algorithm - Big-O analysis for a loop -

i've got analyze loop, among others, , determine running time using big-o notation.

for ( int = 0; < n; += 4 )     ( int j = 0; j < n; j++ )         ( int k = 1; k < j*j; k *= 2 )` 

here's have far:

for ( int = 0; < n; += 4 ) = n 

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for ( int j = 0; j < n; j++ ) = n 

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for ( int k = 1; k < j*j; k *= 2 ) = log^2 n 

now problem i'm coming final running time of loop. best guess o(n^2), uncertain if correct. can help?

edit: sorry oh -> o thing. textbook uses "big-oh"

first note outer loop independent remaining 2 - adds (n/4)* multiplier. consider later.

now let's consider complexity of

for ( int j = 0; j < n; j++ )     ( int k = 1; k < j*j; k *= 2 ) 

we have following sum:

0 + log₂(1) + log₂(2 * 2) + ... + log₂(n*n)

it note log₂(n^2) = 2 * log₂(n). re-factor sum to:

2 * (0 + log₂(1) + log₂(2) + ... + log₂(n))

it not easy analyze sum take @ this post. using sterling's approximation 1 can belongs o(n*log(n)). overall complexity o((n/4)*2*n*log(n))= o(n^2*log(n))