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%I #58 Apr 22 2020 16:14:29
%S 0,1,2,2,-8,-56,-112,848,9088,25216,-310528,-4334848,-14701568,
%T 270029824,4554426368,17536821248,-458243735552,-8926669144064,
%U -37024075153408,1341521605885952,29290212127670272,125297096967061504,-6224109737622372352
%N E.g.f. is the logarithmic derivative of e.g.f. for Pell numbers [1, 0, 1, 2, 5, ...].
%C It looks like the signs have a repeating pattern of +, +, +, -, -, -, etc.; however, a(42) breaks this pattern by being positive after a string of only two negative terms. The sequence oscillates because it has two dominant singularities; the oscillation is irregular and unpredictable because the arguments of the singularities are not rational multiples of Pi; the oscillation first appears regular with period 6 because the arguments of the singularities are +- 1.01*Pi/3, which is very close to 1/6 of the circle. All of this can be proved with the standard techniques of analytic combinatorics (see the Flajolet and Sedgewick reference). - _Justin M. Troyka_, Jun 20 2019
%D P. Flajolet and R. Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, Cambridge, 2009, pages 258-259 ("Expansion of meromorphic functions") and 264-266 ("Nonperiodic fluctuations").
%H Justin M. Troyka, <a href="https://arxiv.org/abs/1907.06681">Period mimicry: A note on the (-1)-evaluation of the peak polynomials</a>, arXiv:1907.06681 [math.CO], 2019.
%F G.f.: 1-2/Q(0) where Q(k) = 1 + 1/(1 + 2*(k+1)/(-1/x +(2*k+2)/Q(k+1))); (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Sep 23 2012
%F G.f.: -1/Q(0), where Q(k) = 2*k+2 - 1/x + (k+1)*(k+2)/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, Apr 15 2013
%F G.f.: 1/Q(0), where Q(k)= 1/(x*(k+1)) - 2 + 1/Q(k+1); (continued fraction). - _Sergei N. Gladkovskii_, May 07 2013
%F G.f.: x/Q(0), m=-2, where Q(k) = 1 - 2*x*(2*k+1) - m*x^2*(k+1)*(2*k+1)/( 1 - 2*x*(2*k+2) - m*x^2*(k+1)*(2*k+3)/Q(k+1) ) ; (continued fraction). - _Sergei N. Gladkovskii_, Sep 24 2013
%F E.g.f.: 2*x/Q(0), where Q(k) = 8*k+2 - 2*x/(1 + 2*x/(4*k+3 - 2*x/(1 + 2*x/Q(k+1) ))); (continued fraction). - _Sergei N. Gladkovskii_, Dec 19 2013
%F E.g.f.: sqrt(2)/(sqrt(2) - tanh(sqrt(2)*x)) -1 = W(0) -1, where W(k) = 1 + x/(4*k+1 - 1*x/(1 + 2*x/(4*k+3 - 2*x/W(k+1) ))); (continued fraction). - _Sergei N. Gladkovskii_, Nov 18 2014
%F a(n) ~ (-2) cos((n+1)*t) R^(-(n+1)) n!, where t = arctan(y/x) and R = sqrt(x^2+y^2), where x = log(3+2*sqrt(2))/(2*sqrt(2)) and y = Pi/(2*sqrt(2)) (note that R is approximately 1.274 and t is approximately 1.012*Pi/3). This formula follows from the standard techniques of analytic combinatorics (see the Flajolet and Sedgewick reference). - _Justin M. Troyka_, Jun 20 2019
%F a(n+1) = 2*a(n) - Sum_{k=1..n-1} binomial(n,k) * a(k) * a(n-k) if n>=1. - _Michael Somos_, Apr 22 2020
%e G.f. = x + 2*x^2 + 2*x^3 - 8*x^4 - 56*x^5 - 112*x^6 + 848*x^7 + 9088*x^8 + ...
%p # After Sergei N. Gladkovskii.
%p seq(k!*coeff(series(1/(sqrt(2)*coth(sqrt(2)*x)-1), x=0, k+2), x, k), k=0..21); # _Peter Luschny_, Nov 18 2014
%t a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1 / (1 - Tanh[ x Sqrt[2]] / Sqrt[2]) - 1, {x, 0, n}]]; (* _Michael Somos_, Nov 22 2014 *)
%o (PARI) {a(n) = my(w=quadgen(8)); if( n<0, 0, n! * polcoeff( 1 / (1 - tanh(w*x + x * O(x^n)) / w) - 1, n))}; /* _Michael Somos_, Nov 22 2014 */
%o (PARI) {a(n) = n--; if(n < 1, n==0, 2*a(n) - sum(k=1, n-1, binomial(n, k) * a(k) * a(n-k)))}; /* _Michael Somos_, Apr 22 2020 */
%Y Cf. A000129.
%K sign
%O 0,3
%A _N. J. A. Sloane_
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