%I
%S 2,3,4,2,8,14,2,98,194,2,18818,37634,2,708158978,1416317954,2,
%T 1002978273411373058,2005956546822746114,2,
%U 2011930833870518011412817828051050498,4023861667741036022825635656102100994
%N A modified Engel expansion for sqrt(3)  1.
%C This is the case p = 2 of a family of quadratic irrationals of the form (p  1)*sqrt(p^2  1)  (p^2  p  1) whose modified Engel expansion, as defined below, has a predictable form. For other cases see A220336 (p = 3), A220337 (p = 4) and A220338 (p = 5).
%C The Engel expansion of a positive real number x in the halfopen interval (0,1] is the unique nondecreasing sequence {e(1), e(2), e(3), ...} of positive integers such that x = 1/e(1) + 1/(e(1)*e(2)) + 1/(e(1)*e(2)*e(3)) + .... The terms in the Engel expansion of x are obtained from the iterates of the map g(x) = x*(1 + floor(1/x))  1 by means of the formula e(n) = 1 + floor(1/g^(n1)(x)). Here g^(n)(x) = g(g^(n1)(x)) denotes the nth iterate of g(x) with the convention g^(0)(x) = x.
%C In a similar way, the modified Engel expansion of x belonging to (0,1] is a sequence {E(1), E(2), E(3), ...} of positive integers such that x = 1/E(1) + 1/(E(1)*E(2)) + 1/(E(1)*E(2)*E(3)) + ... whose terms are obtained from the iterates of the harmonic sawtooth map h(x) = floor(1/x)*g(x). The general formula is E(1) = 1 + floor(1/x) and for n >= 1, E(n) = floor(1/h^(n2)(x))*(1 + floor(1/h^(n1)(x))). For further details see the Bala link.
%H P. Bala, <a href="/A220335/a220335.pdf">A modified Engel expansion for certain quadratic irrationals</a>
%H S. Crowley, <a href="http://arxiv.org/abs/1210.5652">Integral transforms of the harmonic sawtooth map, the Riemann zeta function, fractal strings, and a finite reflection formula</a>, arXiv:1210.5652 [math.NT]
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%F Let x = sqrt(3)  1. Then a(1) = ceiling(1/x) and for n >= 2, a(n) = floor(1/h^(n2)(x))*ceiling(1/h^(n1)(x)), where h^(n)(x) denotes the nth iterate of the harmonic sawtooth map h(x), with the convention h^(0)(x) = x.
%F a(3*n+2) = 1/2*{2 + (2 + sqrt(3))^(2^n) + (2  sqrt(3))^(2^n)} and
%F a(3*n+3) = (2 + sqrt(3))^(2^n) + (2  sqrt(3))^(2^n), both for n >= 0.
%F For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*(A002812(n1))^2 and a(3*n+3) = 4*(A002812(n1))^2  2.
%F Recurrence equations:
%F For n >= 1, a(3*n+2) = 2*{a(3*n1)^2  2*a(3*n1) + 1} and
%F a(3*n+3) = 2*a(3*n+2)  2.
%F Put P(n) = product(k = 1..n} a(k). Then we have the infinite Egyptian fraction representation sqrt(3)  1 = sum {n >=1} 1/P(n) = 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*2) + ....
%Y Cf. A002812, A220336 (p = 3), A220337 (p = 4), A220338 (p = 5).
%K nonn,easy
%O 1,1
%A _Peter Bala_, Dec 12 2012
