cp's OEIS Frontend

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

%I A078371 #100 Jul 26 2024 21:16:31
%S A078371 5,21,45,77,117,165,221,285,357,437,525,621,725,837,957,1085,1221,
%T A078371 1365,1517,1677,1845,2021,2205,2397,2597,2805,3021,3245,3477,3717,
%U A078371 3965,4221,4485,4757,5037,5325,5621,5925,6237,6557,6885,7221,7565,7917,8277,8645
%N A078371 a(n) = (2*n+5)*(2*n+1).
%C A078371 This is the generic form of D in the (nontrivially) solvable Pell equation x^2 - D*y^2 = +4. See A077428 and A078355.
%C A078371 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
%C A078371 From _Vincenzo Librandi_, Aug 08 2010: (Start)
%C A078371 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.
%C A078371 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)
%C A078371 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
%C A078371 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
%C A078371 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
%C A078371 a(n) is the sum of 2*n+5 consecutive integers starting from n-1. - _Bruno Berselli_, Jan 16 2018
%C A078371 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
%H A078371 Bruno Berselli, <a href="/A078371/b078371.txt">Table of n, a(n) for n = 0..1000</a>
%H A078371 Soren Laing Aletheia-Zomlefer, Lenny Fukshansky, and Stephan Ramon Garcia, <a href="https://arxiv.org/abs/1807.08899">The Bateman-Horn Conjecture: Heuristics, History, and Applications</a>, arXiv:1807.08899 [math.NT], 2018-2019. See Example 6.6.5 p. 34.
%H A078371 A. M. S. Ramasamy, <a href="http://www.insa.nic.in/writereaddata/UpLoadedFiles/IJPAM/20006851_577.pdf">Polynomial solutions for the Pell's equation</a>, Indian Journal of Pure and Applied Mathematics 25 (1994), p. 578.
%H A078371 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F A078371 a(n) = 8*(binomial(n+2, 2)-1)+5, hence subsequence of A004770 (5 (mod 8) numbers).
%F A078371 G.f.: (5 + 6*x - 3*x^2)/(1-x)^3.
%F A078371 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
%F A078371 a(n) = 8*(n+1) + a(n-1) for n > 0, a(0)=5. - _Vincenzo Librandi_, Aug 08 2010
%F A078371 From _Ilya Gutkovskiy_, May 22 2016: (Start)
%F A078371 E.g.f.: (5 + 4*x*(4 + x))*exp(x).
%F A078371 Sum_{n>=0} 1/a(n) = 1/3. (End)
%F A078371 Sum_{n>=0} (-1)^n/a(n) = 1/6. - _Amiram Eldar_, Oct 08 2023
%p A078371 seq((2*n+5)*(2*n+1), n=0..48); # _Emeric Deutsch_, Feb 24 2005
%t A078371 Table[(2 n + 5) (2 n + 1), {n, 0, 100}] (* _Wesley Ivan Hurt_, Nov 19 2013 *)
%t A078371 LinearRecurrence[{3,-3,1},{5,21,45},50] (* _Harvey P. Dale_, Oct 18 2020 *)
%o A078371 (PARI) lista(nn) = {for (n=0, nn, print1((2*n+1)*(2*n+5), ", "));} \\ _Michel Marcus_, Nov 21 2013
%o A078371 (Magma) [(2*n+5)*(2*n+1): n in [0..100]]; // _G. C. Greubel_, Sep 19 2018
%Y A078371 Subsequence of A077425 (D values (not a square) for which Pell x^2 - D*y^2 = +4 is solvable in positive integers).
%Y A078371 Cf. A017113, A046092, A061037, A077428, A078355, A078370.
%Y A078371 Supersequence of A143206.
%K A078371 nonn,easy
%O A078371 0,1
%A A078371 _Wolfdieter Lang_, Nov 29 2002
%E A078371 More terms from _Emeric Deutsch_, Feb 24 2005