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Search: id:A076632
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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 values of y. |
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+0
2
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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
(list; graph; listen)
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OFFSET
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1,4
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COMMENT
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Euler (unpublished) showed there is a unique positive solution (x,y) for every positive n.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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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 (dean(AT)math.ucdavis.edu), Oct 19, 2002.
a(n) = (1/sqrt(7))*2^(n/2)*abs(sin(n*t))+1/2, where t=arctan(sqrt(7)). - Paul Boddington (psb(AT)maths.warwick.ac.uk), Jan 23 2004
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PROGRAM
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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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Cf. A076631.
Sequence in context: A026249 A130527 A026366 this_sequence A105646 A059126 A059128
Adjacent sequences: A076629 A076630 A076631 this_sequence A076633 A076634 A076635
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KEYWORD
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nonn,easy
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AUTHOR
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Ed Pegg Jr. (ed(AT)mathpuzzle.com), Oct 17 2002
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EXTENSIONS
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More terms from Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 24 2002
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