login
A220338
A modified Engel expansion for 8*sqrt(6) - 19.
9
2, 6, 10, 2, 50, 98, 2, 4802, 9602, 2, 46099202, 92198402, 2, 4250272665676802, 8500545331353602, 2, 36129635465198759610694779187202, 72259270930397519221389558374402, 2, 2610701117696295981568349760414651575095962187244375364404428802
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(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 see A220335 (p = 2), A220336 (p = 3) and A220337 (p = 4).
FORMULA
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) + ....
CROSSREFS
Cf. A084765, A220335 (p = 2), A220336 (p = 3), A220337 (p = 4).
Sequence in context: A333185 A236106 A095105 * A052194 A392505 A320383
KEYWORD
nonn,easy,changed
AUTHOR
Peter Bala, Dec 12 2012
STATUS
approved