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%I #21 Nov 30 2019 09:07:51
%S 1,2,3,8,5,6,7,4,27,10,11,24,13,14,15,32,17,54,19,40,21,22,23,12,125,
%T 26,9,56,29,30,31,16,33,34,35,216,37,38,39,20,41,42,43,88,135,46,47,
%U 96,343,250,51,104,53,18,55,28,57,58,59,120,61,62,189,64,65,66,67,136,69,70,71,108,73,74,375,152,77,78,79,160,243
%N a(1) = 1, for n > 1, a(n) = A020639(n)^A193231(A067029(n)) * a(A028234(n)).
%C Self-inverse permutation of natural numbers obtained by mapping the exponent of each prime in the prime factorization of n through involution A193231.
%C Multiplicative with p^e -> p^A193231(e), p prime and e > 0.
%H Antti Karttunen, <a href="/A268385/b268385.txt">Table of n, a(n) for n = 1..16384</a>
%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F a(1) = 1, and for n > 1, a(n) = A020639(n)^A193231(A067029(n)) * a(A028234(n)).
%F a(1) = 1, and for n > 1, a(n) = A000079(A193231(A007814(n))) * A003961(a(A064989(n))).
%F a(A059897(n,k)) = A059897(a(n), a(k)). - _Peter Munn_, Nov 27 2019
%e For n = 4 = 2^2, A193231(2) = 3, thus a(4) = 2^3 = 8.
%e For n = 9 = 3^2, A193231(2) = 3, thus a(9) = 3^3 = 27.
%e For n = 72 = 2^3 * 3^2, as A193231(2) = 3 and vice versa A193231(3) = 2, we have a(72) = 2^2 * 3^3 = 108. Note also how a(72) = a(8*9) = a(8) * a(9) = 4*27.
%e For n = 81 = 3^4, A193231(4) = 5, thus a(81) = 3^5 = 243.
%o (Scheme, two variants, both using memoization-macro definec)
%o (definec (A268385 n) (cond ((= 1 n) 1) (else (* (expt (A020639 n) (A193231 (A067029 n))) (A268385 (A028234 n))))))
%o (definec (A268385 n) (if (= 1 n) 1 (* (A000079 (A193231 (A007814 n))) (A003961 (A268385 (A064989 n))))))
%Y Cf. A000079, A003961, A007814, A020639, A028234, A064989, A067029, A193231.
%Y Cf. also A268386, A268387, A268675.
%K nonn,mult
%O 1,2
%A _Antti Karttunen_, Feb 10 2016
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