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.

1, 1, 3, 13, 61, 291, 1393, 6673, 31971, 153181, 733933, 3516483, 16848481, 80725921, 386781123, 1853179693, 8879117341, 42542407011, 203832917713, 976622181553, 4679277990051, 22419767768701, 107419560853453, 514678036498563, 2465970621639361, 11815175071698241, 56609904736851843, 271234348612560973

Offset: 1

Author

M. Benito, O. Ciaurri and E. Fernandez (oscar.ciaurri(AT)dmc.unirioja.es), Jan 13 2005

Keywords

• nonn
• easy

Comments

Also, integers m such that 21*(3*m-1)^2 - 48 is a square. - Max Alekseyev, May 23 2022

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

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

Examples

			a(5) = 61 because (1 + a(4) + a(4)^2)/a(3) = (1 + 13 + 169)/3 = 61.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..1471
• Esther Banaian and Archan Sen, A Generalization of Markov Numbers, arXiv:2210.07366 [math.CO], 2022. See Table 1 p. 12.
• T. Cai, Z. Shen and L. Jia, A congruence involving harmonic sums modulo p^alpha q^beta, arXiv preprint arXiv:1503.02798 [math.NT], 2015.
• W. W. Chao, Problem 2981, Crux Mathematicorum, 30 (2004), p. 430.
• Yasuaki Gyoda, Positive integer solutions to (x+y)^2+(y+z)^2+(z+x)^2=12xyz, arXiv:2109.09639 [math.NT], 2021. See Remark 3.3 p. 6.
• W. H. Mills, A system of quadratic Diophantine equations. Pacific J. Math. 3:1 (1953), 209-220.
• Index entries for linear recurrences with constant coefficients, signature (6,-6,1).

Crossrefs

Cf. A001519, A004253, A061646, A276160.

Programs

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

Magma

[n le 2 select 1 else 5*Self(n-1)-Self(n-2)-1: n in [1..30]]; // Vincenzo Librandi, Dec 25 2018

Maple

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

Mathematica

Rest@ CoefficientList[Series[x (1 - 5 x + 3 x^2)/((1 - x) (1 - 5 x + x^2)), {x, 0, 28}], x] (* or *)
RecurrenceTable[{a[n] == (1 + a[n - 1] + a[n - 1]^2)/a[n - 2], a[1] == a[2] == 1}, a, {n, 1, 28}] (* or *)
RecurrenceTable[{a[n] == 5 a[n - 1] - a[n - 2] - 1, a[1] == a[2] == 1}, a, {n, 1, 28}] (* or *)
LinearRecurrence[{6, -6, 1}, {1, 1, 3}, 28] (* Michael De Vlieger, Aug 28 2016 *)

PARI

Vec(x*(1-5*x+3*x^2)/((1-x)*(1-5*x+x^2)) + O(x^30)) \\ Michel Marcus, Aug 03 2016

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

Formula

Recurrence: a(1)=a(2)=1 and a(n+1)=(1+a(n)+a(n)^2)/a(n-1) for n>2.

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

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

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

For n>1, a(n) = (2 * A004253(n-1) + 1) / 3. - Max Alekseyev, May 23 2022