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Transform of n^3 by the Riordan array (1/(1-x^2), x).
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%I #64 Aug 25 2022 04:36:04

%S 0,1,8,28,72,153,288,496,800,1225,1800,2556,3528,4753,6272,8128,10368,

%T 13041,16200,19900,24200,29161,34848,41328,48672,56953,66248,76636,

%U 88200,101025,115200,130816,147968,166753,187272,209628,233928,260281,288800,319600

%N Transform of n^3 by the Riordan array (1/(1-x^2), x).

%C Recurrence a(n) = a(n-2) + n^3, starting with a(0)=0, a(1)=1. Also, in physics, a(n)/4 is the trace of the spin operator |S_z|^3 for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of the absolute values of their 3rd powers is 2*28/8 = a(3)/4. - _Stanislav Sykora_, Nov 07 2013

%C Also the number of 3-cycles in the (n+1)-triangular honeycomb queen graph. - _Eric W. Weisstein_, Jul 14 2017

%C With zero prepended and offset 1, the sequence starts 0,0,1,8,28,... for n=1,2,3,... Call this b(n). Consider the partitions of n into two parts (p,q). Then b(n) is the total volume of the family of cubes with side length |q - p|. - _Wesley Ivan Hurt_, Apr 14 2018

%H Stanislav Sykora, <a href="/A105636/b105636.txt">Table of n, a(n) for n = 0..9999</a>

%H Stanislav Sýkora, <a href="http://www.ebyte.it/stan/blog12to14.html#14Dec31">Magnetic Resonance on OEIS</a>, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GraphCycle.html">Graph Cycle</a>.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,0,5,-4,1).

%F G.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^5).

%F a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6).

%F a(n) = (2*n^4 + 8*n^3 + 8*n^2 - 1 + (-1)^n)/16.

%F a(n) = Sum_{k=0..floor((n-1)/2)} (n-2*k)^3.

%F a(n+1) = Sum_{k=0..n} k^3*(1 - (-1)^(n+k-1))/2.

%F a(n) = ((((x^2 - (x mod 2) - 4)/4)^2 - (((x^2 - (x mod 2) - 4)/4) mod 2))/8) = floor(((floor(x^2/4) - 1)^2)/8) where x = 2*n + 2. Replace x with 2*n - 1 to obtain A050534(n) = 3*A000332(n+1). Note that a(2*n) = A060300(n)/2 and a(2*n + 1) = A002593(n+1). - _Raphie Frank_, Jan 30 2014

%F a(n) = floor(1/(exp(2/n^2) - 1)^2)/2. Also a(n) = A007590(n+1)*A074148(n-1)/2. - _Richard R. Forberg_, Oct 26 2014

%F Sum_{n>=1} 1/a(n) = -cot(Pi/sqrt(2))*Pi/sqrt(2) - 1/2. - _Amiram Eldar_, Aug 25 2022

%t LinearRecurrence[{4, -5, 0, 5, -4, 1}, {0, 1, 8, 28, 72, 153}, 60] (* _Vladimir Joseph Stephan Orlovsky_, Feb 08 2012 *)

%t CoefficientList[Series[x (1 + 4 x + x^2)/((1 + x) (1 - x)^5), {x, 0, 50}], x] (* _Vincenzo Librandi_, Jun 26 2012 *)

%t Table[((-1)^n + 2 n^2 (n + 2)^2 - 1)/16, {n, 0, 20}] (* _Eric W. Weisstein_, Jul 14 2017 *)

%o (Magma) [(2*n^4+8*n^3+8*n^2-1)/16+(-1)^n/16: n in [0..50]]; // _Vincenzo Librandi_, Oct 27 2014

%o (PARI) my(x='x+O('x^99)); concat(0, Vec(x*(1+4*x+x^2)/((1+x)*(1-x)^5))) \\ _Altug Alkan_, Apr 16 2018

%o (Sage) [(2*n^4 +8*n^3 +8*n^2 -1+(-1)^n)/16 for n in range(30)] # _G. C. Greubel_, Dec 16 2018

%o (GAP) List([0..30], n -> (2*n^4 +8*n^3 +8*n^2 -1+(-1)^n)/16); # _G. C. Greubel_, Dec 16 2018

%Y Cf. A002620, A000292, A000578, A011934, A231303.

%Y Cf. A289705 (4-cycles), A289706 (5-cycles), A289707 (6-cycles).

%K nonn,easy

%O 0,3

%A _Paul Barry_, Apr 16 2005