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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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# A[n] = Θ(lg n)?

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