cp's OEIS Frontend

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

%I A267318 #27 Feb 16 2025 08:33:29
%S A267318 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 A267318 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 A267318 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 A267318 Continued fraction expansion of e^(1/5).
%C A267318 e^(1/5) is a transcendental number.
%C A267318 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 A267318 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/eContinuedFraction.html">e Continued Fraction</a>
%H A267318 <a href="/index/Con#confC">Index entries for continued fractions for constants</a>
%H A267318 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).
%F A267318 G.f.: (1 + 4*x + x^2 - x^3 + 6*x^4 - x^5)/(1 - x^3)^2.
%F A267318 a(n) = 1 + (3 + 10*floor(n/3))*(1 - (n-1)^2 mod 3). [_Bruno Berselli_, Feb 04 2016]
%e A267318 e^(1/5) = 1 + 1/(4 + 1/(1 + 1/(1 + 1/(14 + 1/(1 + 1/...))))).
%t A267318 ContinuedFraction[Exp[1/5], 82]
%t A267318 LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 4, 1, 1, 14, 1}, 82]
%t A267318 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 A267318 Table[1 + (3 + 10 Floor[n/3]) (1 - Mod[(n - 1)^2, 3]), {n, 0, 90}] (* _Bruno Berselli_, Feb 04 2016 *)
%o A267318 (Magma) [1+(3+10*Floor(n/3))*(1-(n-1)^2 mod 3): n in [0..90]]; // _Bruno Berselli_, Feb 04 2016
%Y A267318 Cf. A092514.
%Y A267318 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 A267318 nonn,cofr,easy
%O A267318 0,2
%A A267318 _Ilya Gutkovskiy_, Jan 13 2016
%E A267318 Edited by _Bruno Berselli_, Feb 04 2016