[seqfan] A079416 = A082896

[israel at math.ubc.ca]

A082896 has a Comment "Checked for the first 120000 terms to be the same as 
A079416." In fact, I can prove that they are the same (see below). Shall we 
merge the two sequences?

A082896 is "a(n)=A082893[n]/n", while A082893 is "the closest number to 
n-th prime which is divisible by n".

There is a tie between k*n and (k+1)*n for closest number divisible by n to 
prime(n) iff prime(n) = (k+1/2)*n. But then prime(n) = (2*k+1)*(n/2) (where 
n must be even), and this can't be prime if n > 2 and k > 0. Indeed if n=2, 
there is a tie between 2 and 4 for the closest number to p(2)=3 which is 
divisible by 2; A082893 chooses 4, so A082896(2)=2 which is also 
A079416(2). Of course k=0 would mean prime(n)=n/2, which can't be true 
because prime(n)>n.
 
Thus for n > 2, A082896(n)=k where either k*n < prime(n) and (k+1)*n - 
prime(n) > prime(n) - k*n, i.e. prime(n)/n > k > prime(n)/n - 1/2 , or k*n 
> prime(n) and prime(n) - (k-1)*n > k*n - prime(n), i.e. prime(n)/n + 1/2 > 
k > prime(n)/n. In either case, A082896(n) is an integer in the interval 
(prime(n)/n - 1/2, prime(n)/n + 1/2). Since that interval has length 1, it 
is the only integer therein.

On the other hand, A079416 is "round(prime(n)/n)". Again, that is an 
integer in the interval [prime(n)/n - 1/2, prime(n)/n + 1/2]. Since 
A082896(n) is an integer in the interval (prime(n)/n - 1/2, prime(n)/n + 
1/2), the two must be equal.

Cheers,
Robert