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a(n) = Product_{d|n, d>1} prime(A286561(n,d)), where A286561(n,d) gives the highest exponent of d dividing n.
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%I #17 Nov 13 2017 13:21:34

%S 1,2,2,6,2,8,2,20,6,8,2,48,2,8,8,84,2,48,2,48,8,8,2,320,6,8,20,48,2,

%T 128,2,264,8,8,8,864,2,8,8,320,2,128,2,48,48,8,2,2688,6,48,8,48,2,320,

%U 8,320,8,8,2,3072,2,8,48,1560,8,128,2,48,8,128,2,11520,2,8,48,48,8,128,2,2688,84,8,2,3072,8,8,8,320

%N a(n) = Product_{d|n, d>1} prime(A286561(n,d)), where A286561(n,d) gives the highest exponent of d dividing n.

%H Antti Karttunen, <a href="/A293514/b293514.txt">Table of n, a(n) for n = 1..16384</a>

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

%F a(n) = Product_{d|n, d>1} A000040(A286561(n,d)).

%F Other identities. For all n >= 1:

%F A001222(a(n)) = A032741(n).

%F A007814(a(n)) = A056595(n) [See A046951.]

%F 1+A056239(a(n)) = A169594(n).

%F A064989(a(n)) = A293515(n).

%e For n = 24, its divisors larger than one are: 2, 3, 4, 6, 8, 12, 24. Only 2 has valuation > 1, namely A286561(24,2) = 3 (as 2^3 divides 24), while the other six have valuation 1. Thus a(24) = prime(1)^6 * prime(3) = 64*5 = 320.

%e For n = 64, its divisors larger than one are: 2, 4, 8, 16, 32, 64. We see that 2^6 = 4^3 = 8^2 = 64, while valuation of the last three 16, 32 and 64 is 1. Thus a(64) = prime(1)^3 * prime(2) * prime(3) * prime(6) = 2^3 * 3 * 5 * 13 = 1560.

%o (PARI) A293514(n) = { my(m=1); fordiv(n,d,if(d>1, m *= prime(valuation(n,d)))); m; };

%Y Cf. A000040, A286561.

%Y Cf. also A293515, A293524, A294876, A293442.

%K nonn

%O 1,2

%A _Antti Karttunen_, Nov 11 2017