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%I #30 Oct 06 2023 10:29:31
%S 0,1,12,5,28,9,44,13,60,17,76,21,92,25,108,29,124,33,140,37,156,41,
%T 172,45,188,49,204,53,220,57,236,61,252,65,268,69,284,73,300,77,316,
%U 81,332,85,348,89,364,93,380,97,396,101,412,105,428,109,444,113,460,117,476
%N 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).
%C 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
%H Vincenzo Librandi, <a href="/A110185/b110185.txt">Table of n, a(n) for n = 0..1000</a>
%H D. H. Lehmer, <a href="/A016825/a016825.pdf">Continued fractions containing arithmetic progressions</a>, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,-1).
%F G.f.: x*((1+3*x^2) + 4*x*(3+x^2))/(1-x^2)^2 = sum_{n>=0} a(n)*x^n.
%F From _Carl R. White_, Feb 11 2010: (Start)
%F a(n) = sign(n) * (2*n+1) * (3*cos(Pi*n)+5)/2.
%F a(2n+1) = a(2n-1) + 4, a(2n+2) = a(2n) + 16, with a(0)=0, a(1)=1, a(2)=12. (End)
%F 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
%o (PARI) a(n)=polcoeff(x*(1+12*x+3*x^2+4*x^3)/(1-x^2)^2+x*O(x^n),n)
%Y Cf. continued fraction expansions: A004273 ( tanh(1) ), A204877 ( 3*tanh(1/3) ), A130824 ( tanh(1/2) ).
%K nonn,cofr,easy
%O 0,3
%A _Paul D. Hanna_, Jul 14 2005
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