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A076631
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Solve 2^n - 2 = 7(x^2 - x) + (y^2 - y) for (x,y) with x>0, y>0; a(n) = value of y.
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1
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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
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OFFSET
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1,2
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COMMENTS
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Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.
Proof of uniqueness: Let R be the ring of integers of Q(sqrt(-7)) and let a=(1+sqrt(-7))/2, b=(1-sqrt(-7)/2. It is easy to see that any element of R of even norm (=squared absolute value) can be divided by one of a or b to get back an element of R. Thus since ab=2, the only elements in R of norm 2^n and of the form (p+q*sqrt(-7))/2 with p,q odd are a^n, b^n, -a^n, -b^n - precisely one of which lies in the first quadrant. Finally apply Dean Hickerson's remarks. - Paul Boddington, Jan 23 2004
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REFERENCES
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A. Engel, Problem-Solving Strategies, Springer-Verlag, New York, 1998.
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LINKS
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FORMULA
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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
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MATHEMATICA
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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 *)
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PROG
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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, y=1; while(frac(real(component(polroots(p(n, x, y)), 2)))>0, y++); y)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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More terms from Lambert Klasen (lambert.klasen(AT)gmx.de), Jan 14 2005
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STATUS
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approved
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