A147844 Difference between the number of distinct prime divisors of (2*n)!/n!^2 and pi(2*n), where pi(x) is the prime counting function.

0, 0, 1, 1, 1, 1, 2, 1, 2, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 2, 3, 4, 5, 5, 5, 5, 6, 4, 3, 5, 6, 5, 4, 5, 5, 6, 7, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 6, 7, 7, 8, 9, 8, 8, 10, 10, 11, 10, 10, 9, 9, 9, 9, 9, 9, 9, 8, 9, 10, 11, 11, 10, 10, 10, 10, 11, 10, 10, 11, 10, 10, 11, 11, 12, 12, 11, 12, 12, 12, 13, 13

Offset: 1

Author

Cino Hilliard, Nov 15 2008

Keywords

• nonn

Comments

The expression (2*n)!/n!^2 is taken from C(2*n+1,n+1) - C(2*n,n) = ((2*n)!/n!^2)*(n/(n+1)) = Sum_{k=1..n} C(n,k)*C(n,k-1). This was posed in the Yahoo Group MathForFun, see link.

Examples

			(2*10)!/10!^2 = 184756 = 2*2*11*13*17*19 which has 5 distinct divisors. Pi(2*10) = 8. 8-5=3 = a(10).
		

Links

• MathForFun, Binomial Identity.
• Kenneth Ramsey, Cino Hilliard and others, Binomial Identity ..Hard to Prove this?, digest of 6 messages in mathforfun Yahoo group, Oct 30 - Nov 10, 2008. [Cached copy]

Crossrefs

Cf. A001221, A001791, A099802.

Programs

Magma

[#PrimesUpTo(2*n) - #PrimeDivisors( Factorial(2*n) div Factorial(n)^2):n in [1..91]]; // Marius A. Burtea, Nov 16 2019

Mathematica

Table[PrimePi[2n]-PrimeNu[(2n)!/(n!)^2],{n,100}] (* Harvey P. Dale, Oct 30 2021 *)

PARI

g2(n) = for(x=1,n,ct=omega((2*x)!/x!^2);print1(primepi(2*x)-ct","))