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A293514
a(n) = Product_{d|n, d>1} prime(A286561(n,d)), where A286561(n,d) gives the highest exponent of d dividing n.
10
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, 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, 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
OFFSET
1,2
FORMULA
a(n) = Product_{d|n, d>1} A000040(A286561(n,d)).
Other identities. For all n >= 1:
A001222(a(n)) = A032741(n).
A007814(a(n)) = A056595(n) [See A046951.]
1+A056239(a(n)) = A169594(n).
A064989(a(n)) = A293515(n).
EXAMPLE
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.
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.
PROG
(PARI) A293514(n) = { my(m=1); fordiv(n, d, if(d>1, m *= prime(valuation(n, d)))); m; };
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Nov 11 2017
STATUS
approved