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 A267318 Continued fraction expansion of e^(1/5). 0

%I

%S 1,4,1,1,14,1,1,24,1,1,34,1,1,44,1,1,54,1,1,64,1,1,74,1,1,84,1,1,94,1,

%T 1,104,1,1,114,1,1,124,1,1,134,1,1,144,1,1,154,1,1,164,1,1,174,1,1,

%U 184,1,1,194,1,1,204,1,1,214,1,1,224,1,1,234,1,1,244,1,1,254,1,1,264,1,1

%N Continued fraction expansion of e^(1/5).

%C e^(1/5) is a transcendental number.

%C In general, the ordinary generating function for the continued fraction expansion of e^(1/k), with k = 1, 2, 3..., is (1 + (k - 1)*x + x^2 - (k + 1)*x^3 + 7*x^4 - x^5)/(1 - x^3)^2.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/eContinuedFraction.html">e Continued Fraction</a>

%H <a href="/index/Con#confC">Index entries for continued fractions for constants</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).

%F G.f.: (1 + 4*x + x^2 - x^3 + 6*x^4 - x^5)/(1 - x^3)^2.

%F a(n) = 1 + (3 + 10*floor(n/3))*(1 - (n-1)^2 mod 3). [_Bruno Berselli_, Feb 04 2016]

%e e^(1/5) = 1 + 1/(4 + 1/(1 + 1/(1 + 1/(14 + 1/(1 + 1/...))))).

%t ContinuedFraction[Exp[1/5], 82]

%t LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 4, 1, 1, 14, 1}, 82]

%t CoefficientList[Series[(1 + 4 x + x^2 - x^3 + 6 x^4 - x^5) / (x^3 - 1)^2, {x, 0, 70}], x] (* _Vincenzo Librandi_, Jan 13 2016 *)

%t Table[1 + (3 + 10 Floor[n/3]) (1 - Mod[(n - 1)^2, 3]), {n, 0, 90}] (* _Bruno Berselli_, Feb 04 2016 *)

%o (MAGMA) [1+(3+10*Floor(n/3))*(1-(n-1)^2 mod 3): n in [0..90]]; // _Bruno Berselli_, Feb 04 2016

%Y Cf. A092514.

%Y Cf. continued fraction expansion of e^(1/k): A003417 (k=1), A058281 (k=2), A078689 (k=3), A078688 (k=4), this sequence (k=5).

%K nonn,cofr,easy

%O 0,2

%A _Ilya Gutkovskiy_, Jan 13 2016

%E Edited by _Bruno Berselli_, Feb 04 2016

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Last modified January 17 05:26 EST 2019. Contains 319207 sequences. (Running on oeis4.)