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Four times second pentagonal numbers: a(n) = 2*n*(3*n+1).
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%I #92 Jul 26 2022 21:59:52

%S 0,8,28,60,104,160,228,308,400,504,620,748,888,1040,1204,1380,1568,

%T 1768,1980,2204,2440,2688,2948,3220,3504,3800,4108,4428,4760,5104,

%U 5460,5828,6208,6600,7004,7420,7848,8288,8740,9204,9680,10168,10668,11180,11704,12240

%N Four times second pentagonal numbers: a(n) = 2*n*(3*n+1).

%C Subsequence of A062717: A010052(6*a(n)+1) = 1. - _Reinhard Zumkeller_, Feb 21 2011

%C Sequence found by reading the line from 0, in the direction 0, 8,..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Opposite numbers to the members of A139267 in the same spiral - _Omar E. Pol_, Sep 09 2011

%C a(n) is the number of edges of the octagonal network O(n,n); O(m,n) is defined by Fig. 1 of the Siddiqui et al. reference. - _Emeric Deutsch_ May 13 2018

%C The partial sums of this sequence give A035006. - _Leo Tavares_, Oct 03 2021

%H Ivan Panchenko, <a href="/A033580/b033580.txt">Table of n, a(n) for n = 0..1000</a>

%H M. K. Siddiqui, M. Naeem, N. A. Rahman, and M. Imran, <a href="https://joam.inoe.ro/articles/computing-topological-indices-of-certain-networks/">Computing topological indices of certain networks</a>, J. of Optoelectronics and Advanced Materials, 18, No. 9-10 (2016), pp. 884-892.

%H Leo Tavares, <a href="/A033580/a033580_9.jpg">Illustration: Crossed Stars</a>

%H Leo Tavares, <a href="/A033580/a033580_10.jpg">Illustration: Four Quarter Star Crosses</a>

%H Leo Tavares, <a href="/A033580/a033580_11.jpg">Illustration: Triangulated Star Crosses</a>

%H Leo Tavares, <a href="/A033580/a033580_12.jpg">Illustration: Oblong Star Crosses</a>

%H Leo Tavares, <a href="/A033580/a033580_13.jpg">Illustration: Crossed Diamond Stars</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F a(n) = a(n-1) +12*n -4 (with a(0)=0). - _Vincenzo Librandi_, Aug 05 2010

%F G.f.: 4*x*(2+x)/(1-x)^3. - _Colin Barker_, Feb 13 2012

%F a(-n) = A033579(n). - _Michael Somos_, Jun 09 2014

%F E.g.f.: 2*x*(4 + 3*x)*exp(x). - _G. C. Greubel_, Oct 09 2019

%F From _Amiram Eldar_, Jan 14 2021: (Start)

%F Sum_{n>=1} 1/a(n) = 3/2 - Pi/(4*sqrt(3)) - 3*log(3)/4.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = -3/2 + Pi/(2*sqrt(3)) + log(2). (End)

%F From _Leo Tavares_, Oct 12 2021: (Start)

%F a(n) = A003154(n+1) - A016813(n). See Crossed Stars illustration.

%F a(n) = 4*A005449(n). See Four Quarter Star Crosses illustration.

%F a(n) = 2*A049451(n).

%F a(n) = A046092(n-1) + A033996(n). See Triangulated Star Crosses illustration.

%F a(n) = 4*A000217(n-1) + 8*A000217(n).

%F a(n) = 4*A000217(n-1) + 4*A002378. See Oblong Star Crosses illustration.

%F a(n) = A016754(n) + 4*A000217(n). See Crossed Diamond Stars illustration.

%F a(n) = 2*A001105(n) + 4*A000217(n).

%F a(n) = A016742(n) + A046092(n).

%F a(n) = 4*A000290(n) + 4*A000217(n). (End)

%p seq(2*n*(3*n+1), n=0..50); # _G. C. Greubel_, Oct 09 2019

%t 4*Binomial[3*Range[50]-2, 2]/3 (* _G. C. Greubel_, Oct 09 2019 *)

%o (PARI) a(n)=2*n*(3*n+1) \\ _Charles R Greathouse IV_, Sep 28 2015

%o (Magma) [2*n*(3*n+1): n in [0..50]]; // _G. C. Greubel_, Oct 09 2019

%o (Sage) [2*n*(3*n+1) for n in (0..50)] # _G. C. Greubel_, Oct 09 2019

%o (GAP) List([0..50], n-> 2*n*(3*n+1)); # _G. C. Greubel_, Oct 09 2019

%Y Cf. A033579, A049451, A045945, A186423.

%Y Cf. sequences listed in A254963.

%Y Cf. A003154, A016813.

%Y Cf. A035006.

%Y Cf. A005449, A046092, A033996, A002378, A016742, A000290, A016754, A001105.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_