A016838 a(n) = (4n + 3)^2.

9, 49, 121, 225, 361, 529, 729, 961, 1225, 1521, 1849, 2209, 2601, 3025, 3481, 3969, 4489, 5041, 5625, 6241, 6889, 7569, 8281, 9025, 9801, 10609, 11449, 12321, 13225, 14161, 15129, 16129, 17161, 18225

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

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

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

Using (n,n+1) to generate a Pythagorean triangle with sides of lengths xJ. M. Bergot, Jul 17 2013

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..860
• Milan Janjic, Two Enumerative Functions
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Crossrefs

Cf. A000290, A001539, A016742, A016754, A016802, A016814, A016826, A068465, A087197.

Programs

Magma

[(4*n+3)^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

Maple

A016838:=n->(4*n + 3)^2; seq(A016838(n), n=0..50); # Wesley Ivan Hurt, Feb 24 2014

Mathematica

Table[(4*n + 3)^2, {n, 0, 40}] (* Vaclav Kotesovec, Nov 14 2017 *)

PARI

a(n)=(4*n+3)^2 \\ Charles R Greathouse IV, Jun 17 2017

Formula

Denominators of first differences Zeta[2,(4n-1)/4]-Zeta[2,(4(n+1)-1)/4]. - Artur Jasinski, Mar 03 2010

From George F. Johnson, Oct 03 2012: (Start)

G.f.: (9+22*x+x^2)/(1-x)^3.

a(n+1) = a(n) + 16 + 8*sqrt(a(n)).

a(n+1) = 2*a(n) - a(n-1) + 32 = 3*a(n) - 3*a(n-1) + a(n-2).

a(n-1)*a(n+1) = (a(n)-16)^2; a(n+1) - a(n-1) = 16*sqrt(a(n)).

a(n) = A016754(2*n+1) = (A004767(n))^2.

(End)

Sum_{n>=0} 1/a(n) = Pi^2/16 - G/2, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020

Product_{n>=0} (1 - 1/a(n)) = Gamma(3/4)^2/sqrt(Pi) = A068465^2 * A087197. - Amiram Eldar, Feb 01 2021