A076631 Solve 2^n - 2 = 7(x^2 - x) + (y^2 - y) for (x,y) with x>0, y>0; a(n) = value of y.
1, 2, 3, 1, 6, 5, 7, 16, 3, 29, 34, 24, 91, 44, 138, 225, 51, 500, 399, 601, 1398, 197, 2599, 2992, 2206, 8189, 3778, 12600, 20155, 5045, 45354, 35265, 55443, 125972, 15087, 236857, 267030, 206684, 740743, 327376, 1154110, 1808861, 499359, 4117080
Offset: 1
References
- A. Engel, Problem-Solving Strategies, Springer-Verlag, New York, 1998.
Crossrefs
Cf. A076632 (values of x).
Programs
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Mathematica
Table[Reduce[{2^n-2==7(x^2-x)+(y^2-y),x>0,y>0},{x,y},Integers][[-1,-1]], {n,50}] (* Harvey P. Dale, Dec 15 2018 *) -
PARI
p(n,x,y)=2^n-2-7*(x^2-x)-(y^2-y) a(n)=if(n<0,0,y=1; while(frac(real(component(polroots(p(n,x,y)),2)))>0,y++); y)
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) = 2^(n/2)*abs(cos(n*t))+1/2, where t=arctan(sqrt(7)). - Paul Boddington, Jan 23 2004
Extensions
More terms from Benoit Cloitre, Oct 24 2002
More terms from Lambert Klasen (lambert.klasen(AT)gmx.de), Jan 14 2005
Definition corrected by Harvey P. Dale, Dec 15 2018
Comments