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a(n) = 4*n^2 - 1.
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%I #190 Sep 08 2022 08:44:28

%S -1,3,15,35,63,99,143,195,255,323,399,483,575,675,783,899,1023,1155,

%T 1295,1443,1599,1763,1935,2115,2303,2499,2703,2915,3135,3363,3599,

%U 3843,4095,4355,4623,4899,5183,5475,5775,6083,6399,6723,7055,7395

%N a(n) = 4*n^2 - 1.

%C Sum_{n>=1} (-1)^n*a(n)/n! = 1 - 1/e = A068996. - _Gerald McGarvey_, Nov 06 2007

%C Sequence arises from reading the line from -1, in the direction -1, 15, ... and the same line from 3, in the direction 3, 35, ..., in the square spiral whose nonnegative vertices are the squares A000290. - _Omar E. Pol_, May 24 2008

%C a(n) is the product of the consecutive odd integers 2n-1 and 2n+1 (cf. A005408). - _Doug Bell_, Mar 08 2009

%C For n>0: a(n) = A176271(2*n,n); cf. A016754, A053755. - _Reinhard Zumkeller_, Apr 13 2010

%C a(n+1) gives the curvature c(n) of the n-th circle touching the two equal semicircles of the symmetric arbelos (1/2, 1/2) and the (n-1)-st circle, with input c(0) = 3 = A059100(1) (referring to the second circle of the Pappus chain), for n >= 0. - _Wolfdieter Lang_ and _Kival Ngaokrajang_, Jul 03 2015

%C After 3, a(n) is pseudoprime to base 2n. For example: (2*2)^(a(2)-1) == 1 (mod a(2)), in fact 4^14 = 15*17895697+1. - _Bruno Berselli_, Sep 24 2015

%C Numbers m such that m+1 and (m+1)/4 are squares. - _Bruno Berselli_, Mar 03 2016

%C After -1, the least common multiple of 2*m+1 and 2*m-1. - _Colin Barker_, Feb 11 2017

%C This sequence contains all products of the twin prime pairs (see A037074). - _Charles Kusniec_, Oct 03 2019

%D T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.

%D L. B. W. Jolley, Summation of Series, Dover, 2nd ed., 1961.

%D Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), pp. 980-981.

%D A. Languasco and A. Zaccagnini, Manuale di Crittografia, Ulrico Hoepli Editore (2015), p. 259.

%H Vincenzo Librandi, <a href="/A000466/b000466.txt">Table of n, a(n) for n = 0..900</a>

%H Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, <a href="https://www.emis.de/journals/JIS/VOL21/Falcao/falcao2.html">Combinatorial Identities Associated with a Multidimensional Polynomial Sequence</a>, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.

%H Milan Janjic and Boris Petkovic, <a href="http://arxiv.org/abs/1301.4550">A Counting Function</a>, arXiv 1301.4550 [math.CO], 2013.

%H Kival Ngaokrajang, <a href="/A000466/a000466.pdf">Illustration of the Pappus chain (downwards direction)</a>.

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

%F O.g.f.: ( 1-6*x-3*x^2 ) / (x-1)^3 . - _R. J. Mathar_, Mar 24 2011

%F E.g.f.: (-1 + 4*x + 4*x^2)*exp(x). - _Ilya Gutkovskiy_, May 26 2016

%F Sum_{n>=1} 1/a(n) = 1/2 [Jolley eq. 233]. - _Benoit Cloitre_, Apr 05 2002

%F Sum_{n>=1} 2/a(n) = 1 = 2/3 + 2/15 + 2/35 + 2/63 + 2/99 + 2/143, ..., with partial sums: 2/3, 4/5, 6/7, 8/9, 10/11, 12/13, 14/15, ... - _Gary W. Adamson_, Jun 16 2003

%F 1/3 + Sum_{n>=2} 4/a(n) = 1 = 1/3 + 4/15 + 4/35 + 4/63, ..., with partial sums: 1/3, 3/5, 5/7, 7/9, 9/11, ..., (2n+1)/(2n+3). - _Gary W. Adamson_, Jun 18 2003

%F Sum_{n>=0} 2/a(2*n+1) = Pi/4 = 2/3 + 2/35 + 2/99, ... = (1 - 1/3) + (1/5 - 2/7) + (1/9 - 1/11) + ... = Sum_{n>=0} (-1)^n/(2*n+1). - _Gary W. Adamson_, Jun 22 2003

%F Product(n>=1, (a(n)+1)/a(n)) = Pi/2 (Wallis formula). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004

%F a(n)+2 = A053755(n). - _Zak Seidov_, Jan 16 2007

%F a(n)^2 + A008586(n)^2 = A053755(n)^2 (Pythagorean triple). - _Zak Seidov_, Jan 16 2007

%F a(n) = a(n-1) + 8*n - 4 for n > 0, a(0)=-1. - _Vincenzo Librandi_, Dec 17 2010

%F Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/4 - 1/2 = (A019669-1)/2. [Jolley eq (366)]. - _R. J. Mathar_, Mar 24 2011

%F For n>0, a(n) = 2/(Integral_{x=0..Pi/2} (sin(x))^3*(cos(x))^(2*n-2)). - _Francesco Daddi_, Aug 02 2011

%F Nonlinear recurrence for c(n) = a(n+1) (see the arbelos comment above) from Descartes' three circle theorem (see the links under A259555): c(n) = 4 + c(n-1) + 4*sqrt(c(n-1) + 1), with input c(0) = 3 = A059100(1), for n >= 0. The appropriate solution of this recurrence is c(n-1) + 1 = 4*n^2. - _Wolfdieter Lang_, Jul 03 2015

%F a(n) = 3*Pochhammer(5/2,n-1)/Pochhammer(1/2,n-1). Hence, the e.g.f. for a(n+1), i.e., dropping the first term, is 3* 1F1(5/2;1/2;x), with 1F1 being the confluent hypergeometric function (also known as Kummer's). - _Stanislav Sykora_, May 26 2016

%F Product_{n>=1} (1 - 1/a(n)) = sin(Pi/sqrt(2))/sqrt(2). - _Amiram Eldar_, Feb 04 2021

%p A000466:=n->4*n^2-1; seq(A000466(n), n=0..100); # _Wesley Ivan Hurt_, Nov 19 2013

%t 4 Range[0,50]^2-1 (* _Harvey P. Dale_, Jan 23 2011 *)

%o (Magma) [4*n^2-1: n in [0..50]]; // _Vincenzo Librandi_, Apr 26 2011

%o (PARI) a(n)=4*n^2-1 \\ _Charles R Greathouse IV_, Oct 27 2011

%o (Maxima) makelist(4*n^2-1, n, 0, 50); /* _Martin Ettl_, Nov 12 2012 */

%o (Sage) [4*n^2-1 for n in (0..50)] # _Bruno Berselli_, Sep 24 2015

%Y Cf. A000290, A001539, A016286, A016742.

%Y Factor of A160466. Superset of A037074.

%Y Cf. A059100 (curvatures for a Pappus chain).

%K sign,easy

%O 0,2

%A Chan Siu Kee (skchan5(AT)hkein.ie.cuhk.hk)