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Search: id:A001068
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| 0, 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 65, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 83, 85, 86, 87, 88
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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Contribution from M. F. Hasler (MHasler(AT)univ-ag.fr), Oct 21 2008: (Start)
Also, for n>0, the 4th term (after [0,n,3n]) in the continued fraction expansion of arctan(1/n). (Observation by V. Reshetnikov.) Proof:
arctan(1/n) = (1/n) / (1 + (1/n)^2/( 3 + (2/n)^2/( 5 + (3/n)^2/( 7 + ...)...)
= 1 / ( n + 1/( 3n + 4/( 5n + 9/( 7n + 25/(...)...)
= 1 / ( n + 1/( 3n + 1/( 5n/4 + (9/4)/( 7n + 25/(...)...),
and the term added to 5n/4, (9/4)/(7n+...)=(1/4)*9/(7n+...) is less than 1/4 for all n>=2. (End)
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REFERENCES
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P. Erdos, Some recent problems and results in graph theory, Discr. Math., 164 (1997), 81-85.
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LINKS
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Wikipedia: Continued fraction for arctangent. [From M. F. Hasler (MHasler(AT)univ-ag.fr), Oct 21 2008]
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FORMULA
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Numbers that are congruent to {0, 1, 2, 3} mod 5.
contfrac( arctan( 1/n )) = 0 + 1/( n + 1/( 3n + 1/( a(n) + 1/(...)))) [From M. F. Hasler (MHasler(AT)univ-ag.fr), Oct 21 2008]
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PROGRAM
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(PARI) a(n)=5*n\4 /* or, cf. comment: */ a(n)=contfrac(atan(1/n))[4] [From M. F. Hasler (MHasler(AT)univ-ag.fr), Oct 21 2008]
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CROSSREFS
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Cf. A110255-A110260. [From M. F. Hasler (MHasler(AT)univ-ag.fr), Oct 21 2008]
Sequence in context: A039193 A028798 A138309 this_sequence A039145 A038129 A062071
Adjacent sequences: A001065 A001066 A001067 this_sequence A001069 A001070 A001071
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KEYWORD
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nonn,easy
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
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More terms from James A. Sellers (sellersj(AT)math.psu.edu), Sep 19 2000
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