This site is supported by donations to The OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 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 (list; graph; refs; listen; history; text; internal format)
 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). LINKS 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 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 = 0..n} a(k). Then we have the infinite Egyptian fraction representation 4*sqrt(2) - 5 = sum {n >=0} 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 A061504 A077179 Adjacent sequences:  A220333 A220334 A220335 * A220337 A220338 A220339 KEYWORD nonn,easy AUTHOR Peter Bala, Dec 12 2012 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified November 19 06:03 EST 2019. Contains 329310 sequences. (Running on oeis4.)