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Search: id:A001204
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| A001204 |
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Continued fraction for e^2.
(Formerly M4322 N1811)
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+0
1
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| 7, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, 12, 54, 14, 1, 1, 15, 66, 17, 1, 1, 18, 78, 20, 1, 1, 21, 90, 23, 1, 1, 24, 102, 26, 1, 1, 27, 114, 29, 1, 1, 30, 126, 32, 1, 1, 33, 138, 35, 1, 1, 36, 150, 38, 1, 1, 39, 162, 41, 1, 1, 42, 174, 44, 1, 1, 45, 186, 47, 1, 1
(list; graph; listen)
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OFFSET
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0,1
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REFERENCES
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N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
O. Perron, Die Lehre von den Kettenbr\"{u}chen, 2nd ed., Teubner, Leipzig, 1929, p. 138.
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LINKS
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Harry J. Smith, Table of n, a(n) for n=0,...,20000
G. Xiao, Contfrac
Index entries for continued fractions for constants
K. Matthews, Finding the continued fraction of e^(l/m)
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FORMULA
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G.f.: (x^10-x^8-x^7+x^6+4x^5+3x^4+x^3+x^2+2x+7)/(x^5-1)^2. - Ralf Stephan (ralf(AT)ark.in-berlin.de), Mar 23 2003
For n>0, a(5n)=12n+6, a(5n+1)=3n+2, a(5n+2)=a(5n+3)=1 and a(5n+4)=3n+3. - Dean Hickerson (dean.hickerson(AT)yahoo.com), Mar 25 2003
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EXAMPLE
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7.389056098930650227230427460... = 7 + 1/(2 + 1/(1 + 1/(1 + 1/(3 + ...)))) [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Apr 30 2009]
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MATHEMATICA
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ContinuedFraction[ E^2, 100]
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PROGRAM
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(PARI) contfrac(exp(2))
(PARI) { allocatemem(932245000); default(realprecision, 95000); x=contfrac(exp(2)); for (n=1, 20001, write("b001204.txt", n-1, " ", x[n])); } [From Harry J. Smith (hjsmithh(AT)sbcglobal.net), Apr 30 2009]
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CROSSREFS
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Sequence in context: A153589 A010505 A020844 this_sequence A021585 A103713 A089129
Adjacent sequences: A001201 A001202 A001203 this_sequence A001205 A001206 A001207
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KEYWORD
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easy,nonn,cofr,nice
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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 Robert G. Wilson v (rgwv(AT)rgwv.com), Dec 07 2000
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