2*m+1)*(-1)^m) = (-1)^m*A047221(m). On the other hand, a(n) is divisible by 5 when n is a member of A047221.
a(n) = (1/5)*(18*(-1)^n + ((3-sqrt(5))/2)^n + ((3+sqrt(5))/2)^n). - Colin Barker, Sep 20 2017
Sum_{n>=4} 1/a(n) = 143/960. - Amiram Eldar, Oct 05 2020
Sum_{n>=4} (-1)^n/a(n) = 3/(4*phi) - 407/960, where phi is the golden ratio (A001622). - Amiram Eldar, Oct 06 2020

A192919 Constant term in the reduction by (x^2 -> x+1) of the polynomial F(n+4)*x^n, where F=A000045 (Fibonacci sequence).

Original entry on oeis.org

3, 0, 8, 13, 42, 102, 275, 712, 1872, 4893, 12818, 33550, 87843, 229968, 602072, 1576237, 4126650, 10803702, 28284467, 74049688, 193864608, 507544125, 1328767778, 3478759198, 9107509827, 23843770272, 62423801000, 163427632717, 427859097162, 1120149658758
Offset: 0

Views

Author

Clark Kimberling, Jul 12 2011

Keywords

Comments

See A192872.

Links

Crossrefs

Cf. A000045, A001622, A192232, A192744, A192872, A192883, A192920, A290565.

Programs

  • GAP
    F:=Fibonacci;; List([0..30], n-> F(n-1)*F(n+4)); # G. C. Greubel, Jul 28 2019
  • Magma
    F:=Fibonacci; [F(n-1)*F(n+4): n in [0..30]]; // G. C. Greubel, Jul 28 2019
  • Maple
    with(combinat):seq(fibonacci(n-1)*fibonacci(n+4), n=0..27);
  • Mathematica
    (* First program *)
q = x^2; s = x + 1; z = 28;
p[0, x_]:= 3; p[1, x_]:= 5 x;
p[n_, x_]:= p[n-1, x]*x + p[n-2, x]*x^2;
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}] (* A192919 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192920 *)
(* Second program *)
With[{F=Fibonacci}, Table[F[n-1]*F[n+4], {n,0,30}]] (* G. C. Greubel, Jul 28 2019 *)
  • PARI
    a(n) = round((2^(-n)*(11*(-2)^n-(3-sqrt(5))^n*(-2+sqrt(5))+(2+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Oct 01 2016
  • PARI
    Vec((3+2*x^2-6*x)/((1+x)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Oct 01 2016
  • PARI
    vector(30, n, n--; f=fibonacci; f(n-1)*f(n+4)) \\ G. C. Greubel, Jul 28 2019
  • Sage
    f=fibonacci; [f(n-1)*f(n+4) for n in (0..30)] # G. C. Greubel, Jul 28 2019

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = Fibonacci(n-1)*Fibonacci(n+4). - Gary Detlefs, Oct 19 2011
G.f.: (3 -6*x +2*x^2)/((1+x)*(1-3*x+x^2)). - R. J. Mathar, May 08 2014
a(n) + a(n+1) = A001906(n+1). - R. J. Mathar, May 08 2014
a(n) = (2^(-n)*(11*(-2)^n-(3-sqrt(5))^n*(-2+sqrt(5))+(2+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Oct 01 2016
From Amiram Eldar, Oct 06 2020: (Start)
Sum_{n>=2} 1/a(n) = (1/5) * A290565 - 17/150.
Sum_{n>=2} (-1)^n/a(n) = 1/phi - 83/150, where phi is the golden ratio (A001622). (End)

A192920 Coefficient of x in the reduction by (x^2 -> x+1) of the polynomial F(n+4)*x^n, where F=A000045 (Fibonacci sequence).

Original entry on oeis.org

0, 5, 8, 26, 63, 170, 440, 1157, 3024, 7922, 20735, 54290, 142128, 372101, 974168, 2550410, 6677055, 17480762, 45765224, 119814917, 313679520, 821223650, 2149991423, 5628750626, 14736260448, 38580030725, 101003831720, 264431464442
Offset: 0

Views

Author

Clark Kimberling, Jul 12 2011

Keywords

Comments

See A192872.

Links

Crossrefs

Cf. A000045, A192232, A192744, A192872, A192883, A192919.

Programs

  • GAP
    a:=[0,5,8];; for n in [4..30] do a[n]:=2*a[n-1]+2*a[n-2]-a[n-3]; od; a; # G. C. Greubel, Feb 06 2019
  • Magma
    [Fibonacci(n+2)^2 -(-1)^n: n in [0..30]]; // G. C. Greubel, Feb 06 2019, modified Jul 28 2019
  • Mathematica
    (* First program *)
q = x^2; s = x + 1; z = 28;
p[0, x_]:= 3; p[1, x_]:= 5 x;
p[n_, x_]:= p[n-1, x]*x + p[n-2, x]*x^2;
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}] (* A192919 *)
u2 = Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A192920 *)
(* Second program *)
LinearRecurrence[{2,2,-1}, {0,5,8}, 30] (* G. C. Greubel, Feb 06 2019 *)
  • PARI
    vector(30, n, n--; fibonacci(n+2)^2 -(-1)^n) \\ G. C. Greubel, Feb 06 2019, modified Jul 28 2019
  • Sage
    [fibonacci(n+2)^2 -(-1)^n for n in (0..30)] # G. C. Greubel, Feb 06 2019, modified Jul 28 2019

Formula

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
a(n) = A192883(n+1).
G.f.: x*(5-2*x)/((1+x)*(1-3*x+x^2)). - R. J. Mathar, Aug 01 2011
a(n) = (A005248(n+2) - 7*(-1)^n)/5. - R. J. Mathar, Aug 01 2011
a(n) = Fibonacci(n+2)^2 - (-1)^n. - G. C. Greubel, Feb 06 2019
Sum_{n>=1} 1/a(n) = 7/18. - Amiram Eldar, Oct 05 2020

A295687 a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 2, a(3) = 1.

Original entry on oeis.org

1, 2, 2, 1, 4, 8, 11, 16, 28, 47, 74, 118, 193, 314, 506, 817, 1324, 2144, 3467, 5608, 9076, 14687, 23762, 38446, 62209, 100658, 162866, 263521, 426388, 689912, 1116299, 1806208, 2922508, 4728719, 7651226, 12379942, 20031169, 32411114, 52442282, 84853393
Offset: 0

Views

Author

Clark Kimberling, Nov 29 2017

Keywords

Comments

a(n)/a(n-1) -> (1 + sqrt(5))/2 = golden ratio (A001622), so that a( ) has the growth rate of the Fibonacci numbers (A000045).

Links

Crossrefs

Cf. A000045, A001622, A001654, A002878, A005248, A192883, A295690.

Programs

  • Mathematica
    LinearRecurrence[{1, 0, 1, 1}, {1, 2, 2, 1}, 100]

Formula

a(n) = a(n-1) + a(n-3) + a(n-4), where a(0) = 1, a(1) = 2, a(2) = 2, a(3) = 1.
G.f.: (-1 - x + 2*x^3)/(-1 + x + x^3 + x^4).
From Peter Bala, Nov 27 2021: (Start)
a(2*n) = 3*a(2*n-2) - a(2*n-4) - (-1)^n, for n >= 2;
a(2*n+1) = 3*a(2*n-1) - a(2*n-3) + 7*(-1)^n, for n >= 2.
a(2*n) = Lucas(2*n-1) - Fibonacci(n-3)*Fibonacci(n-2) = A002878(n-1) - A001654(n-3);
a(2*n+1) = Lucas(2*n) - Fibonacci(n-4)*Fibonacci(n) = A005248(n) - A192883(n-3).
a(4*n-1) = Fibonacci(2*n+1)^2 - Fibonacci(2*n)^2 + Fibonacci(2*n-1)^2 - Fibonacci(2*n-2)^2 - 3, for n >= 1;
a(4*n+1) = Fibonacci(2*n+2)^2 - Fibonacci(2*n+1)^2 + Fibonacci(2*n)^2 - Fibonacci(2*n-1)^2 + 3, for n >= 0.
Conjecture: a(2*n+7) = Fibonacci(n)^3*Sum_{k >= 1} k^2 * Fibonacci(n*k)/ Fibonacci(n+2)^(k+1), n >= 1. (End)
Showing 1-4 of 4 results.

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