# maths

the non-
decreasing sequence of odd integers {a1, a2, a3, .
. .} = {1,3,3,3,5,5,5,5,5,...} each positive odd
integer k appears k times. it is a fact that there
are integers b, c, and d such that, for all positive
integers n, añ = b[√(n+c)] +d. Where [x] denotes the largest integer not exceeding x. Find the sum of b+c+d.

Show step too hard

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1. since the terms of the sequence are all odd numbers (2k+1), and there are k^2 terms <= the kth odd number, look for something like

2[√n]+1

That sequence is

3,3,3, 5,5,5,5,5,5,5, ...

Looks like we have to adjust it by 1 to get

2[√(n-1)]+1

So, b+c+d = 2

No, I don't have an actual proof, but maybe you can work from here. I'll think on it some.

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