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

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, 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, 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

Offset: 0

Author

Ilya Gutkovskiy, Jan 13 2016

Keywords

• nonn
• cofr
• easy

Comments

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

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.

Examples

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

Links

• Eric Weisstein's World of Mathematics, e Continued Fraction
• Index entries for continued fractions for constants
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Crossrefs

Cf. A092514.

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

Programs

Magma

[1+(3+10*Floor(n/3))*(1-(n-1)^2 mod 3): n in [0..90]]; // Bruno Berselli, Feb 04 2016

Mathematica

ContinuedFraction[Exp[1/5], 82]
LinearRecurrence[{0, 0, 2, 0, 0, -1}, {1, 4, 1, 1, 14, 1}, 82]
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 *)
Table[1 + (3 + 10 Floor[n/3]) (1 - Mod[(n - 1)^2, 3]), {n, 0, 90}] (* Bruno Berselli, Feb 04 2016 *)

Formula

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

a(n) = 1 + (3 + 10*floor(n/3))*(1 - (n-1)^2 mod 3). [Bruno Berselli, Feb 04 2016]

Extensions

Edited by Bruno Berselli, Feb 04 2016