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A129204 The denominator of 2/n^3.

%I A129204 #43 Aug 16 2025 06:20:09
%S A129204 1,1,4,27,32,125,108,343,256,729,500,1331,864,2197,1372,3375,2048,
%T A129204 4913,2916,6859,4000,9261,5324,12167,6912,15625,8788,19683,10976,
%U A129204 24389,13500,29791,16384,35937,19652,42875,23328,50653,27436,59319,32000
%N A129204 The denominator of 2/n^3.
%C A129204 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
%C A129204 Multiplicative because both A000578 and A040001 are. - _Andrew Howroyd_, Jul 26 2018
%H A129204 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,4,0,-6,0,4,0,-1).
%F A129204 G.f.: (1+x+23x^2+22x^4+23x^5+x^7+x^8)/(1-x^2)^4.
%F A129204 a(n) = 0^n + n^3*(3/4 - (-1)^n/4).
%F A129204 a(n+1) = A129196(n)*(5/3 + (4/3)*cos(2*Pi*(n+1)/3)).
%F A129204 a(2n) = 4n^3, a(2n+1) = (2n+1)^3.
%F A129204 a(n) = A000578(n) / A040001(n). - _Andrew Howroyd_, Jul 25 2018
%F A129204 From _Amiram Eldar_, Aug 25 2022: (Start)
%F A129204 Sum_{n>=0} 1/a(n) = 1 + 9*zeta(3)/8.
%F A129204 Sum_{n>=0} (-1)^n/a(n) = 1 - 5*zeta(3)/8. (End)
%F A129204 From _Peter Bala_, Jan 21 2024: (Start)
%F A129204 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.
%F A129204 Dirichlet g.f. for sequence without the a(0) term: (1 - 4/2^s)*zeta(s-3). (End)
%t A129204 Join[{1}, Table[Denominator[2 / n^3], {n, 100}]] (* _Vincenzo Librandi_, Jul 26 2018 *)
%o A129204 (PARI) a(n) = if(n < 1, n==0, lcm(2, n^3)/2) \\ _Andrew Howroyd_, Jul 25 2018
%o A129204 (Magma) [1] cat [Denominator(2/n^3): n in [1..40]]; // _Vincenzo Librandi_, Jul 26 2018
%Y A129204 Cf. A000578, A040001, A129194, A309337.
%K A129204 nonn,easy,mult
%O A129204 0,3
%A A129204 _Paul Barry_, Apr 03 2007
%E A129204 More terms from _Vincenzo Librandi_, Jul 26 2018