A078371 a(n) = (2*n+5)*(2*n+1).

5, 21, 45, 77, 117, 165, 221, 285, 357, 437, 525, 621, 725, 837, 957, 1085, 1221, 1365, 1517, 1677, 1845, 2021, 2205, 2397, 2597, 2805, 3021, 3245, 3477, 3717, 3965, 4221, 4485, 4757, 5037, 5325, 5621, 5925, 6237, 6557, 6885, 7221, 7565, 7917, 8277, 8645

Offset: 0

Author

Wolfdieter Lang, Nov 29 2002

Keywords

• nonn
• easy

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).

Crossrefs

Subsequence of A077425 (D values (not a square) for which Pell x^2 - D*y^2 = +4 is solvable in positive integers).

Cf. A017113, A046092, A061037, A077428, A078355, A078370.

Supersequence of A143206.

Programs

Magma

[(2*n+5)*(2*n+1): n in [0..100]]; // G. C. Greubel, Sep 19 2018

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 *)

PARI

lista(nn) = {for (n=0, nn, print1((2*n+1)*(2*n+5), ", "));} \\ Michel Marcus, Nov 21 2013

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

Extensions

More terms from Emeric Deutsch, Feb 24 2005