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A014642 Even octagonal numbers: a(n) = 4*n*(3*n-1).

%I A014642 #72 Apr 15 2025 12:58:42
%S A014642 0,8,40,96,176,280,408,560,736,936,1160,1408,1680,1976,2296,2640,3008,
%T A014642 3400,3816,4256,4720,5208,5720,6256,6816,7400,8008,8640,9296,9976,
%U A014642 10680,11408,12160,12936,13736,14560,15408,16280,17176,18096,19040,20008,21000,22016
%N A014642 Even octagonal numbers: a(n) = 4*n*(3*n-1).
%C A014642 8 times pentagonal numbers. - _Omar E. Pol_, Dec 11 2008
%C A014642 Sequence found by reading the line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - _Omar E. Pol_, Jul 18 2012
%C A014642 The sequence forms the even nesting cube-frames (see illustrations in A000567), which separate and appear according to formula along the axes on the zero-centered and one-centered hexagonal number spirals, as well as the axes of the zero-centered and one-centered square number spirals. See illustrations in links. - _John Elias_, Jul 20 2022
%H A014642 G. C. Greubel, <a href="/A014642/b014642.txt">Table of n, a(n) for n = 0..1000</a>
%H A014642 John Elias, <a href="/A014642/a014642_1.png">Octagonal Nesting Cubes on the Hexagonal Number Spiral</a> <a href="/A014642/a014642_2.png">Octagonal Nesting Cubes on the Square Number Spiral</a>
%H A014642 Craig Knecht, <a href="/A014642/a014642.png">Number of positions the remaining tiles can occupy in a 4*n length polyiamond bilayer when one tile is missing</a>.
%H A014642 Yaohui Zhu, Kaiming Sun, Zhengdong Luo, and Lingfeng Wang, <a href="https://doi.org/10.1609/aaai.v39i2.32162">Progressive Self-Learning for Domain Adaptation on Symbolic Regression of Integer Sequences</a>, Proc. 39th AAAI Conf. Artif. Intel. (2025) Vol. 39, No. 1, 1692-1699. See p. 1698.
%H A014642 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A014642 a(n) = A000326(n)*8. - _Omar E. Pol_, Dec 11 2008
%F A014642 a(n) = A049450(n)*4 = A033579(n)*2. - _Omar E. Pol_, Dec 13 2008
%F A014642 a(n) = a(n-1) + 24*n - 16 (with a(0)=0). - _Vincenzo Librandi_, Nov 20 2010
%F A014642 G.f.: x*(8+16*x)/(1-3*x+3*x^2-x^3). - _Colin Barker_, Jan 06 2012
%F A014642 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - _G. C. Greubel_, Jun 07 2017
%F A014642 E.g.f.: 4*x*(2 + 3*x)*exp(x). - _G. C. Greubel_, Oct 09 2019
%F A014642 From _Amiram Eldar_, Mar 24 2021: (Start)
%F A014642 Sum_{n>=1} 1/a(n) = 3*log(3)/8 - Pi/(8*sqrt(3)).
%F A014642 Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2 - Pi/(4*sqrt(3)). (End)
%p A014642 seq(8*binomial(3*n,2)/3, n=0..50); # _G. C. Greubel_, Oct 09 2019
%t A014642 LinearRecurrence[{3,-3,1},{0,8,40}, 50] (* _G. C. Greubel_, Jun 07 2017 *)
%t A014642 PolygonalNumber[8,Range[0,90,2]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Aug 19 2020 *)
%o A014642 (PARI) vector(51, n, 8*binomial(3*(n-1),2)/3 ) \\ _G. C. Greubel_, Jun 07 2017
%o A014642 (Magma) [8*Binomial(3*n,2)/3: n in [0..50]]; // _G. C. Greubel_, Oct 09 2019
%o A014642 (Sage) [8*binomial(3*n,2)/3 for n in (0..50)] # _G. C. Greubel_, Oct 09 2019
%o A014642 (GAP) List([0..50], n-> 8*Binomial(3*n,2)/3); # _G. C. Greubel_, Oct 09 2019
%Y A014642 Cf. A000567, A000326, A001082, A014641, A014793, A014794, A033579, A049450.
%K A014642 nonn,easy
%O A014642 0,2
%A A014642 _Mohammad K. Azarian_
%E A014642 More terms from _Patrick De Geest_