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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.

%I A076632 #21 May 08 2019 07:11:00
%S A076632 1,1,1,2,1,3,4,2,9,6,12,23,1,46,45,47,136,43,229,314,144,771,484,1058,
%T A076632 2025,91,4140,3959,4321,12238,3597,20879,28072,13686,69829,42458,
%U A076632 97200,182115,12285,376514,351945,401083,1104972,302807
%N 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.
%C A076632 Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.
%D A076632 Engel, Problem-Solving Strategies.
%F A076632 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
%F A076632 a(n) = (1/sqrt(7))*2^(n/2)*abs(sin(n*t))+1/2, where t=arctan(sqrt(7)). - _Paul Boddington_, Jan 23 2004
%F A076632 a(n) = (1+2*A077020(n+2))/2. - _R. J. Mathar_, May 08 2019
%o A076632 (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)
%Y A076632 Cf. A076631 (values of y).
%K A076632 nonn,easy
%O A076632 1,4
%A A076632 _Ed Pegg Jr_, Oct 17 2002
%E A076632 More terms from _Benoit Cloitre_, Oct 24 2002
%E A076632 Definition corrected by _Harvey P. Dale_, Dec 15 2018