A056813 - OEIS

%I #22 Oct 30 2013 00:08:02

%S 1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,5,5,5,5,5,5,5,5,5,5,

%T 5,5,5,5,5,5,5,5,5,5,5,5,5,5,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,

%U 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7

%N Largest non-unitary prime factor of LCM(1,...,n); that is, the largest prime which occurs to power > 1 in prime factorization of LCM(1,..,n).

%C For n>0, prime(n) appears {(prime(n+1))^2 - (prime(n))^2} times [from n=(prime(n))^2 to n=(prime(n+1))^2 - 1], that is, A000040(n) appears A069482(n) times (from n=A001248(n) to n=A084920(n+1)). - _Lekraj Beedassy_, Mar 31 2005

%C a(n) is the largest prime factor of A045948(n). [_Matthew Vandermast_, Oct 29 2008]

%C Alternative definition: a(n) = largest prime <= sqrt(n) (considering 1 as prime for this occasion, see A008578 for the 19th century definition of primes). - _Jean-Christophe Hervé_, Oct 29 2013

%H Jean-Christophe Hervé, <a href="/A056813/b056813.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = prime(w) if prime(w)^2 <= n < prime(w+1)^2.

%F To get the sequence, repeat 1 three times, and then for any k >= 1, repeat A000040(k) A069482(k) times; or equivalently, for any k >= 1, repeat A008578(k) a number of times equal to A008578(k+1)^2 - A008578(k)^2. - _Jean-Christophe Hervé_, Oct 29 2013

%e The j-th prime appears at the position of its square, at n = prime(j)^2.

%t Table[f = Transpose[FactorInteger[LCM @@ Range[n]]]; pos = Position[f[[2]], _?(# > 1 &)]; If[pos == {}, 1, f[[1, pos[[-1]]]][[1]]], {n, 100}] (* _T. D. Noe_, Oct 30 2013 *)

%Y Cf. A054041, A056170.

%Y Cf. A000040, A001248, A008578, A069482.

%K nonn,easy

%O 1,4

%A _Labos Elemer_, Aug 28 2000

%E Corrected offset by _Jean-Christophe Hervé_, Oct 29 2013