Proving a[n] = Θ(lg n): A Case Study

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The discussion focuses on proving that the sequence a[n] is Θ(lg n) under the condition that a[n] is an increasing sequence where a[n] = a[n/m] + d for any divisor m of n. Participants explore the relationship between a[n] and logarithmic growth, with one member successfully demonstrating that a[n] is O(lg n). The challenge remains to establish that a[n] is also Ω(lg n), which is crucial for the Θ notation. Concerns are raised about the implications of the proof when n is odd, highlighting the need for clarity in the reasoning. The conversation reflects a collaborative effort to resolve the proof's complexities.
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a[n] = Θ(lg n)!?

Suppose {a[n]} is an increasing sequence and whenever m divides n, then

a[n] = a[n/m] + d

where m is a positive integer and d is a real number. Show that a[n] = Θ(lg n).

a[n/m] + d ≤ a[n/2] + d, and with this I can show that a[n] = O(lg n). All I need to do is show that a[n] = Ω(lg n). This is where I'm stuck. Any hints?
 
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To show that something is in omega(log x), it suffices to find some sequence {x_i} such that a[x_i] grows like log x.

By the way, I'm disconcerted with your reasoning that it is in O(log x); what if n is odd? I suppose you're just abbreviating the complete proof you have.
 
Hurkyl said:
By the way, I'm disconcerted with your reasoning that it is in O(log x); what if n is odd? I suppose you're just abbreviating the complete proof you have.
Sorry. What I meant was

a_n = a_{\lfloor n/2 \rfloor} + d

My head is a total blank. I can't think of anything.
 

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