A199323 Number of primes of the form n*(n+1)+2*k-3 with k from 1 to n+1.

1, 2, 3, 2, 3, 4, 3, 4, 5, 3, 5, 5, 5, 5, 6, 5, 6, 6, 6, 8, 8, 5, 7, 10, 7, 8, 7, 10, 7, 9, 11, 10, 8, 9, 13, 5, 11, 12, 14, 8, 12, 11, 8, 13, 14, 10, 13, 15, 9, 11, 19, 13, 12, 12, 12, 13, 16, 14, 16, 16, 13, 18, 15, 16, 12, 17, 16

Offset: 1

Author

Pierre CAMI, Nov 06 2011

Keywords

• nonn

Comments

n/log(n) is a good approximation of a(n) as n increases.

Examples

			Write the odd numbers like this :
1 3
5 7 9
11 13 15 17
19 21 23 25 27
29 31 33 35 37 39
41 43 45 47 49 51 53
.......................
line n : n*(n+1)+2*k-3 , k from 1 to n+1
line 1 , n=1 1*(1+1)*2*k-3  k=1 & k=2 , 3 prime a(1)=1
line 2 , n=2 for k=1 to 3 , 5 & 7 prime so a(2)=2
n=3 11 13 & 17 prime a(3)=3
n=4 19 & 23 prime a(4)=2
a(5)=3 , a(6)=4
		

Links

• Pierre CAMI, Table of n, a(n) for n = 1..10000

Programs

Mathematica

npf[n_]:=Count[Table[n(n+1)+2k-3,{k,n+1}],?PrimeQ]; Array[npf,70] (* _Harvey P. Dale, Jul 04 2013 *)

PARI

a(n)=if(n<4,n,primepi(n^2+3*n-1)-primepi(n^2+n-2)) \\ Charles R Greathouse IV, Nov 11 2011

PARI

a(n)=my(s);forstep(a=n^2+n-1,n^2+3*n-1,2,s+=isprime(a)); s \\ Charles R Greathouse IV, Nov 11 2011

Formula

a(n) = pi(n^2 + 3n - 1) - pi(n^2 + n - 2) for n > 1. [Charles R Greathouse IV, Nov 11 2011]