This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A220336 #15 Jun 19 2025 03:25:57
%S A220336 2,4,6,2,18,34,2,578,1154,2,665858,1331714,2,886731088898,
%T A220336 1773462177794,2,1572584048032918633353218,3145168096065837266706434,
%U A220336 2,4946041176255201878775086487573351061418968498178,9892082352510403757550172975146702122837936996354
%N A220336 A modified Engel expansion for 4*sqrt(2) - 5.
%C A220336 For a brief description of the modified Engel expansion of a real number see A220335.
%C A220336 Let p >= 2 be an integer and set Q(p) = (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1), so Q(3) = 4*sqrt(2) - 5. Iterating the identity Q(p) = 1/2 + 1/(2*(p+1)) + 1/(2*(p+1)*(2*p)) + 1/(2*(p+1)*(2*p))*Q(2*p^2-1) leads to a representation for Q(p) as an infinite series of unit fractions. The sequence of denominators of these unit fractions can be used to find the modified Engel expansion of Q(p). For further details see the Bala link. The present sequence is the case p = 3. For other cases see A220335 (p = 2), A220337 (p = 4) and A220338 (p = 5).
%H A220336 Peter Bala, <a href="/A220335/a220335.pdf">A modified Engel expansion for certain quadratic irrationals</a>
%H A220336 Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%F A220336 Define the map h(x) := floor(1/x)*(x*ceiling(1/x) - 1). Let x = 4*sqrt(2) - 5. Then a(1) = ceiling(1/x) and for n >= 2, a(n) = floor(1/h^(n-2)(x))*ceiling(1/h^(n-1)(x)), where h^(n)(x) denotes the n-th iterate of the map h(x), with the convention h^(0)(x) = x.
%F A220336 a(3*n+2) = 1/2*{2 + (1+sqrt(2))^(2^(n+1)) + (1-sqrt(2))^(2^(n+1))},
%F A220336 a(3*n+3) = {(1 + sqrt(2))^(2^(n+1)) + (1 - sqrt(2))^(2^(n+1))}, both
%F A220336 for n >= 0.
%F A220336 For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*A001601(n)^2 and a(3*n+3) = 4*A001601(n)^2 - 2.
%F A220336 Recurrence equations:
%F A220336 For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
%F A220336 a(3*n+3) = 2*a(3*n+2) - 2.
%F A220336 Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation 4*sqrt(2) - 5 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*4) + 1/(2*4*6) + 1/(2*4*6*2) + 1/(2*4*6*2*18) + ....
%Y A220336 Cf. A001601, A028257, A220335 (p = 2), A220337 (p = 4), A220338 (p = 5).
%K A220336 nonn,easy
%O A220336 1,1
%A A220336 _Peter Bala_, Dec 12 2012