%I #21 Dec 21 2021 23:42:22
%S 1,1,1,1,1,2,1,1,1,2,1,4,1,2,3,1,1,2,1,4,3,2,1,8,1,2,1,4,1,6,1,1,3,2,
%T 5,4,1,2,3,8,1,6,1,4,9,2,1,16,1,2,3,4,1,2,5,8,3,2,1,12,1,2,9,1,5,6,1,
%U 4,3,10,1,8,1,2,3,4,7,6,1,16,1,2,1,12,5,2,3,8,1,18,7,4,3,2,5,32,1,2,9,4,1
%N n/p^k, where p = largest prime dividing n and p^k = highest power of p dividing n.
%H T. D. Noe, <a href="/A051119/b051119.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = n/A053585(n).
%e a(36) = 4 because 36/3^2 = 4, 3^2 is highest power dividing 36 of largest prime dividing 36.
%e a(50) = 50 / 5^2 = 2.
%t f[n_]:=Module[{c=Last[FactorInteger[n]]},n/First[c]^Last[c]]; Array[ f, 110] (* _Harvey P. Dale_, Oct 14 2011 *)
%o (Python)
%o from sympy import factorint, primefactors
%o def a053585(n):
%o if n==1: return 1
%o p = primefactors(n)[-1]
%o return p**factorint(n)[p]
%o def a(n): return n/a053585(n) # _Indranil Ghosh_, May 19 2017
%Y Cf. A053585(n).
%K nonn,easy,nice
%O 1,6
%A _Leroy Quet_
%E More terms from _James A. Sellers_, Jan 21 2000