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%I #48 Sep 08 2022 08:44:41
%S 9,49,121,225,361,529,729,961,1225,1521,1849,2209,2601,3025,3481,3969,
%T 4489,5041,5625,6241,6889,7569,8281,9025,9801,10609,11449,12321,13225,
%U 14161,15129,16129,17161,18225
%N a(n) = (4n + 3)^2.
%C If Y is a fixed 2-subset of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - _Milan Janjic_, Oct 21 2007
%C A bisection of A016754. Sequence arises from reading the line from 9, in the direction 9, 49, ... in the square spiral whose vertices are the squares A000290. - _Omar E. Pol_, May 24 2008
%C Using (n,n+1) to generate a Pythagorean triangle with sides of lengths x<y<z, 3*z+4*x+5*y+2 = (2*x+1)^2 will give a(n) starting at n=1. - _J. M. Bergot_, Jul 17 2013
%H Vincenzo Librandi, <a href="/A016838/b016838.txt">Table of n, a(n) for n = 0..860</a>
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Two Enumerative Functions</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F Denominators of first differences Zeta[2,(4n-1)/4]-Zeta[2,(4(n+1)-1)/4]. - _Artur Jasinski_, Mar 03 2010
%F From _George F. Johnson_, Oct 03 2012: (Start)
%F G.f.: (9+22*x+x^2)/(1-x)^3.
%F a(n+1) = a(n) + 16 + 8*sqrt(a(n)).
%F a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).
%F a(n-1)*a(n+1) = (a(n)-16)^2; a(n+1) - a(n-1) = 16*sqrt(a(n)).
%F a(n) = A016754(2*n+1) = (A004767(n))^2.
%F (End)
%F Sum_{n>=0} 1/a(n) = Pi^2/16 - G/2, where G is the Catalan constant (A006752). - _Amiram Eldar_, Jun 28 2020
%F Product_{n>=0} (1 - 1/a(n)) = Gamma(3/4)^2/sqrt(Pi) = A068465^2 * A087197. - _Amiram Eldar_, Feb 01 2021
%p A016838:=n->(4*n + 3)^2; seq(A016838(n), n=0..50); # _Wesley Ivan Hurt_, Feb 24 2014
%t Table[(4*n + 3)^2, {n, 0, 40}] (* _Vaclav Kotesovec_, Nov 14 2017 *)
%o (Magma) [(4*n+3)^2: n in [0..50]]; // _Vincenzo Librandi_, Apr 26 2011
%o (PARI) a(n)=(4*n+3)^2 \\ _Charles R Greathouse IV_, Jun 17 2017
%Y Cf. A000290, A001539, A016742, A016754, A016802, A016814, A016826, A068465, A087197.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_
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