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.

A129203 a(n) = numerator(3/(n+1)^3)*(3/2 + (-1)^n/2).

Original entry on oeis.org

6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1
Offset: 0

Views

Author

Paul Barry, Apr 03 2007

Keywords

Comments

(1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*(n+1)*t)*((t-Pi)/i)^3 = (A129202(n)*Pi^2 - a(n))/A129196(n), i=sqrt(-1).
Periodic with period 6. - Alois P. Heinz, Oct 03 2012

Links

Crossrefs

Cf. sequences listed in Comments section of A283393.

Programs

  • GAP
    List(List([0..10],n->3/(n+1)^3*(3/2+(-1)^n/2)),NumeratorRat); # Muniru A Asiru, Jul 01 2018
  • Magma
    [Numerator(3/(n+1)^3)*(3/2 + (-1)^n/2): n in [1..100]]; // Vincenzo Librandi, Jul 01 2018
  • Magma
    &cat [[6, 3, 2, 3, 6, 1]^^20]; // Vincenzo Librandi, Jul 01 2018
  • Maple
    a:= n-> [6, 3, 2, 3, 6, 1][irem(n, 6)+1]:
seq(a(n), n=0..119);  # Alois P. Heinz, Oct 03 2012
  • Mathematica
    Array[Numerator[3/(# + 1)^3] (3/2 + (-1)^#/2) &, 105, 0] (* or *)
PadRight[{}, 105, {6, 3, 2, 3, 6, 1}] (* Michael De Vlieger, Jun 30 2018 *)
CoefficientList[ Series[-(x^5 + 6x^4 + 3x^3 + 2x^2 + 3x + 6)/(x^6 - 1), {x, 0,
   105}], x] (* or *)
LinearRecurrence[{0, 0, 0, 0, 0, 1}, {6, 3, 2, 3, 6, 1}, 105] (* Robert G. Wilson v, Jul 28 2018 *)
  • PARI
    a(n)=[6, 3, 2, 3, 6, 1][n%6+1] \\ Charles R Greathouse IV, Oct 28 2014
  • Sage
    def Gauss_factorial(N, n): return mul(j for j in (1..N) if gcd(j, n) == 1)
def A129203(n): return Gauss_factorial(3, n+1)
[A129203(j) for j in (0..71)] # Peter Luschny, Oct 01 2012

Formula

G.f.: (6 + 3*x + 2*x^2 + 3*x^3 + 6*x^4 + x^5)/(1 - x^6).
a(n) = cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3) + cos(Pi*n/3)/3 - sqrt(3)*sin(Pi*n/3)/3 + 7*cos(Pi*n)/6 + 7/2.
a(n) = numerator(6/(n+1)^2). - Paul Barry, Apr 03 2007
a(n) = denominator of coefficient of x^6 in the Maclaurin expansion of -exp(-(n+1)*x^2). - Francesco Daddi, Aug 04 2011
a(n) = 3_(n+1)! = Gauss_factorial(3, n+1) = Product_{1<=j<=3, gcd(j,n+1)=1} j. - Peter Luschny, Oct 01 2012
a(n) = denominator((n+1)/6). - Jon Hearn, Nov 10 2013
a(n) = denominator of 2*(1/(12*n)+1)*n^n; related to gamma function approximation for positive integers less the factor sqrt(Pi/2)/(exp(n)*sqrt(n)). - Thomas Blankenhorn, Jun 21 2018
a(n) = 6/gcd(n+1,6). - Ridouane Oudra, Jul 29 2022

A129197 a(n) = numerator( 3*(3+(-1)^n)/(n+1)^3 ).

Original entry on oeis.org

12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1, 12, 3, 4, 3, 12, 1
Offset: 0

Views

Author

Paul Barry, Apr 02 2007

Keywords

Comments

Denominator of 3*(3+(-1)^n)/(n+1)^3 is A129196.

Links

Crossrefs

Cf. sequences listed in Comments section of A283393.

Programs

  • Mathematica
    Table[3 (3+(-1)^n)/(n+1)^3,{n,0,80}]//Numerator (* Harvey P. Dale, Sep 26 2020 *)

Formula

G.f.: (12 + 3*x + 4*x^2 + 3*x^3 + 12*x^4 + x^5)/(1 - x^6).
Period 6: repeat [12, 3, 4, 3, 12, 1].

A129202 Denominator of 3*(3+(-1)^n) / (n+1)^2.

Original entry on oeis.org

1, 2, 3, 8, 25, 6, 49, 32, 27, 50, 121, 24, 169, 98, 75, 128, 289, 54, 361, 200, 147, 242, 529, 96, 625, 338, 243, 392, 841, 150, 961, 512, 363, 578, 1225, 216, 1369, 722, 507, 800, 1681, 294, 1849, 968, 675, 1058, 2209, 384, 2401, 1250, 867, 1352, 2809, 486
Offset: 0

Views

Author

Paul Barry, Apr 03 2007

Keywords

Comments

A divisibility sequence, that is, if n divides m then a(n) divides a(m). - Peter Bala, Feb 27 2019

Links

Crossrefs

Cf. A026741, A051176, A129196, A129197 (numerators), A060789.

Programs

  • Magma
    [Denominator(3*(3+(-1)^n)/(n+1)^2): n in [0..50]]; // G. C. Greubel, Oct 26 2017
  • Maple
    A129202:=n->numer((n+1)/2)*numer((n+1)/3): seq(A129202(n), n=0..100); # Wesley Ivan Hurt, Jul 18 2014
  • Mathematica
    Table[Numerator[(n + 1)/2] Numerator[(n + 1)/3], {n, 0, 100}] (* Wesley Ivan Hurt, Jul 18 2014 *)
LinearRecurrence[{0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, -3, 0, 0, 0, 0, 0, 1}, {1, 2, 3, 8, 25, 6, 49, 32, 27, 50, 121, 24, 169, 98, 75, 128, 289, 54}, 60] (* Harvey P. Dale, Nov 20 2016 *)
  • PARI
    for(n=0,50, print1(denominator(3*(3+(-1)^n)/(n+1)^2), ", ")) \\ G. C. Greubel, Oct 26 2017

Formula

a(n) = A129196(n)/(n+1).
(1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*(n+1)*t)*((t-Pi)/i)^3 dt = (a(n)*Pi^2-A129203(n))/A129196(n), i=sqrt(-1).
a(n) = ( Numerator of (n+1)/2 ) * ( Numerator of (n+1)/3 ) = A026741(n+1) * A051176(n+1). - Wesley Ivan Hurt, Jul 18 2014
G.f.: -(x^16 +2*x^15 +3*x^14 +8*x^13 +25*x^12 +6*x^11 +46*x^10 +26*x^9 +18*x^8 +26*x^7 +46*x^6 +6*x^5 +25*x^4 +8*x^3 +3*x^2 +2*x +1) / ((x -1)^3*(x +1)^3*(x^2 -x +1)^3*(x^2 +x +1)^3). - Colin Barker, Jul 18 2014
a(n+18) = 3*a(n+12)-3*a(n+6)+a(n). - Robert Israel, Jul 18 2014
a(n) = 2*(n+1)^2 * (7-4*cos(2*Pi*(n+1)/3)) / (9*(3-(-1)^n)). - Vaclav Kotesovec, Jul 20 2014
From Peter Bala, Feb 27 2019: (Start)
The following remarks assume an offset of 1.
a(n) = n^2/gcd(n,6) = n*A060789(n).
a(n) = n^2/b(n), where b(n) is the purely periodic sequence [1,2,3,2,1,6,...] with period 6. Thus a(n) is a quasi-polynomial in n:
a(6*n+1) = (6*n + 1)^2;
a(6*n+2) = 2*(3*n + 1)^2;
a(6*n+3) = 3*(2*n + 1)^2;
a(6*n+4) = 2*(3*n + 2)^2;
a(6*n+5) = (6*n + 5)^2;
a(6*n) = 6*n^2.
O.g.f.: F(x) - 2*F(x^2) - 6*F(x^3) + 12*F(x^6), where F(x) = x*(1 + x)/(1 - x)^3 is the generating function for the squares. (End)
Sum_{n>=0} 1/a(n) = 55*Pi^2/216. - Amiram Eldar, Sep 27 2022

Extensions

More terms from Wesley Ivan Hurt, Jul 18 2014

A129204 The denominator of 2/n^3.

Original entry on oeis.org

1, 1, 4, 27, 32, 125, 108, 343, 256, 729, 500, 1331, 864, 2197, 1372, 3375, 2048, 4913, 2916, 6859, 4000, 9261, 5324, 12167, 6912, 15625, 8788, 19683, 10976, 24389, 13500, 29791, 16384, 35937, 19652, 42875, 23328, 50653, 27436, 59319, 32000
Offset: 0

Views

Author

Paul Barry, Apr 03 2007

Keywords

Comments

Take two consecutive triangular numbers t1 and t2 and create a triangle using (0,0), (t1,t2) and (t2,t1). The area of this triangle will be ((n+1)^3)/2 for t1 = n*(n+1)/2. - J. M. Bergot, May 08 2012
Multiplicative because both A000578 and A040001 are. - Andrew Howroyd, Jul 26 2018

Links

Crossrefs

Cf. A000578, A040001, A129194, A309337.

Programs

  • Magma
    [1] cat [Denominator(2/n^3): n in [1..40]]; // Vincenzo Librandi, Jul 26 2018
  • Mathematica
    Join[{1}, Table[Denominator[2 / n^3], {n, 100}]] (* Vincenzo Librandi, Jul 26 2018 *)
  • PARI
    a(n) = if(n < 1, n==0, lcm(2, n^3)/2) \\ Andrew Howroyd, Jul 25 2018

Formula

G.f.: (1+x+23x^2+22x^4+23x^5+x^7+x^8)/(1-x^2)^4.
a(n) = 0^n + n^3*(3/4 - (-1)^n/4).
a(n+1) = A129196(n)*(5/3 + (4/3)*cos(2*Pi*(n+1)/3)).
a(2n) = 4n^3, a(2n+1) = (2n+1)^3.
a(n) = A000578(n) / A040001(n). - Andrew Howroyd, Jul 25 2018
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + 9*zeta(3)/8.
Sum_{n>=0} (-1)^n/a(n) = 1 - 5*zeta(3)/8. (End)
From Peter Bala, Jan 21 2024: (Start)
For n >= 1, a(n) = n*A129194(n) = n*Sum_{d divides n} (-1)^(d+1)*J(2,n/d), where the Jordan totient function J_2(n) = A007434(n). Cf. A309337.
Dirichlet g.f. for sequence without the a(0) term: (1 - 4/2^s)*zeta(s-3). (End)

Extensions

More terms from Vincenzo Librandi, Jul 26 2018
Showing 1-4 of 4 results.

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