"Infinitely Nested" Tree-Log

Calculus Level 5

Given that n > 1 n > 1 , a 1 > 1 a_1 > 1 and an odd positive integer b b , let:

a n = b log 10 ( a 1 + a n 1 ) . \large a_n = b\log_{10}(a_1 + a_{n-1}).

If the following conditions are true:

  • b b and lim n a n \displaystyle \lim_{n \to \infty} a_n can be arbitrarily chosen.
  • lim n a n \displaystyle\lim_{n \to \infty} a_n exists and is finite.
  • lim n a n \displaystyle\lim_{n \to \infty} a_n is a positive integer.

What is the least possible value of a 1 a_1 if it must be a perfect square ?


The answer is 4.

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1 solution

Louie Dy
Jul 17, 2016

First, it is best to express lim n a n \large \lim \limits_{n \to \infty} a_n in a simpler way. Let S \large S denote this expression.

S = lim n a n \large S = \lim \limits_{n \to \infty} a_n

The limit can be expressed as an infinitely nested logarithmic expression:

S = b log ( a 1 + b log ( a 1 + b log ( a 1 + . . . ) . . . ) ) \large S = b\log(a_1 + b\log(a_1 + b\log(a_1 + ...)...))

...which can be expressed as:

S = b log ( a 1 + S ) \large S = b\log(a_1 + S)

Since b \large b is an odd positive integer, the equation can be rearranged as:

a 1 + S = 1 0 S b \large a_1 + S = 10^\frac{S}{b}

By arduous trial and error, we come up with some possible "practical" solutions:

a 1 = 1 , b = 9 , S = 9 \large a_1 = 1, b = 9, S = 9 a 1 = 4 , b = 2499 , S = 9996 \large a_1 = 4, b = 2499, S = 9996 a 1 = 9 , b = 1 , S = 1 \large a_1 = 9, b = 1, S = 1

Since a 1 > 1 \large a_1 > 1 , the least possible perfect square value for a 1 \large a_1 is 4 \boxed{4} .

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