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 A220338 #16 Jun 19 2025 03:25:48
%S A220338 2,6,10,2,50,98,2,4802,9602,2,46099202,92198402,2,4250272665676802,
%T A220338 8500545331353602,2,36129635465198759610694779187202,
%U A220338 72259270930397519221389558374402,2,2610701117696295981568349760414651575095962187244375364404428802
%N A220338 A modified Engel expansion for 8*sqrt(6) - 19.
%C A220338 For a brief description of the modified Engel expansion of a real number see A220335.
%C A220338 Let p >= 2 be an integer and set Q(p) = (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1), so Q(5) = 8*sqrt(6) - 19. 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 = 5. For other cases see A220335 (p = 2), A220336 (p = 3) and A220337 (p = 4).
%H A220338 Peter Bala, <a href="/A220335/a220335.pdf">A modified Engel expansion for certain quadratic irrationals</a>
%H A220338 Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%F A220338 Define the harmonic sawtooth map h(x) := floor(1/x)*(x*ceiling(1/x) - 1). Let x = 8*sqrt(6) - 19. 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 A220338 a(3*n+2) = 1/2*{2 + (5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} and
%F A220338 a(3*n+3) = {(5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} both for n >= 0.
%F A220338 For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*A084765(n)^2 and a(3*n+3) = 4*A085765(n)^2 - 2.
%F A220338 Recurrence equations:
%F A220338 For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
%F A220338 a(3*n+3) = 2*a(3*n+2) - 2.
%F A220338 Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation 8*sqrt(6) - 19 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*6) + 1/(2*6*10) + 1/(2*6*10*2) + 1/(2*6*10*2*50) + ....
%Y A220338 Cf. A084765, A220335 (p = 2), A220336 (p = 3), A220337 (p = 4).
%K A220338 nonn,easy
%O A220338 1,1
%A A220338 _Peter Bala_, Dec 12 2012