a(n) = n^(1/k) with the smallest k>1 such that n is a k-th power, taking k=1 if no such k>1 exists.
0, 1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 12, 13, 14, 15, 4, 17, 18, 19, 20, 21, 22, 23, 24, 5, 26
6 seqfan posts
Sat Sep 4 18:09:17 CEST 2010 [seqfan] Re: A175781: What if n is 6th power (as 64) - n^(1/2)?
Sat Sep 4 17:58:26 CEST 2010 [seqfan] Re: A175781: What if n is 6th power (as 64) - n^(1/2)?
Sat Sep 4 17:58:26 CEST 2010 [seqfan] Re: A175781: What if n is 6th power (as 64) - n^(1/2)?
Sat Sep 4 15:12:03 CEST 2010 [seqfan] Re: A175781: What if n is 6th power (as 64) - n^(1/2)?
Sat Sep 4 14:14:58 CEST 2010 [seqfan] Re: A175781: What if n is 6th power (as 64) - n^(1/2)?
Sat Sep 4 12:05:15 CEST 2010 [seqfan] A175781: What if n is 6th power (as 64) - n^(1/2)?
Index of A-numbers in seqfan: by ascending order
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