A293514 a(n) = Product_{d|n, d>1} prime(A286561(n,d)), where A286561(n,d) gives the highest exponent of d dividing n.

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

Links

Antti Karttunen, Table of n, a(n) for n = 1..16384

Index entries for sequences computed from exponents in factorization of n

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

Cf. A000040, A286561.

Cf. also A293515, A293524, A294876, A293442.

Sequence in context: A165460 A242649 A294876 * A028330 A293524 A363831

Adjacent sequences: A293511 A293512 A293513 * A293515 A293516 A293517

Keyword

nonn

Author

Antti Karttunen, Nov 11 2017

Status

approved