login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

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.
14
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
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.
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
KEYWORD
nonn,easy
AUTHOR
M. Benito, O. Ciaurri and E. Fernandez (oscar.ciaurri(AT)dmc.unirioja.es), Jan 13 2005
STATUS
approved