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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 OFFSET 1,4 COMMENTS Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n. REFERENCES Engel, Problem-Solving Strategies.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 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

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Last modified January 20 21:49 EST 2021. Contains 340332 sequences. (Running on oeis4.)