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A modified Engel expansion of exp(1).
5

%I #16 Jul 26 2021 02:06:35

%S 1,1,2,3,4,5,8,2,10,99,20,2,2,2,2,2,2,3,6,4,8,14,2,2,4,6,10,252,81,30,

%T 28,31,60,4,6,3,4,2,2,2,2,19,54,8,6,22,63,4,2,4,6,2,2,5,12,4,2,2,2,2,

%U 6,15,10,348,172,2,2,4,6,4,30,207,220

%N A modified Engel expansion of exp(1).

%C See A220393 for a description of the modified Engel expansion of a positive real number. For further details see the Bala link.

%C The Engel expansion for exp(1) is the sequence of positive integers A000027.

%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) where x = exp(1) - 2, then a(1) = a(2) = 1, a(3) = ceiling(1/x) and, for n >= 1, a(n+3) = floor(1/h^(n-1)(x))*(1 + floor(1/h^(n)(x))).

%F Put P(n) = Product_{k = 1..n} a(k). Then we have the Egyptian fraction series expansion exp(1) = Sum_{n>=1} 1/P(n) = 1/1 + 1/1 + 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*5) + 1/(2*3*4*5*8) + .... For n >= 3, the error made in truncating this series to n terms is less than the n-th term.

%Y Cf. A000027, A220335, A220336, A220337, A220338, A220393, A220395, A220396, A220397, A220398.

%K nonn,easy

%O 1,3

%A _Peter Bala_, Dec 13 2012