A268385 a(1) = 1, for n > 1, a(n) = A020639(n)^A193231(A067029(n)) * a(A028234(n)).

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, 26, 9, 56, 29, 30, 31, 16, 33, 34, 35, 216, 37, 38, 39, 20, 41, 42, 43, 88, 135, 46, 47, 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

Offset

1,2

Comments

Self-inverse permutation of natural numbers obtained by mapping the exponent of each prime in the prime factorization of n through involution A193231.

Multiplicative with p^e -> p^A193231(e), p prime and e > 0.

Links

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

Index entries for sequences that are permutations of the natural numbers

Formula

a(1) = 1, and for n > 1, a(n) = A020639(n)^A193231(A067029(n)) * a(A028234(n)).

a(1) = 1, and for n > 1, a(n) = A000079(A193231(A007814(n))) * A003961(a(A064989(n))).

a(A059897(n,k)) = A059897(a(n), a(k)). - Peter Munn, Nov 27 2019

Example

For n = 4 = 2^2, A193231(2) = 3, thus a(4) = 2^3 = 8.
For n = 9 = 3^2, A193231(2) = 3, thus a(9) = 3^3 = 27.
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.
For n = 81 = 3^4, A193231(4) = 5, thus a(81) = 3^5 = 243.

Prog

(Scheme, two variants, both using memoization-macro definec)
(definec (A268385 n) (cond ((= 1 n) 1) (else (* (expt (A020639 n) (A193231 (A067029 n))) (A268385 (A028234 n))))))
(definec (A268385 n) (if (= 1 n) 1 (* (A000079 (A193231 (A007814 n))) (A003961 (A268385 (A064989 n))))))

Crossrefs

Cf. A000079, A003961, A007814, A020639, A028234, A064989, A067029, A193231.

Cf. also A268386, A268387, A268675.

Sequence in context: A222243 A264978 A268675 * A093928 A356191 A135874

Adjacent sequences: A268382 A268383 A268384 * A268386 A268387 A268388

Keyword

nonn,mult

Author

Antti Karttunen, Feb 10 2016

Status

approved