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Search: id:A101368
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| A101368 |
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The sequence solves the following problem: find all the pairs (i,j) such that i divides 1+j+j^2 and j divides 1+i+i^2. In fact, the pairs (a(n),a(n+1)), n>0, are all the solutions. |
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2
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| 1, 1, 3, 13, 61, 291, 1393, 6673, 31971, 153181, 733933, 3516483, 16848481, 80725921, 386781123, 1853179693, 8879117341, 42542407011, 203832917713, 976622181553
(list; graph; listen)
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OFFSET
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1,3
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REFERENCES
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W. W. Chao, Problem 2981, Crux Mathematicorum, 30 (2004), p. 430.
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FORMULA
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Recurrence: a(1)=a(2)=1 and a(n+1)=(1+a(n)+a(n)^2)/a(n-1) for n>2.
G.f.: x(1 - 5x + 3x^2) / [(1-x)(1 - 5x + x^2)]; a(n) = 2 * A089817(n-3) + 1, n>2. - Conjectured by Ralf Stephan, Jan 14 2005, proved by Max Alekseyev, Aug 03, 2006.
a(n) = 6a(n-1)-6a(n-2)+a(n-3), a(n) = 5a(n-1)-a(n-2)-1. - Floor en Lyanne van Lamoen (fvanlamoen(AT)planet.nl), Aug 01 2006
a(n) = (4/3 - (2/7)*sqrt(21))*((5 + sqrt(21))/2)^n + (4/3 + (2/7)*sqrt(21))*((5 - sqrt(21))/2)^n + 1/3. - Floor en Lyanne van Lamoen (fvanlamoen(AT)planet.nl), Aug 04 2006
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EXAMPLE
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a(5)=61 because (1+a(4)+a(4)^2)/a(3)=(1+13+169)/3=61
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CROSSREFS
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Cf. A001519.
Sequence in context: A112568 A104089 A108143 this_sequence A026704 A046748 A074548
Adjacent sequences: A101365 A101366 A101367 this_sequence A101369 A101370 A101371
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KEYWORD
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nonn
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AUTHOR
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M. Benito, O. Ciaurri and E. Fernandez (oscar.ciaurri(AT)dmc.unirioja.es), Jan 13 2005
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