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A220336
A modified Engel expansion for 4*sqrt(2) - 5.
9
2, 4, 6, 2, 18, 34, 2, 578, 1154, 2, 665858, 1331714, 2, 886731088898, 1773462177794, 2, 1572584048032918633353218, 3145168096065837266706434, 2, 4946041176255201878775086487573351061418968498178, 9892082352510403757550172975146702122837936996354
OFFSET
1,1
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).
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) + ....
CROSSREFS
Cf. A001601, A028257, A220335 (p = 2), A220337 (p = 4), A220338 (p = 5).
Sequence in context: A097009 A214213 A085593 * A021410 A356216 A336843
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Dec 12 2012
STATUS
approved