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A051119 a(n) = n/p^k, where p = largest prime dividing n and p^k = highest power of p dividing n.

%I A051119 #29 Jul 02 2025 16:01:58
%S A051119 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 A051119 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 A051119 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 A051119 a(n) = n/p^k, where p = largest prime dividing n and p^k = highest power of p dividing n.
%H A051119 T. D. Noe, <a href="/A051119/b051119.txt">Table of n, a(n) for n = 1..10000</a>
%F A051119 a(n) = n/A053585(n).
%e A051119 a(36) = 4 because 36/3^2 = 4, 3^2 is highest power dividing 36 of largest prime dividing 36.
%e A051119 a(50) = 50/5^2 = 2.
%t A051119 f[n_]:=Module[{c=Last[FactorInteger[n]]},n/First[c]^Last[c]]; Array[ f, 110] (* _Harvey P. Dale_, Oct 14 2011 *)
%o A051119 (Python)
%o A051119 from sympy import factorint, primefactors
%o A051119 def a053585(n):
%o A051119     if n==1: return 1
%o A051119     p = primefactors(n)[-1]
%o A051119     return p**factorint(n)[p]
%o A051119 def a(n): return n/a053585(n) # _Indranil Ghosh_, May 19 2017
%o A051119 (PARI) a(n) = if(n>1, my(f=factor(n)); n/f[#f~, 1]^f[#f~, 2], 1); \\ _Michel Marcus_, Jan 10 2025
%Y A051119 Cf. A053585.
%K A051119 nonn,easy,nice
%O A051119 1,6
%A A051119 _Leroy Quet_
%E A051119 More terms from _James Sellers_, Jan 21 2000