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A101368 The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j+j^2 and j divides 1+i+i^2. In fact, the pairs (a(n),a(n+1)), n>0, are all the solutions.

%I A101368 #45 Oct 17 2022 04:31:30
%S A101368 1,1,3,13,61,291,1393,6673,31971,153181,733933,3516483,16848481,
%T A101368 80725921,386781123,1853179693,8879117341,42542407011,203832917713,
%U A101368 976622181553,4679277990051,22419767768701,107419560853453,514678036498563,2465970621639361,11815175071698241,56609904736851843,271234348612560973
%N A101368 The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j+j^2 and j divides 1+i+i^2. In fact, the pairs (a(n),a(n+1)), n>0, are all the solutions.
%C A101368 Also, integers m such that 21*(3*m-1)^2 - 48 is a square. - _Max Alekseyev_, May 23 2022
%C A101368 a(n) is prime exactly for n = 3, 4, 5, 8, 16, 20, 22, 23, 58, 302, 386, 449, 479, 880 up to 1000. - _Tomohiro Yamada_, Dec 23 2018
%C A101368 Similarly, positive integers m,k with m|(1+k+^2) and k|(1-m+m^2) are consecutive terms of A061646, where m has an even index. - _Max Alekseyev_, May 23 2022
%H A101368 Seiichi Manyama, <a href="/A101368/b101368.txt">Table of n, a(n) for n = 1..1471</a>
%H A101368 Esther Banaian and Archan Sen, <a href="https://arxiv.org/abs/2210.07366">A Generalization of Markov Numbers</a>, arXiv:2210.07366 [math.CO], 2022. See Table 1 p. 12.
%H A101368 T. Cai, Z. Shen and L. Jia, <a href="http://arxiv.org/abs/1503.02798">A congruence involving harmonic sums modulo p^alpha q^beta</a>, arXiv preprint arXiv:1503.02798 [math.NT], 2015.
%H A101368 W. W. Chao, <a href="https://cms.math.ca/crux/v30/n7/page429-434.pdf">Problem 2981</a>, Crux Mathematicorum, 30 (2004), p. 430.
%H A101368 Yasuaki Gyoda, <a href="https://arxiv.org/abs/2109.09639">Positive integer solutions to (x+y)^2+(y+z)^2+(z+x)^2=12xyz</a>, arXiv:2109.09639 [math.NT], 2021. See Remark 3.3 p. 6.
%H A101368 W. H. Mills, <a href="http://projecteuclid.org/euclid.pjm/1103051516">A system of quadratic Diophantine equations</a>. Pacific J. Math. 3:1 (1953), 209-220.
%H A101368 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-6,1).
%F A101368 Recurrence: a(1)=a(2)=1 and a(n+1)=(1+a(n)+a(n)^2)/a(n-1) for n>2.
%F A101368 G.f.: x(1 - 5x + 3x^2) / [(1-x)(1 - 5x + x^2)]; a(n) = 2 * A089817(n-3) + 1, n>2. - Conjectured by _Ralf Stephan_, Jan 14 2005, proved by _Max Alekseyev_, Aug 03 2006
%F A101368 a(n) = 6a(n-1)-6a(n-2)+a(n-3), a(n) = 5a(n-1)-a(n-2)-1. - _Floor van Lamoen_, Aug 01 2006
%F A101368 a(n) = (4/3 - (2/7)*sqrt(21))*((5 + sqrt(21))/2)^n + (4/3 + (2/7)*sqrt(21))*((5 - sqrt(21))/2)^n + 1/3. - _Floor van Lamoen_, Aug 04 2006
%F A101368 For n>1, a(n) = (2 * A004253(n-1) + 1) / 3. - _Max Alekseyev_, May 23 2022
%e A101368 a(5) = 61 because (1 + a(4) + a(4)^2)/a(3) = (1 + 13 + 169)/3 = 61.
%p A101368 seq(coeff(series(x*(1-5*x+3*x^2)/((1-x)*(1-5*x+x^2)),x,n+1), x, n), n = 1 .. 30); # _Muniru A Asiru_, Dec 28 2018
%t A101368 Rest@ CoefficientList[Series[x (1 - 5 x + 3 x^2)/((1 - x) (1 - 5 x + x^2)), {x, 0, 28}], x] (* or *)
%t A101368 RecurrenceTable[{a[n] == (1 + a[n - 1] + a[n - 1]^2)/a[n - 2], a[1] == a[2] == 1}, a, {n, 1, 28}] (* or *)
%t A101368 RecurrenceTable[{a[n] == 5 a[n - 1] - a[n - 2] - 1, a[1] == a[2] == 1}, a, {n, 1, 28}] (* or *)
%t A101368 LinearRecurrence[{6, -6, 1}, {1, 1, 3}, 28] (* _Michael De Vlieger_, Aug 28 2016 *)
%o A101368 (PARI) Vec(x*(1-5*x+3*x^2)/((1-x)*(1-5*x+x^2)) + O(x^30)) \\ _Michel Marcus_, Aug 03 2016
%o A101368 (PARI) a(n)=([0,1,0;0,0,1;1,-6,6]^n*[3;1;1])[1,1] \\ _Charles R Greathouse IV_, Aug 28 2016
%o A101368 (Magma) [n le 2 select 1 else 5*Self(n-1)-Self(n-2)-1: n in [1..30]]; // _Vincenzo Librandi_, Dec 25 2018
%o A101368 (GAP) a:=[1,1];; for n in [3..30] do a[n]:=5*a[n-1]-a[n-2]-1; od; Print(a); # _Muniru A Asiru_, Dec 28 2018
%Y A101368 Cf. A001519, A004253, A061646, A276160.
%K A101368 nonn,easy
%O A101368 1,3
%A A101368 M. Benito, O. Ciaurri and E. Fernandez (oscar.ciaurri(AT)dmc.unirioja.es), Jan 13 2005