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
References
- Engel, Problem-Solving Strategies.
Crossrefs
Cf. A076631 (values of y).
Programs
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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)
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
Extensions
More terms from Benoit Cloitre, Oct 24 2002
Definition corrected by Harvey P. Dale, Dec 15 2018
Comments