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a(n) = 6*n^2 - 1.
12

%I #90 May 24 2022 08:10:50

%S -1,5,23,53,95,149,215,293,383,485,599,725,863,1013,1175,1349,1535,

%T 1733,1943,2165,2399,2645,2903,3173,3455,3749,4055,4373,4703,5045,

%U 5399,5765,6143,6533,6935,7349,7775,8213,8663,9125,9599,10085,10583,11093,11615

%N a(n) = 6*n^2 - 1.

%C Also: The numerators in the j=2 column of the array a(i,j) defined in A140825, where the columns j=0 and j=1 are represented by A000012 and A005408. This could be extended to column j=3: 1, -1, 9, 55, 161, ... The common feature of these sequences derived from a(i,j) is that their j-th differences are constant sequences defined by A091137(j).

%C a(n) is the set of all k such that 6k+6 is a perfect square. - _Gary Detlefs_, Mar 04 2010

%C The identity (6*n^2 - 1)^2 - (9*n^2 - 3)*(2*n)^2 = 1 can be written as a(n+1)^2 - A157872(n)*A005843(n+1)^2 = 1. - _Vincenzo Librandi_, Feb 05 2012

%C Apart from first term, sequence found by reading the line from 5, in the direction 5, 23, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - _Omar E. Pol_, Jul 18 2012

%C From _Paul Curtz_, Sep 17 2018: (Start)

%C Terms from center to right in the following spiral:

%C .

%C 65--63--61--59

%C / \

%C 67 31--29--27 57

%C / / \ \

%C 69 33 9---7 25 55

%C / / / \ \ \

%C 71 35 11 -1===5==23==53==>

%C / / / / / /

%C 37 13 1---3 21 51

%C \ \ / /

%C 39 15--17--19 49

%C \ /

%C 41--43--45--47 (End)

%D P. Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Note 12, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969, 132 pages, pp. 28-36. CCSA, then CELAR. Now DGA Maitrise de l'Information 35131 Bruz.

%H Vincenzo Librandi, <a href="/A140811/b140811.txt">Table of n, a(n) for n = 0..1000</a>

%H Leo Tavares, <a href="/A140811/a140811_1.jpg">Illustration: Barred 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) = 2*a(n-1) - a(n-2) + 12.

%F First differences: a(n+1) - a(n) = A017593(n).

%F Second differences: A071593(n+1) - A071593(n) = 12.

%F G.f.: (1-8*x-5*x^2)/(x-1)^3. - _Jaume Oliver Lafont_, Aug 30 2009

%F a(n) = a(n-1) + 12*n -6. - _Vincenzo Librandi_, Feb 05 2012

%F a(n) = 3*a(n-1) -3*a(n-2) + a(n-3). - _Vincenzo Librandi_, Feb 05 2012

%F a(n) = A033581(n) - 1. - _Omar E. Pol_, Jul 18 2012

%F a(n) = A032528(2n) - 1. - _Adriano Caroli_, Jul 21 2013

%F For n > 0, a(n) = floor(3/(cosh(1/n) - 1)) = floor(1/(n*sinh(1/n) - 1)); for similar formulas for cosine and sine, see A033581. - _Clark Kimberling_, Oct 19 2014, corrected by _M. F. Hasler_, Oct 21 2014

%F a(-n) = a(n). - _Paul Curtz_, Sep 17 2018

%F From _Amiram Eldar_, Feb 04 2021: (Start)

%F Sum_{n>=1} 1/a(n) = (1 - (Pi/sqrt(6))*cot(Pi/sqrt(6)))/2.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = ((Pi/sqrt(6))*csc(Pi/sqrt(6)) - 1)/2.

%F Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(6))*csc(Pi/sqrt(6)).

%F Product_{n>=1} (1 - 1/a(n)) = csc(Pi/sqrt(6))*sin(Pi/sqrt(3))/sqrt(2). (End)

%F a(n) = A003154(n+1) - 2*A016777(n). - _Leo Tavares_, May 13 2022

%t LinearRecurrence[{3,-3,1},{-1,5,23},40] (* _Vincenzo Librandi_, Feb 05 2012 *)

%t CoefficientList[Series[(1-8*x-5*x^2)/(x-1)^3 , {x, 0, 40}], x] (* _Stefano Spezia_, Sep 17 2018 *)

%o (PARI) a(n)=6*n^2-1 \\ _Charles R Greathouse IV_, Jun 01 2011

%o (Magma) [6*n^2 - 1: n in [0..50]]; // _Vincenzo Librandi_, Jun 02 2011

%Y Cf. A005843, A157872. A060747, A103115(n+1), A141417 (array).

%Y Cf. A003154, A016777.

%K sign,easy

%O 0,2

%A _Paul Curtz_, Jul 16 2008

%E Edited and extended by _R. J. Mathar_, Aug 06 2008

%E Better description _Ray Chandler_, Feb 03 2009