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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, 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 (list; graph; refs; listen; history; text; internal format)
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Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.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
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
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

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Last modified March 19 04:58 EDT 2024. Contains 370952 sequences. (Running on oeis4.)