A108309 - OEIS

A108309

Consider the triangle of odd numbers where the n-th row has the next n odd numbers. The sequence is the number of primes in the n-th row.

0, 2, 2, 3, 2, 3, 3, 4, 4, 5, 3, 4, 6, 4, 6, 6, 4, 6, 7, 6, 8, 7, 5, 8, 9, 8, 7, 8, 9, 8, 9, 10, 10, 8, 10, 12, 5, 12, 12, 13, 9, 11, 11, 9, 13, 14, 9, 14, 14, 10, 10, 19, 14, 12, 12, 12, 12, 16, 15, 16, 15, 13, 18, 16, 16, 12, 16, 17, 15, 16, 18, 14, 15, 20, 18, 19, 14, 19, 20, 18, 16

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OFFSET

1,2

COMMENTS

Except for the initial term, a(n)>=2 because in the interval 2n-1 of odd numbers there are always at least two primes.

For n>2, this is the same as the number of primes between n^2-n and n^2+n, which is the sum of A089610 and A094189. - T. D. Noe, Sep 16 2008

a(n) = SUM(A010051(A176271(n,k)): 1<=k<=n). - Reinhard Zumkeller, Apr 13 2010

From Pierre CAMI, Sep 03 2014: (Start)

For n>1 a(n)~floor(1/2 + n/log(n)).

The number of primes < n^2 is ~ n^2/2/log(n) by the prime number theorem, as a(n) ~ floor(1/2 + n/log(n)) we have:

n^2/2/log(n) ~ 1 + floor(1/2 + 2/log(2)) + floor(1/2 + 3/log(3)) + floor(1/2 + 4/log(4)) + ... + floor(1/2 + (n-1)/log(n-1)) + floor(1/2 + n/log(n)).

For n=16000 the number of primes < n^2 is 13991985, the sum: 1 + floor(1/2 + 2/log(2)) + floor(1/2 + 3/log(3)) + floor(1/2 + 4/log(4))+ ... + floor(1/2 + (n-1)/log(n-1)) + floor(1/2 + n/log(n)) is 13991101 and (n^2)/(2*log(n)) is 13222671.

So between n^2+n and n^2+3*n there are n odd numbers and ~floor(1/2 + n/log(n)) prime numbers.

The twin primes are of the form T1=n^2+n-1 and T2=n^2+n+1, or n^2+n+T1 and n^2+n+T2 with T1<=2*n-1, or n^2+n+P and n^2+n+P(-2 or +2) with P prime <=2*n-1.

(End)

LINKS

T. D. Noe, Table of n, a(n) for n=1..10000

Pierre CAMI, Table of n, a(n) and Floor(1/2+n/log(n))for n=1..10000

EXAMPLE

Triangle begins:

1: 1 -> 0 primes,

2: 3,5 -> 2 primes,

3: 7,9,11 -> 2 primes,

4: 13,15,17,19 -> 3 primes.

MAPLE

seq(numtheory:-pi(n^2+n-1)-numtheory:-pi(n^2-n), n=1..100); # Robert Israel, Sep 03 2014