OFFSET
1,2
COMMENTS
Conjecture: For each n there is at least one prime p such that a(n) < p < a(n+1). From the conjecture follows that the prime gaps g(n) = p(n+1) - p(n) = O(sqrt(p(n))/log(p(n))). Legendre's hypothesis is that g(n) = O(sqrt(p(n))). - Thomas Ordowski, Aug 11 2012
LINKS
T. D. Noe, Table of n, a(n) for n = 1..10000
Wikipedia, Legendre's conjecture
FORMULA
a(n) ~ (n log n)*(n/log n) = n^2. a(n) > n^2 for n > 4. - Thomas Ordowski, Aug 09 2012
MATHEMATICA
Table[Prime[n] * PrimePi[n], {n, 50} (* Harvey P. Dale, Mar 17 2011 *)
PROG
(PARI) g(n) = for(x=1, n, y=prime(x)*primepi(x); print1(y", "))
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Cino Hilliard, Apr 23 2007
STATUS
approved