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a(2^k) = 2^k, and for other numbers, if n = 2^e1 * 3^e2 * 5^e3 * ... p_k^e_k, then a(n) = 2^(e_k - 1) * 3^(e_{k-1}) * ... * p_{k-1}^e2 * p_k^(e1+1). Here p_k is the greatest prime factor of n (A006530), and e_k is its exponent (A071178), and the exponents e1, ..., e_{k-1} >= 0.
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%I #34 Dec 29 2022 11:19:14

%S 1,2,3,4,5,9,7,8,6,25,11,27,13,49,15,16,17,18,19,125,35,121,23,81,10,

%T 169,12,343,29,75,31,32,77,289,21,54,37,361,143,625,41,245,43,1331,45,

%U 529,47,243,14,50,221,2197,53,36,55,2401,323,841,59,375,61,961,175,64

%N a(2^k) = 2^k, and for other numbers, if n = 2^e1 * 3^e2 * 5^e3 * ... p_k^e_k, then a(n) = 2^(e_k - 1) * 3^(e_{k-1}) * ... * p_{k-1}^e2 * p_k^(e1+1). Here p_k is the greatest prime factor of n (A006530), and e_k is its exponent (A071178), and the exponents e1, ..., e_{k-1} >= 0.

%C For other numbers than the powers of 2 (that are fixed), this permutation reverses the sequence of exponents in the prime factorization of n from the exponent of 2 to that of the largest prime factor, except that the exponents of 2 and the greatest prime factor present are adjusted by one. Note that some of the exponents might be zeros.

%C Self-inverse permutation of natural numbers, composition of A122111 & A241909 in either order: a(n) = A122111(A241909(n)) = A241909(A122111(n)).

%C This permutation preserves both bigomega and the (index of) largest prime factor: for all n it holds that A001222(a(n)) = A001222(n) and A006530(a(n)) = A006530(n) [equally: A061395(a(n)) = A061395(n)].

%C From the above it follows, that this fixes both primes (A000040) and powers of two (A000079), among other numbers.

%C Even positions from n=4 onward contain only terms of A070003, and the odd positions only the terms of A102750, apart from 1 which is at a(1), and 2 which is at a(2).

%H Antti Karttunen, <a href="/A241916/b241916.txt">Table of n, a(n) for n = 1..8192</a>

%H A. Karttunen, <a href="/A122111/a122111.txt">A few notes on A122111, A241909 & A241916.</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) = A006530(n) * A137502(n)/2.

%F a(n) = A122111(A241909(n)) = A241909(A122111(n)).

%F If 2n has prime factorization Product_{i=1..k} prime(x_i), then a(n) = Product_{i=1..k-1} prime(x_k-x_i+1). The opposite version is A000027, even bisection of A246277. - _Gus Wiseman_, Dec 28 2022

%t nn = 65; f[n_] := If[n == 1, {0}, Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, f]]@ FactorInteger@ n]; g[w_List] := Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, w]; Table[If[IntegerQ@ #, n/4, g@ Reverse@(# - Join[{1}, ConstantArray[0, Length@ # - 2], {1}] &@ f@ n)] &@ Log2@ n, {n, 4, 4 nn, 4}] (* _Michael De Vlieger_, Aug 27 2016 *)

%o (PARI)

%o A209229(n) = (n && !bitand(n,n-1));

%o A241916(n) = if(1==A209229(n), n, my(f = factor(2*n), nbf = #f~, igp = primepi(f[nbf,1]), g = f); for(i=1,nbf,g[i,1] = prime(1+igp-primepi(f[i,1]))); factorback(g)/2); \\ _Antti Karttunen_, Jul 02 2018

%o (Scheme) (define (A241916 n) (A122111 (A241909 n)))

%Y A241912 gives the fixed points; A241913 their complement.

%Y Cf. A006530, A137502, A070003, A102750, A278525.

%Y {A000027, A122111, A241909, A241916} form a 4-group.

%Y The sum of prime indices of a(n) is A243503(n).

%Y Even bisection of A358195 = Heinz numbers of rows of A358172.

%Y A112798 lists prime indices, length A001222, sum A056239.

%Y Cf. A005940, A019565, A161511, A246277, A253565, A326844, A356958, A358137.

%K nonn

%O 1,2

%A _Antti Karttunen_, May 03 2014

%E Description clarified by _Antti Karttunen_, Jul 02 2018