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A294874 a(n) = Product_{d|n, d>1, d = x^(2k) for some maximal k} prime(k).

%I A294874 #11 Nov 13 2017 13:22:18
%S A294874 1,1,1,2,1,1,1,2,2,1,1,2,1,1,1,6,1,2,1,2,1,1,1,2,2,1,2,2,1,1,1,6,1,1,
%T A294874 1,8,1,1,1,2,1,1,1,2,2,1,1,6,2,2,1,2,1,2,1,2,1,1,1,2,1,1,2,30,1,1,1,2,
%U A294874 1,1,1,8,1,1,2,2,1,1,1,6,6,1,1,2,1,1,1,2,1,2,1,2,1,1,1,6,1,2,2,8,1,1,1,2,1,1,1,8,1,1,1,6,1,1,1,2,2,1,1,2
%N A294874 a(n) = Product_{d|n, d>1, d = x^(2k) for some maximal k} prime(k).
%H A294874 Antti Karttunen, <a href="/A294874/b294874.txt">Table of n, a(n) for n = 1..65537</a>
%H A294874 <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>
%F A294874 a(n) = Product_{d|n, d>1, r = A052409(d) is even} A000040(r/2).
%F A294874 Other identities. For all n >= 1:
%F A294874 A001222(a(n)) = A071325(n).
%F A294874 1 + A001222(a(n)) = A046951(n).
%e A294874 For n = 36, it has three square-divisors: 4 = 2^(2*1), 9 = 3^(2*1) and 36 = 6^(2*1). Thus a(36) = prime(1) * prime(1) * prime(1) = 2*2*2 = 8.
%e A294874 For n = 64, it has three square-divisors: 4 = 2^(2*1), 16 = 2^(2*2) and 64 = 2^(2*3). Thus a(64) = prime(1) * prime(2) * prime(3) = 2*3*5 = 30.
%o A294874 (PARI) A294874(n) = { my(m=1,e); fordiv(n,d, if(d>1, e = ispower(d); if((e>1)&&!(e%2), m *= prime(e/2)))); m; };
%Y A294874 Cf. A000040, A046951, A052409, A071325.
%Y A294874 Cf. A293524, A294873, A294875.
%K A294874 nonn
%O A294874 1,4
%A A294874 _Antti Karttunen_, Nov 11 2017