A129204 The denominator of 2/n^3.

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

Author

Paul Barry, Apr 03 2007

Keywords

• nonn
• easy
• mult

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

• Index entries for linear recurrences with constant coefficients, signature (0,4,0,-6,0,4,0,-1).

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