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A076632
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Solve 2^n - 2 = 7(x^2 - x) + (y^2 - y) for (x,y) with x>0, y>0; sequence gives value of x.
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1
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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
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OFFSET
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1,4
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COMMENTS
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Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.
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REFERENCES
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Engel, Problem-Solving Strategies.
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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) = (1/sqrt(7))*2^(n/2)*abs(sin(n*t))+1/2, where t=arctan(sqrt(7)). - Paul Boddington, Jan 23 2004
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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, x=1; while(frac(real(component(polroots(p(n, x, y)), 2)))>0, x++); x)
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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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STATUS
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approved
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