

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.


1



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
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OFFSET

1,7


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,k1)). This was posed in the Yahoo Group MathForFun, see link.


LINKS

Table of n, a(n) for n=1..91.
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]


EXAMPLE

(2*10)!/10!^2 = 184756 = 2*2*11*13*17*19 which has 5 distinct divisors. Pi(2*10) = 8. 85=3 = a(10).


PROG

(PARI) g2(n) = for(x=1, n, ct=omega((2*x)!/x!^2); print1(primepi(2*x)ct", "))
(MAGMA) [#PrimesUpTo(2*n)  #PrimeDivisors( Factorial(2*n) div Factorial(n)^2):n in [1..91]]; // Marius A. Burtea, Nov 16 2019


CROSSREFS

Cf. A001221, A001791, A099802.
Sequence in context: A191408 A115756 A067731 * A291985 A317192 A053735
Adjacent sequences: A147841 A147842 A147843 * A147845 A147846 A147847


KEYWORD

nonn


AUTHOR

Cino Hilliard, Nov 15 2008


EXTENSIONS

Corrected the link.  Cino Hilliard, Nov 18 2008


STATUS

approved



