A076632 Solve 2^n - 2 = 7(x^2 - x) + (y^2 - y) for (x,y) with x>0, y>0; sequence gives value of x.

1, 1, 1, 2, 1, 3, 4, 2, 9, 6, 12, 23, 1, 46, 45, 47, 136, 43, 229, 314, 144, 771, 484, 1058, 2025, 91, 4140, 3959, 4321, 12238, 3597, 20879, 28072, 13686, 69829, 42458, 97200, 182115, 12285, 376514, 351945, 401083, 1104972, 302807

Offset

1,4

Comments

Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.

References

Engel, Problem-Solving Strategies.

Links

Table of n, a(n) for n=1..44.

Formula

Note that the equation is equivalent to 2^(n+2) = (2y-1)^2 + 7 (2x-1)^2, so it is related to norms of elements of the ring of integers in the quadratic field Q(sqrt(-7)) and Euler's claim presumably follows from unique factorization in that field. From this we can get a formula for the x's and y's: Let a(n) and b(n) be the unique rational numbers such that a(n) + b(n) sqrt(-7) = ((1 + sqrt(-7))/2)^n. I.e., a(n) = (((1 + sqrt(-7))/2)^n + ((1 - sqrt(-7))/2)^n)/2. - Dean Hickerson, Oct 19 2002

a(n) = (1/sqrt(7))*2^(n/2)*abs(sin(n*t))+1/2, where t=arctan(sqrt(7)). - Paul Boddington, Jan 23 2004

a(n) = (1+2*A077020(n+2))/2. - R. J. Mathar, May 08 2019

Prog

(PARI) p(n, x, y)=2^n-2-7*(x^2-x)-(y^2-y) a(n)=if(n<0, 0, x=1; while(frac(real(component(polroots(p(n, x, y)), 2)))>0, x++); x)

Crossrefs

Cf. A076631 (values of y).

Sequence in context: A122164 A210793 A281715 * A105646 A059126 A059128

Adjacent sequences: A076629 A076630 A076631 * A076633 A076634 A076635

Keyword

nonn,easy

Author

Ed Pegg Jr, Oct 17 2002

Extensions

More terms from Benoit Cloitre, Oct 24 2002

Definition corrected by Harvey P. Dale, Dec 15 2018

Status

approved