A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12*x, 5*x, 28*x, 9*x, 44*x, 13*x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).

0, 1, 12, 5, 28, 9, 44, 13, 60, 17, 76, 21, 92, 25, 108, 29, 124, 33, 140, 37, 156, 41, 172, 45, 188, 49, 204, 53, 220, 57, 236, 61, 252, 65, 268, 69, 284, 73, 300, 77, 316, 81, 332, 85, 348, 89, 364, 93, 380, 97, 396, 101, 412, 105, 428, 109, 444, 113, 460, 117, 476

Offset

0,3

Comments

Simple continued fraction expansion of 2*(e - 1)/(e + 1) = 2*tanh(1/2) = 1/(1 + 1/(12 + 1/(5 + 1/(28 + ...)))). - Peter Bala, Oct 01 2023

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Formula

G.f.: x*((1+3*x^2) + 4*x*(3+x^2))/(1-x^2)^2 = sum_{n>=0} a(n)*x^n.

From Carl R. White, Feb 11 2010: (Start)

a(n) = sign(n) * (2*n+1) * (3*cos(Pi*n)+5)/2.

a(2n+1) = a(2n-1) + 4, a(2n+2) = a(2n) + 16, with a(0)=0, a(1)=1, a(2)=12. (End)

a(n) = (5+3*(-1)^n)*(2*n-1)/2, with a(0)=0. Sum_{i=0..n} a(i) = A085787(A042948(n)). - Bruno Berselli, Jan 20 2012

Prog

(PARI) a(n)=polcoeff(x*(1+12*x+3*x^2+4*x^3)/(1-x^2)^2+x*O(x^n), n)

Crossrefs

Cf. continued fraction expansions: A004273 ( tanh(1) ), A204877 ( 3*tanh(1/3) ), A130824 ( tanh(1/2) ).

Sequence in context: A282578 A205141 A122561 * A299516 A278423 A278661

Adjacent sequences: A110182 A110183 A110184 * A110186 A110187 A110188

Keyword

nonn,cofr,easy

Author

Paul D. Hanna, Jul 14 2005

Status

approved