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d^2 = (3t-9)/2

nonononnon, it is "d to the power of 0.66666666666667" or "d to the power of 2/3"

above is sum of time
for mean of time , [2*(d1^(2/3)+d2^(2/3)+....+dn^(2/3)) + 3*n]/n

so, i WILL have to do the whole procedure, then. thanks :)