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A001204 Continued fraction for e^2.
(Formerly M4322 N1811)
6
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
G. Xiao, Contfrac
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 2, 0, 0, 0, 0, -1).
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]
LinearRecurrence[{0, 0, 0, 0, 2, 0, 0, 0, 0, -1}, {7, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30}, 80] (* Harvey P. Dale, Dec 30 2023 *)
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
Sequence in context: A010505 A363360 A020844 * A177969 A021585 A103713
KEYWORD
nonn,easy,cofr,nice
AUTHOR
EXTENSIONS
More terms from Robert G. Wilson v, Dec 07 2000
STATUS
approved

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Last modified July 19 11:53 EDT 2024. Contains 374394 sequences. (Running on oeis4.)