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.

A162990 Triangle of polynomial coefficients related to 3F2([1,n+1,n+1],[n+2,n+2],z).

Original entry on oeis.org

4, 36, 9, 576, 144, 64, 14400, 3600, 1600, 900, 518400, 129600, 57600, 32400, 20736, 25401600, 6350400, 2822400, 1587600, 1016064, 705600, 1625702400, 406425600, 180633600, 101606400, 65028096, 45158400, 33177600, 131681894400
Offset: 1

Views

Author

Johannes W. Meijer, Jul 21 2009

Keywords

Comments

The hypergeometric function 3F2([1,n+1,n+1],[n+2,n+2],z) = (n+1)^2*Li2(z)/z^(n+1) - MN(z;n)/(n!^2*z^n) for n >= 1, with Li2(z) the dilogarithm. The polynomial coefficients of MN(z;n) lead to the triangle given above.
We observe that 3F2([1,1,1],[2,2],z) = Li2(z)/z and that 3F2([1,0,0],[1,1],z) = 1.
The generating function for the EG1[3,n] coefficients of the EG1 matrix, see A162005, is GFEG1(z;m=2) = 1/(1-z)*(3*zeta(3)/2-2*z*log(2)* 3F2([1,1,1],[2,2],z) + sum((2^(1-2*n)* factorial(2*n-1)*z^(n+1)*3F2([1,n+1,n+1],[n+2,n+2],z))/(factorial(n+1)^2), n=1..infinity)).
The zeros of the MN(z;n) polynomials for larger values of n get ever closer to the unit circle and resemble the full moon, hence we propose to call the MN(z;n) the moon polynomials.

Examples

			The first few rows of the triangle are:
  [4]
  [36, 9]
  [576, 144, 64]
  [14400, 3600, 1600, 900]
The first few MN(z;n) polynomials are:
  MN(z;n=1) = 4
  MN(z;n=2) = 36 + 9*z
  MN(z;n=3) = 576 + 144*z + 64*z^2
  MN(z;n=4) = 14400 + 3600*z + 1600*z^2 + 900*z^3
		

References

  • Lewin, L., Polylogarithms and Associated Functions. New York, North-Holland, 1981.

Crossrefs

A162995 is a scaled version of this triangle.
A001819(n)*(n+1)^2 equals the row sums for n>=1.
A162991 and A162992 equal the first and second right hand columns.
A001048, A052747, A052759, A052778, A052794 are related to the square root of the first five right hand columns.
A001044, A162993 and A162994 equal the first, second and third left hand columns.
A000142, A001710, A002301, A133799, A129923, A001715 are related to the square root of the first six left hand columns.
A027451(n+1) equals the denominators of M(z, n)/(n!)^2.
A129202(n)/A129203(n) = (n+1)^2*Li2(z=1)/(Pi^2) = (n+1)^2/6.
Cf. A002378 and A035287.

Programs

  • Maple
    a := proc(n, m): ((n+1)!/m)^2 end: seq(seq(a(n, m), m=1..n), n=1..7); # Johannes W. Meijer, revised Nov 29 2012
  • Mathematica
    Table[((n+1)!/m)^2, {n, 10}, {m, n}] (* Paolo Xausa, Mar 30 2024 *)

Formula

a(n,m) = ((n+1)!/m)^2 for n >= 1 and 1 <= m <= n.

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

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

A129196 a(n) = denominator(3*(3+(-1)^n)/(n+1)^3).

Original entry on oeis.org

1, 4, 9, 32, 125, 36, 343, 256, 243, 500, 1331, 288, 2197, 1372, 1125, 2048, 4913, 972, 6859, 4000, 3087, 5324, 12167, 2304, 15625, 8788, 6561, 10976, 24389, 4500, 29791, 16384, 11979, 19652, 42875, 7776, 50653, 27436, 19773, 32000, 68921, 12348, 79507, 42592
Offset: 0

Views

Author

Paul Barry, Apr 02 2007, Apr 03 2007

Keywords

Comments

Numerator of 3*(3+(-1)^n)/(n+1)^3 is A129197.
(1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*(n+1)*t)*((t-Pi)/i)^3 dt = (A129202(n)*Pi^2-A129203(n))/A129196(n) with i=sqrt(-1).

Crossrefs

Programs

  • Mathematica
    a[n_] := Denominator[3*(3 + (-1)^n)/(n + 1)^3]; Array[a, 50, 0] (* Amiram Eldar, Sep 11 2022 *)

Formula

a(n) = A129204(n+1)/(5/3+(4/3)*cos(2*Pi*(n+1)/3)).
a(n) = denominator((1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*(n+1)*t)*((t-Pi)/i)^3 dt) with i=sqrt(-1).
a(n) = denominator((Pi^2*(n+1)^2-6)/(n+1)^3).
a(n) = ((n+1)^3/(gcd(n+1,2)*gcd(n+1,3))). - Paul Barry, Oct 09 2007
a(n) = numerator of coefficient of x^6 in the Maclaurin expansion of -exp(-(n+1)*x^2). - Francesco Daddi, Aug 04 2011
Sum_{n>=0} 1/a(n) = 29*zeta(3)/24. - Amiram Eldar, Sep 11 2022
Showing 1-3 of 3 results.