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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 x.
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.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 (psb(AT)maths.warwick.ac.uk), Jan 23 2004
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REFERENCES
| A. Engel, Problem-Solving Strategies, Springer-Verlag, New York, 1998.
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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 (dean.hickerson(AT)yahoo.com), Oct 19, 2002.
a(n)=2^(n/2)*abs(cos(n*t))+1/2, where t=arctan(sqrt(7)). - Paul Boddington (psb(AT)maths.warwick.ac.uk), Jan 23 2004
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PROG
| (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
| Cf. A076632.
Sequence in context: A116468 A110237 A189970 * A035485 A074306 A036039
Adjacent sequences: A076628 A076629 A076630 * A076632 A076633 A076634
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KEYWORD
| nonn,easy
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AUTHOR
| Ed Pegg Jr. (ed(AT)mathpuzzle.com), Oct 17 2002
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EXTENSIONS
| More terms from Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 24 2002
More terms from Lambert Klasen (lambert.klasen(AT)gmx.de), Jan 14 2005
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