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For a brief description of the modified Engel expansion of a real number seeA220335.

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 seeA220335 (p = 2),A220336 (p = 3) andA220337 (p = 4).

P. Bala,A modified Engel expansion for certain quadratic irrationals

S. Crowley,Integral transforms of the harmonic sawtooth map, the Riemann zeta function, fractal strings, and a finite reflection formula, arXiv:1210.5652 [math.NT]

Wikipedia,Engel Expansion

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.

a(3*n+2) = 1/2*{2 + (5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} and

a(3*n+3) = {(5 + 2*sqrt(6))^(2^n) + (5 - 2*sqrt(6))^(2^n)} both for n >= 0.

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.

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 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) + ....

Cf.A084765,A220335 (p = 2),A220336 (p = 3),A220337 (p = 4).

Sequence in context:A333185A236106A095105 *A052194A320383A073662

Adjacent sequences:A220335A220336A220337 *A220339A220340A220341

nonn,easy

Peter Bala, Dec 12 2012

approved