Sum of series with binary parity in the numerator I'm now stuck with this question, and I don't eve

Izabella Ponce 2022-06-28 Answered
Sum of series with binary parity in the numerator
I'm now stuck with this question, and I don't even know where to start: Find sum of series
1 f ( n ) n ( n + 1 )
, where f(n) - number of ones in binary representation of n.
I wish I could post some moves, that I've tried but I don't know what to do.
Thanks!
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Answers (2)

iceniessyoy
Answered 2022-06-29 Author has 27 answers
Maybe there is a quicker way, but here is one. Let the sum be S. We have
f ( n ) n ( n + 1 ) = f ( n ) n f ( n ) n + 1
f ( 2 n ) ( 2 n ) ( 2 n + 1 ) + f ( 2 n + 1 ) ( 2 n + 1 ) ( 2 n + 2 ) = f ( 2 n ) 2 n f ( 2 n ) f ( 2 n + 1 ) 2 n + 1 f ( 2 n + 1 ) 2 n + 2
Now f ( 2 n + 1 ) = f ( 2 n ) + 1 , f ( 2 n ) = f ( n ), so we can write:
f ( 2 n ) ( 2 n ) ( 2 n + 1 ) + f ( 2 n + 1 ) ( 2 n + 1 ) ( 2 n + 2 ) = f ( 2 n ) 2 n + 1 2 n + 1 f ( 2 n ) + 1 2 n + 2 = 1 2 ( f ( n ) n f ( n ) n + 1 ) + ( 1 2 n + 1 1 2 n + 2 )
S = 1 2 + 1 2 n = 1 f ( n ) n ( n + 1 ) + n = 1 ( 1 2 n + 1 1 2 n + 2 )
2 S = 1 + S + 2 log 2 1 S = 2 log 2 1.386
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polivijuye
Answered 2022-06-30 Author has 16 answers
Code:
double sum=0;
for(int i=1;i<9999999;i++){
double s2=Integer.bitCount(i);
sum+=(s2)/(i*(i+1));
}
Output Data
n 9 1.065079365079365 99 1.3394382621894894 999 1.3800972409478014 9999 1.3854974129587205 99999 1.3852676077956714 ( limit of data type, therefore decreased value )
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