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A001204
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Continued fraction for e^2.
(Formerly M4322 N1811)
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10
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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
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OFFSET
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0,1
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COMMENTS
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Note that e^2 = 7 + 2/(5 + 1/(7 + 1/(9 + 1/(11 + ...)))) (follows from the fact that A004273 is the continued fraction expansion of tanh(1) = (e^2 - 1)/(e^2 + 1)). - Peter Bala, Jan 15 2022
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REFERENCES
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O. Perron, Die Lehre von den Kettenbrüchen, 2nd ed., Teubner, Leipzig, 1929, p. 138.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Harry J. Smith, Table of n, a(n) for n = 0..20000
K. Matthews, Finding the continued fraction of e^(l/m)
G. Xiao, Contfrac
Index entries for continued fractions for constants
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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, 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, Mar 25 2003
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EXAMPLE
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7.389056098930650227230427460... = 7 + 1/(2 + 1/(1 + 1/(1 + 1/(3 + ...)))).
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MATHEMATICA
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ContinuedFraction[ E^2, 100]
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PROG
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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])); \\ Harry J. Smith, Apr 30 2009
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CROSSREFS
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Cf. A003417, A004273, A005131, A058282, A072334.
Sequence in context: A340617 A010505 A020844 * A177969 A021585 A103713
Adjacent sequences: A001201 A001202 A001203 * A001205 A001206 A001207
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KEYWORD
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nonn,easy,cofr,nice,changed
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AUTHOR
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N. J. A. Sloane
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EXTENSIONS
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More terms from Robert G. Wilson v, Dec 07 2000
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STATUS
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approved
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