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As N increases, (Sum_{n=1..N} a(n))/(Sum_{n=1..N} n) appears to tend to log(2), as can be seen by plotting the first 10000 terms.
This observation is in conformity consistent with the prime number theorem as the probability that k*2^n+1 is prime is 1/(n*log(2)+log(k)) so ~ 1/(n*log(2)) as n increases, if k ~ n*log(2) then k/(n*log(2)) ~ 1.
proposed
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editing
proposed
As N increases sum, (Sum_{n =1..N, } a(n)})/sum(Sum_{n =1..N, } n} )appears to tend to log(2) , as it can be seen by plotting the first 10000 terms.
This observation is in conformity with the prime number theorem as the probability of that k*2^n+1 to be is prime is 1/(n*log(2)+log(k)) so ~ 1/(n*log(2)) as n increases, if k ~ n*log(2) then k/(n*log(2)) ~ 1.
1*2^(2*0+1)+1=3 is prime , so a(0)=1.
1*2^(2*1+1)+1=9 and 3*2^(2*1+1)+1=25 are composite, 5*2^(2*1+1)+1=41 is prime , so a(1)=5.
proposed
editing
editing
proposed
As N increases sum{n 1..N, a(n)}/sum{n 1..N, n} tends appears to tend to log(2) as it can be seen plotting the first 10000 terms.
proposed
editing
editing
proposed
As N increases sum{n 1..N, a(n)}/sum{n 1..N, n} tends to log(2) as it can be seen ploting plotting the first 10000 terms.
This observation is in oonformity conformity with the prime number theorem as the probability of k*2^n+1 to be prime is 1/(n*log(2)+log(k)) so ~ 1/(n*log(2)) as n increases, if k ~ n*log(2) then k/(n*log(2)) ~ 1.
As N increases sum{a(n), n= 1 to ..N, a(n)}/sum{n, n= 1 to ..N, n} tends to log(2) as it can be seen ploting the first 10000 terms.
proposed
editing
editing
proposed