algorithm - Solving runtime of T(n) = n^(1/2)T(n^(1/2)) + n -


I am trying to find the recursion order of the following: 

  T ) = N ^ (1/2) T (N ^ (1/2)) + n

But I can not find the zodiac or even an equation that (N) Does anyone have to help me with the equation for the recap of the conference? 

  t (n) = n ^ (1/2) t (n ^ (1 / 2) + n = n ^ (1/2) [n ^ (1/2 ^ 2) t (n ^ (1/2 ^ 2) + n ^ (1/2)] + n = n ^ (3 / 2 ^ 2) T (n ^ (1/2 ^ 2) + 2n = n ^ (3/2 ^ 2) [n ^ 1/2 ^ 3 t (n ^ (1/2 ^ 3) + n ^ (1/2 ^ 3)] + 2n = and so on n ^ (1/2 ^ k) = 2 1/2 ^ k logn = 1 logn = 2k log logn = k = n ^ (1-1 / 2 ^ K) T (n ^ (1/2 ^ k) + cube = n / n (1/2 ^ k) t (2) + cube = n / 2 * 2 + nloglogn = n + nloglogn = o (nloglogn )

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