A147844 - OEIS

%I #18 Sep 08 2022 08:45:38

%S 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,

%T 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,

%U 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

%N 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.

%C 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.

%H MathForFun, <a href="http://groups.yahoo.com/group/mathforfun/message/13576">Binomial Identity</a>.

%H Kenneth Ramsey, Cino Hilliard and others, <a href="/A147844/a147844.txt">Binomial Identity ..Hard to Prove this?</a>, digest of 6 messages in mathforfun Yahoo group, Oct 30 - Nov 10, 2008. [Cached copy]

%e (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).

%t Table[PrimePi[2n]-PrimeNu[(2n)!/(n!)^2],{n,100}] (* _Harvey P. Dale_, Oct 30 2021 *)

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

%o (Magma) [#PrimesUpTo(2*n) - #PrimeDivisors( Factorial(2*n) div Factorial(n)^2):n in [1..91]]; // _Marius A. Burtea_, Nov 16 2019

%Y Cf. A001221, A001791, A099802.

%K nonn

%O 1,7

%A _Cino Hilliard_, Nov 15 2008

%E Corrected the link. - _Cino Hilliard_, Nov 18 2008