How does the complexity of this code would be 0(N LOG LOG N)?

#1

how does the complexity of this code would be 0(N LOG LOG N)

?

for(int j=3;j<=100;j=j+2){

if(p[i]===1){

for(int j=j*j;j<=100;j=j+2){

   p[j]=0;

}
}

   }
#2

hello @manish99

it time complexity is just summation of
n/2 + n/3+ n/5 + n/7 + n/11 …

take n common
n( 1/2 + 1/3 + 1/5 + 1/7 + 1/11 + 1/13…1/p )

1/2 + 1/3 + 1/5 + 1/7 … 1/p= log(logn)
for proof refer this -> link

=n * log (log n )

#3

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#4

I am not able to understand it

#5

to undersand the prooof u must know about harmonic series,euler product forumula and tyler series.

also i have already shared prrof link with u , do check that out