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%I #16 Jul 26 2021 01:15:38
%S 1,2,5,8,3,4,4,6,2,162,322,2,51842,103682,2,5374978562,10749957122,2,
%T 57780789062419261442,115561578124838522882,2,
%U 6677239169351578707225356193679818792962,13354478338703157414450712387359637585922,2
%N A modified Engel expansion of the golden ratio (1/2)*(1 + sqrt(5)) (A001622).
%C See A220393 for a definition of the modified Engel expansion of a positive real number. For further details see the Bala link.
%H Peter Bala, <a href="/A220393/a220393.pdf">A modified Engel expansion</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], 2012-2020
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%F Let h(x) = x*(floor(1/x) + (floor(1/x))^2) - floor(1/x). Let x = 1/2*(1 + sqrt(5)) - 1. Then a(1) = 1, a(2) = ceiling(1/x) and, for n >= 1, a(n+2) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).
%F Recurrence equations: For n >= 3, a(3*n) = 2. For n >= 4 we have a(3*n+2) = 2*a(3*n+1) - 2 and a(3*n+1) = 2*(a(3*n-2) - 1)^2.
%F Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion sqrt(2) = Sum_{n>=1} 1/P(n) = 1 + 1/2 + 1/(2*5) + 1/(2*5*8) + 1/(2*5*8*3) + 1/(2*5*8*3*4) + .... For n >= 2, the error made in truncating this series to n terms is less than the n-th term.
%Y Cf. A001622, A028259, A220335, A220336, A220337, A220338, A220393, A220394, A220395, A220396, A220397.
%K nonn,easy
%O 1,2
%A _Peter Bala_, Dec 13 2012
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