OFFSET
0,1
COMMENTS
This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = +4. See A077428 and A078355.
Consider all primitive Pythagorean triples (a,b,c) with c-a=8, sequence gives values of a. (Corresponding values for b are A017113(n), while c follows A078370(n).) - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 19 2004
From Vincenzo Librandi, Aug 08 2010: (Start)
The identity (4*n^3 + 18*n^2 + 24*n + 9)^2 - (4*n^2 + 12*n + 5)*(2*n^2 + 6*n + 4)^2 = 1 (see Ramasamy's paper in link) can be written as A141530(n+2)^2 - a(n)*A046092(n+1)^2 = 1.
a(n)^3 + 6*a(n)^2 + 9*a(n) + 4 is a square: in fact, a(n)^3 + 6*a(n)^2 + 9*a(n) + 4 = (a(n) + 1)^2*(a(n) + 4), where a(n) + 4 = (2*n+3)^2. (End)
Products of two positive odd integers with difference 4 (i.e., 1*5, 3*7, 5*9, 7*11, 9*13, ...). - Wesley Ivan Hurt, Nov 19 2013
Starting with stage 1, the number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 675", based on the 5-celled von Neumann neighborhood. - Robert Price, May 21 2016
The continued fraction expansion of (sqrt(a(n))-1)/2 is [n; {1,2*n+1}] with periodic part of length 2: repeat {1,2*n+1}. - Ron Knott, May 11 2017
a(n) is the sum of 2*n+5 consecutive integers starting from n-1. - Bruno Berselli, Jan 16 2018
The continued fraction expansion of sqrt(a(n)) is [2n+2; {1, n, 2, n, 1, 4n+4}]. For n=0, this collapses to [2; {4}]. - Magus K. Chu, Aug 26 2022
LINKS
Bruno Berselli, Table of n, a(n) for n = 0..1000
Soren Laing Aletheia-Zomlefer, Lenny Fukshansky, and Stephan Ramon Garcia, The Bateman-Horn Conjecture: Heuristics, History, and Applications, arXiv:1807.08899 [math.NT], 2018-2019. See Example 6.6.5 p. 34.
A. M. S. Ramasamy, Polynomial solutions for the Pell's equation, Indian Journal of Pure and Applied Mathematics 25 (1994), p. 578.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
a(n) = 8*(binomial(n+2, 2)-1)+5, hence subsequence of A004770 (5 (mod 8) numbers).
G.f.: (5 + 6*x - 3*x^2)/(1-x)^3.
a(n) = A061037(2*n+1) = (2*n+3)^2 - 4. For A061037: a(2*n+1) = (2*n+1)*(2*n+5) = (2*n+3)^2-4. - Paul Curtz, Sep 24 2008
a(n) = 8*(n+1) + a(n-1) for n > 0, a(0)=5. - Vincenzo Librandi, Aug 08 2010
From Ilya Gutkovskiy, May 22 2016: (Start)
E.g.f.: (5 + 4*x*(4 + x))*exp(x).
Sum_{n>=0} 1/a(n) = 1/3. (End)
Sum_{n>=0} (-1)^n/a(n) = 1/6. - Amiram Eldar, Oct 08 2023
MAPLE
seq((2*n+5)*(2*n+1), n=0..48); # Emeric Deutsch, Feb 24 2005
MATHEMATICA
Table[(2 n + 5) (2 n + 1), {n, 0, 100}] (* Wesley Ivan Hurt, Nov 19 2013 *)
LinearRecurrence[{3, -3, 1}, {5, 21, 45}, 50] (* Harvey P. Dale, Oct 18 2020 *)
PROG
(PARI) lista(nn) = {for (n=0, nn, print1((2*n+1)*(2*n+5), ", ")); } \\ Michel Marcus, Nov 21 2013
(Magma) [(2*n+5)*(2*n+1): n in [0..100]]; // G. C. Greubel, Sep 19 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Nov 29 2002
EXTENSIONS
More terms from Emeric Deutsch, Feb 24 2005
STATUS
approved