A220335 A modified Engel expansion for sqrt(3) - 1.

2, 3, 4, 2, 8, 14, 2, 98, 194, 2, 18818, 37634, 2, 708158978, 1416317954, 2, 1002978273411373058, 2005956546822746114, 2, 2011930833870518011412817828051050498, 4023861667741036022825635656102100994

Offset: 1

Author

Peter Bala, Dec 12 2012

Keywords

• nonn
• easy

Comments

This is the case p = 2 of a family of quadratic irrationals of the form (p - 1)*sqrt(p^2 - 1) - (p^2 - p - 1) whose modified Engel expansion, as defined below, has a predictable form. For other cases see A220336 (p = 3), A220337 (p = 4) and A220338 (p = 5).

The Engel expansion of a positive real number x in the half-open interval (0,1] is the unique nondecreasing sequence {e(1), e(2), e(3), ...} of positive integers such that x = 1/e(1) + 1/(e(1)*e(2)) + 1/(e(1)*e(2)*e(3)) + .... The terms in the Engel expansion of x are obtained from the iterates of the map g(x) = x*(1 + floor(1/x)) - 1 by means of the formula e(n) = 1 + floor(1/g^(n-1)(x)). Here g^(n)(x) = g(g^(n-1)(x)) denotes the n-th iterate of g(x) with the convention g^(0)(x) = x.

In a similar way, the modified Engel expansion of x belonging to (0,1] is a sequence {E(1), E(2), E(3), ...} of positive integers such that x = 1/E(1) + 1/(E(1)*E(2)) + 1/(E(1)*E(2)*E(3)) + ... whose terms are obtained from the iterates of the harmonic sawtooth map h(x) = floor(1/x)*g(x). The general formula is E(1) = 1 + floor(1/x) and for n >= 1, E(n) = floor(1/h^(n-2)(x))*(1 + floor(1/h^(n-1)(x))). For further details see the Bala link.

Links

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

Crossrefs

Cf. A002812, A028257, A220336 (p = 3), A220337 (p = 4), A220338 (p = 5).

Formula

Let x = sqrt(3) - 1. 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 harmonic sawtooth map h(x), with the convention h^(0)(x) = x.

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

a(3*n+3) = (2 + sqrt(3))^(2^n) + (2 - sqrt(3))^(2^n), both for n >= 0.

For n >= 0, a(3*n+1) = 2. For n >= 1, a(3*n+2) = 2*(A002812(n-1))^2 and a(3*n+3) = 4*(A002812(n-1))^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 sqrt(3) - 1 = Sum_{n>=1} 1/P(n) = 1/2 + 1/(2*3) + 1/(2*3*4) + 1/(2*3*4*2) + ....