A016814 a(n) = (4*n + 1)^2.
1, 25, 81, 169, 289, 441, 625, 841, 1089, 1369, 1681, 2025, 2401, 2809, 3249, 3721, 4225, 4761, 5329, 5929, 6561, 7225, 7921, 8649, 9409, 10201, 11025, 11881, 12769, 13689, 14641, 15625, 16641, 17689, 18769, 19881, 21025, 22201, 23409, 24649, 25921, 27225, 28561, 29929
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..10000 (terms 0..200 from Ivan Panchenko).
- Leo Tavares, Illustration: Square trapeziums
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Programs
-
Magma
[(4*n+1)^2: n in [0..40]]; // G. C. Greubel, Dec 28 2022
-
Maple
A016814:=n->(4*n+1)^2; seq(A016814(k), k=0..100); # Wesley Ivan Hurt, Nov 02 2013
-
Mathematica
(4*Range[0,40] +1)^2 (* or *) LinearRecurrence[{3,-3,1}, {1,25,81}, 40] (* Harvey P. Dale, Nov 20 2012 *) Accumulate[32Range[0, 47] - 8] + 9 (* Alonso del Arte, Aug 19 2017 *)
-
PARI
a(n)=(4*n+1)^2 \\ Charles R Greathouse IV, Oct 07 2015
-
SageMath
[(4*n+1)^2 for n in range(41)] # G. C. Greubel, Dec 28 2022
Formula
a(n) = a(n-1) + 32*n - 8, n > 0. - Vincenzo Librandi, Dec 15 2010
From George F. Johnson, Sep 28 2012: (Start)
G.f.: (1 + 22*x + 9*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)).
Sum_{n>=0} 1/a(n) = G/2 + Pi^2/16, where G is the Catalan constant (A006752). - Amiram Eldar, Jun 28 2020
Product_{n>=1} (1 - 1/a(n)) = 2*Gamma(5/4)^2/sqrt(Pi) = 2 * A068467^2 * A087197. - Amiram Eldar, Feb 01 2021
From G. C. Greubel, Dec 28 2022: (Start)
a(2*n) = A017078(n).
a(2*n+1) = A017126(n).
E.g.f.: (1 + 24*x + 16*x^2)*exp(x). (End)
Comments