A272861 Sum of two integers when equal to the product of their prime-counting functions.

12, 16, 18, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 116, 120, 280, 310, 325, 330, 942, 948, 954, 960, 966, 972, 984, 990, 996, 2968, 3003, 8224, 8232, 8240, 8248, 8280, 8288, 8304, 8312, 8360, 8408, 23499, 23508, 23589

Offset: 1

Author

Giuseppe Coppoletta, Jun 18 2016

Keywords

• nonn

Comments

The sums are listed in increasing order. The only term with equal addends is a(2)= 16 = 8 + 8 = pi(8)^2, Indeed j=8 is the only solution to pi(j)^2 = 2*j, which is easily seen using pi(j) > j/log(j) for j>16.

Examples

			32 is a term because 32 = 10 + 22 = 4 * 8 = pi(10) * pi(22).
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..320 (terms < 4*10^9)
• Eric Weisstein's World of Mathematics, Prime Counting Function

Crossrefs

Cf. A272860, A273286, A000720.

Programs

Mathematica

nn = 10^3; Select[Range@ nn, Function[k, IntegerQ@ SelectFirst[Range@ nn, k == PrimePi[#] PrimePi[k - #] &]]] (* Version 10, or *)
Select[Range[10^3], Function[n, Length@ Select[Transpose@ {#, n - #} &@ Range[Floor[n/2]], n == PrimePi[First@ #] PrimePi[Last@ #] &] > 0]] (* Michael De Vlieger, Jun 30 2016 *)

Sage

def g(n): return [k for k in (3..n/2) if n==prime_pi(k)*prime_pi(n-k)]
[n for n in range(2,1000) if len(g(n))>0]

Formula

Positive integers n such that n = pi(j) * pi(n-j) for some j.

Extensions

a(50)-a(53) from Giovanni Resta, Jun 29 2016