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A242728
Sequence a(n) with all (x,y)=(a(2m),a(2m+-1)) satisfying y|x^2+1 and x|y^2+y+1.
2
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.
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
KEYWORD
nonn,easy
AUTHOR
Oboifeng Dira, May 21 2014
STATUS
approved