A242728 Sequence a(n) with all (x,y)=(a(2m),a(2m+-1)) satisfying y|x^2+1 and x|y^2+y+1.

1, 2, 7, 25, 93, 346, 1291, 4817, 17977, 67090, 250383, 934441, 3487381, 13015082, 48572947, 181276705, 676533873, 2524858786, 9422901271, 35166746297, 131244083917, 489809589370, 1827994273563, 6822167504881, 25460675745961, 95020535478962

Offset

0,2

Comments

a(n) with a(1)=2, a(2)=7 is that two-way sequence such that (a(n),a(n+1)) and (a(n),a(n-1)) for n even together with the corresponding pairs of A242725 give all solutions of the two congruences x^2+1 mod y = 0 and y^2+y+1 mod x = 0. The negative part b(n) = a(-n) is given in sequence A242725.

References

T. Bier, Classifications of solutions of certain positive biquadratic division system, submitted May 12 2014.

T. Bier and O. Dira, Construction of integer sequences, submitted May 12 2014.

Links

Table of n, a(n) for n=0..25.

Oboifeng Dira, Sequences solving division systems

Index entries for linear recurrences with constant coefficients, signature (4,0,-4,1).

Formula

a(n+1) = 4*a(n) - a(n-1) - p_n (n>0), where p_n=0 if n is odd and p_n = 1 if n is even.

a(n) = 4*a(n-1) - 4*a(n-3) + a(n-4). - Colin Barker, May 21 2014

G.f.: -(x^3-x^2-2*x+1) / ((x-1)*(x+1)*(x^2-4*x+1)). - Colin Barker, May 21 2014

a(n) = (1/12) * (2*A077136(n) + (-1)^n + 3). - Ralf Stephan, May 24 2014

Example

Considering the pair a(1)=2 and a(2)=7, 2 divides 7^2+1 and 7 divides 2^2+2+1.

Maple

x0:=1: x1:=2: L:=[x0, x1]: for k from 1 to 30 do:if k mod 2 = 1 then z:=4*x1-x0: fi: if k mod 2 = 0 then z:=4*x1-x0-1: fi: L:=[op(L), z]: x0:=x1: x1:=z: od: print(L);

Prog

(PARI) Vec(-x*(x^3-x^2-2*x+1)/((x-1)*(x+1)*(x^2-4*x+1)) + O(x^100)) \\ Colin Barker, May 21 2014

Crossrefs

A101368 gives a similar problem with x^2+x+1 mod y = 0 and y^2+y+1 mod x = 0.

Sequence in context: A108081 A270785 A199247 * A116396 A150453 A150454

Adjacent sequences: A242725 A242726 A242727 * A242729 A242730 A242731

Keyword

nonn,easy

Author

Oboifeng Dira, May 21 2014

Status

approved