

A241909


Selfinverse permutation of natural numbers: a(1)=1, a(p_i) = 2^i, and if n = p_i1 * p_i2 * p_i3 * ... * p_{ik1} * p_ik, where p's are primes, with their indexes are sorted into nondescending order: i1 <= i2 <= i3 <= ... <= i_{k1} <= ik, then a(n) = 2^(i11) * 3^(i2i1) * 5^(i3i2) * ... * p_k^(1+(iki_{k1})). Here k = A001222(n) and ik = A061395(n).


39



1, 2, 4, 3, 8, 9, 16, 5, 6, 27, 32, 25, 64, 81, 18, 7, 128, 15, 256, 125, 54, 243, 512, 49, 12, 729, 10, 625, 1024, 75, 2048, 11, 162, 2187, 36, 35, 4096, 6561, 486, 343, 8192, 375, 16384, 3125, 50, 19683, 32768, 121, 24, 45, 1458, 15625, 65536, 21, 108, 2401
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OFFSET

1,2


COMMENTS

This permutation maps between the partitions as ordered in A112798 and A241918 (the original motivation for this sequence).
For all n > 2, A007814(a(n)) = A055396(n)1, which implies that this selfinverse permutation maps between primes (A000040) and the powers of two larger than one (A000079(n>=1)), and apart from a(1) & a(2), this also maps each even number to some odd number, and vice versa, which means there are no fixed points after 2.
A122111 commutes with this one, that is, a(n) = A122111(a(A122111(n))).
Conjugates between A243051 and A242424 and other rows of A243060 and A243070.


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..1024
A. Karttunen, A few notes on A122111, A241909 & A241916.
Index entries for sequences that are permutations of the natural numbers


FORMULA

If n is a prime with index i (p_i), then a(n) = 2^i, otherwise when n = p_i1 * p_i2 * p_i3 * ... p_ik, where p_i1, p_i2, p_i3, ..., p_ik are the primes present (not necessarily all distinct) in the prime factorization of n, sorted into nondescending order, a(n) = 2^(i11) * 3^(i2i1) * 5^(i3i2) * ... * p_k^(1+(iki_{k1})).
Equally, if n = 2^k, then a(n) = p_k, otherwise, when n = 2^e1 * 3^e2 * 5^e3 * ... * p_k^{e_k}, i.e., where e1 ... e_k are the exponents (some of them possibly zero, except the last) of the primes 2, 3, 5, ... in the prime factorization of n, a(n) = p_{1+e1} * p_{1+e1+e2} * p_{1+e1+e2+e3} * ... * p_{e1+e2+e3+...+e_k}.
From the equivalence of the above two formulas (which are inverses of each other) it follows that a(a(n)) = n, i.e., that this permutation is an involution. For a proof, please see the attached notes.
The first formula corresponds to this recurrence:
a(1) = 1, a(p_k) = 2^k for primes with index k, otherwise a(n) = (A000040(A001222(n))^(A241917(n)+1)) * A052126(a(A052126(n))).
And the latter formula with this recurrence:
a(1) = 1, and for n>1, if n = 2^k, a(n) = A000040(k), otherwise a(n) = A000040(A001511(n)) * A242378(A007814(n), a(A064989(n))).
[Here A242378(k,n) changes each prime p(i) in the prime factorization of n to p(i+k), i.e., it's the result of A003961 iterated k times starting from n.]
We also have:
a(1)=1, and for n>1, a(n) = Product_{i=A203623(n1)+2..A203623(n)+1} A000040(A241918(i)).
For all n >= 1, A001222(a(n)) = A061395(n), and vice versa, A061395(a(n)) = A001222(n).
For all n > 1, a(2n1) = 2*a(A064216(n)).


EXAMPLE

For n = 12 = 2 * 2 * 3 = p_1 * p_1 * p_2, we obtain by the first formula 2^(11) * 3^(11) * 5^(1+(21)) = 5^2 = 25. By the second formula, as n = 2^2 * 3^1, we obtain the same result, p_{1+2} * p_{2+1} = p_3 * p_3 = 25, thus a(12) = 25.
Using the product formula over the terms of row n of table A241918, we see, because 9450 = 2*3*3*3*5*5*7 = p_1^1 * p_2^3 * p_3^2 * p_4^1, that the corresponding row in A241918 is {2,5,7,7}, and multiplying p_2 * p_5 * p_7^2 yields 3 * 11 * 17 * 17 = 9537, thus a(9450) = 9537.
Similarly, for 9537, the corresponding row in A241918 is {1,2,2,2,3,3,4}, and multiplying p_1^1 * p_2^3 * p_3^2 * p_4^1, we obtain 9450 back.


PROG

(Scheme, with Antti Karttunen's IntSeqlibrary)
(definec (A241909 n) (cond ((<= n 1) n) ((prime? n) (A000079 (A049084 n))) (else (* (A052126 (A241909 (A052126 n))) (expt (A000040 (A001222 n)) (+ 1 (A241917 n)))))))
;; Another recursive version:
(definec (A241909 n) (cond ((<= n 2) n) ((pow2? n) (A000040 (A007814 n))) (else (* (A000040 (A001511 n)) (A242378bi (A007814 n) (A241909 (A064989 n))))))) ;; The function A242378bi is given in A242378.
(define (pow2? n) (let loop ((n n) (i 0)) (cond ((zero? n) #f) ((odd? n) (and (= 1 n) i)) (else (loop (/ n 2) (1+ i)))))) ;; Gives nonfalse only when n is a power of two.
(Haskell)
a241909 1 = 1
a241909 n = product $ zipWith (^) a000040_list $ zipWith () is (1 : is)
where is = reverse ((j + 1) : js)
(j:js) = reverse $ map a049084 $ a027746_row n
 Reinhard Zumkeller, Aug 04 2014


CROSSREFS

Cf. A203623, A241918, A242378, A007814, A064989, A064216, A001222, A061395, A125976, A243060, A243070.
{A000027, A122111, A241909, A241916} form a 4group.
Cf. A049084, A027746.
Sequence in context: A129593 A279355 A279356 * A245451 A026166 A186003
Adjacent sequences: A241906 A241907 A241908 * A241910 A241911 A241912


KEYWORD

nonn


AUTHOR

Antti Karttunen, May 03 2014, partly inspired by Marc LeBrun's Jan 11 2006 message on SeqFan mailing list. Typos in the name corrected May 31 2014


STATUS

approved



