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If n = Product (p_j^k_j) then a(n) = max{p_j}^max{k_j}.
2

%I #12 Sep 09 2024 06:20:16

%S 1,2,3,4,5,3,7,8,9,5,11,9,13,7,5,16,17,9,19,25,7,11,23,27,25,13,27,49,

%T 29,5,31,32,11,17,7,9,37,19,13,125,41,7,43,121,25,23,47,81,49,25,17,

%U 169,53,27,11,343,19,29,59,25,61,31,49,64,13,11,67,289,23,7,71,27,73,37,25

%N If n = Product (p_j^k_j) then a(n) = max{p_j}^max{k_j}.

%H Ilya Gutkovskiy, <a href="/A304180/a304180.jpg">Scatter plot of a(n) up to n=100000</a>.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/GreatestPrimeFactor.html">Greatest Prime Factor</a>.

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.

%F a(n) = A006530(n)^A051903(n).

%F a(p^k) = p^k where p is a prime.

%F a(A005117(k)) = A073482(k).

%F a(A002110(k)) = A000040(k).

%e a(40) = 125 because 40 = 2^3*5^1, max{2,5} = 5, max{3,1} = 3 and 5^3 = 125.

%t Table[(FactorInteger[n][[-1, 1]])^(Max @@ Last /@ FactorInteger[n]), {n, 75}]

%o (PARI) a(n) = if(n == 1, 1, my(f = factor(n), p = f[, 1], e = f[, 2]); vecmax(p)^vecmax(e)); \\ _Amiram Eldar_, Sep 08 2024

%Y Cf. A000040, A000961 (fixed points), A002110, A005117, A006530, A034699, A051903, A053585, A073482, A081810, A304181.

%K nonn

%O 1,2

%A _Ilya Gutkovskiy_, May 07 2018