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 A001204 Continued fraction for e^2. (Formerly M4322 N1811) 10
 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; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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 REFERENCES 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). LINKS 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 FORMULA 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 EXAMPLE 7.389056098930650227230427460... = 7 + 1/(2 + 1/(1 + 1/(1 + 1/(3 + ...)))). MATHEMATICA ContinuedFraction[ E^2, 100] PROG (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 CROSSREFS Cf. A003417, A004273, A005131, A058282, A072334. Sequence in context: A340617 A010505 A020844 * A177969 A021585 A103713 Adjacent sequences:  A001201 A001202 A001203 * A001205 A001206 A001207 KEYWORD nonn,easy,cofr,nice,changed AUTHOR EXTENSIONS More terms from Robert G. Wilson v, Dec 07 2000 STATUS approved

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Last modified January 24 00:05 EST 2022. Contains 350515 sequences. (Running on oeis4.)