A220336 A modified Engel expansion for 4*sqrt(2) - 5.

2, 4, 6, 2, 18, 34, 2, 578, 1154, 2, 665858, 1331714, 2, 886731088898, 1773462177794, 2, 1572584048032918633353218, 3145168096065837266706434, 2, 4946041176255201878775086487573351061418968498178, 9892082352510403757550172975146702122837936996354

Offset: 1

Author

Peter Bala, Dec 12 2012

Keywords

• nonn
• easy

Comments

For a brief description of the modified Engel expansion of a real number see A220335.

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).

Links

• Peter Bala, A modified Engel expansion for certain quadratic irrationals
• Wikipedia, Engel Expansion

Crossrefs

Cf. A001601, A028257, A220335 (p = 2), A220337 (p = 4), A220338 (p = 5).

Formula

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.

a(3*n+2) = 1/2*{2 + (1+sqrt(2))^(2^(n+1)) + (1-sqrt(2))^(2^(n+1))},

a(3*n+3) = {(1 + sqrt(2))^(2^(n+1)) + (1 - sqrt(2))^(2^(n+1))}, both

for n >= 0.

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.

Recurrence equations:

For n >= 1, a(3*n+2) = 2*{a(3*n-1)^2 - 2*a(3*n-1) + 1} and

a(3*n+3) = 2*a(3*n+2) - 2.

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) + ....