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 A220337 #16 Jun 19 2025 03:25:52
%S A220337 2,5,8,2,32,62,2,1922,3842,2,7380482,14760962,2,108942999582722,
%T A220337 217885999165442,2,23737154316161495960243527682,
%U A220337 47474308632322991920487055362,2,1126904990058528673830897031906808442930637286502826475522
%N A220337 A modified Engel expansion for 3*sqrt(15) - 11.
%C A220337 For a brief description of the modified Engel expansion of a real number see A220335.
%C A220337 Let p >= 2 be an integer and set Q(p) = (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1), so Q(4) = 3*sqrt(15) - 11. 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 = 4. For other cases see A220335 (p = 2), A220336 (p = 3) and A220338 (p = 5).
%H A220337 Peter Bala, <a href="/A220335/a220335.pdf">A modified Engel expansion for certain quadratic irrationals</a>
%H A220337 Wikipedia, <a href="http://en.wikipedia.org/wiki/Engel_expansion">Engel Expansion</a>
%F A220337 Define the harmonic sawtooth map h(x) := floor(1/x)*(x*ceiling(1/x) - 1). Let x = 3*sqrt(15) - 11. 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 A220337 a(3*n+2) = 1/2*{2 + (4 + sqrt(15))^(2^n) + (4 - sqrt(15))^(2^n)} and
%F A220337 a(3*n+3) = (4 + sqrt(15))^(2^n) + (4 - sqrt(15))^(2^n), both for n >= 0.
%F A220337 For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*(A005828(n-1))^2 and a(3*n+3) = 4*(A005828(n-1))^2 - 2.
%F A220337 Recurrence equations:
%F A220337 For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and
%F A220337 a(3*n+3) = 2*a(3*n+2) - 2.
%F A220337 Put P(n) = Product_{k=1..n} a(k). Then we have the infinite Egyptian fraction representation 3*sqrt(15) - 11 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*5) + 1/(2*5*8) + 1/(2*5*8*2) + 1/(2*5*8*2*32) + ....
%Y A220337 Cf. A005828, A220335 (p = 2), A220336 (p = 3), A220338 (p = 5).
%K A220337 nonn,easy
%O A220337 1,1
%A A220337 _Peter Bala_, Dec 12 2012