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

1,3

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

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).

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

Example

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

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 *)

Prog

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

Crossrefs

Cf. A001519, A004253, A061646, A276160.

Sequence in context: A239995 A319924 A108143 * A341250 A026704 A046748

Adjacent sequences: A101365 A101366 A101367 * A101369 A101370 A101371

Keyword

nonn,easy

Author

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

Status

approved