cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A242725 Sequence with all (x, y) = (a(2m), a(2m+-1)) satisfying x|y^2+y+1 and y|x^2+1.

Original entry on oeis.org

1, 1, 3, 10, 37, 137, 511, 1906, 7113, 26545, 99067, 369722, 1379821, 5149561, 19218423, 71724130, 267678097, 998988257, 3728274931, 13914111466, 51928170933, 193798572265, 723266118127, 2699265900242, 10073797482841, 37595924031121, 140309898641643
Offset: 0

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Author

Oboifeng Dira, May 21 2014

Keywords

Comments

a(n) with a(0)=1, a(1)=1, a(2)=3 is that two way sequence such that (a(n), a(n+1)) and (a(n), a(n-1)) for n even together with corresponding pairs of A242728 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 A242728.

Examples

			a(2)=3, a(3)=10 satisfy: 10 divides 3^2+1 and 3 divides 10^2+10+1.
		

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.

Crossrefs

Cf. A101368 (which considers a similar problem with x^2+x+1 mod y = 0 and y^2+y+1 mod x = 0).

Programs

  • Maple
    x0:=1: x1:=1: 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);
  • Mathematica
    Table[(3 + (-1)^n + (4 + Sqrt[3]) (2 - Sqrt[3])^n + (4 - Sqrt[3]) (2 + Sqrt[3])^n)/12, {n, 0, 30}] (* Bruno Berselli, May 25 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.
G.f.: (1 - 3*x - x^2 + 2*x^3)/((1 - x)*(1 + x)*(1 - 4*x + x^2)). [Bruno Berselli, May 22 2014]
a(n) = 4*a(n-1) - 4*a(n-3) + a(n-4) for n>3. [Bruno Berselli, May 22 2014]
a(n) = (3+(-1)^n+(4+sqrt(3))*(2-sqrt(3))^n+(4-sqrt(3))*(2+sqrt(3))^n)/12. [Bruno Berselli, May 25 2014]

Extensions

Edited by Bruno Berselli, May 25 2014
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