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

0,1

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

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.

Sequence in context: A366346 A031292 A147331 * A265056 A049741 A166010

Adjacent sequences: A078368 A078369 A078370 * A078372 A078373 A078374

Keyword

nonn,easy

Author

Wolfdieter Lang, Nov 29 2002

Extensions

More terms from Emeric Deutsch, Feb 24 2005

Status

approved