

A153212


A permutation of the natural numbers: in the prime factorization of n, swap each prime's index difference (from the previous distinct prime that divides n) and the prime's power.


6



1, 2, 4, 3, 8, 6, 16, 5, 9, 18, 32, 15, 64, 54, 12, 7, 128, 10, 256, 75, 36, 162, 512, 35, 27, 486, 25, 375, 1024, 30, 2048, 11, 108, 1458, 24, 21, 4096, 4374, 324, 245, 8192, 150, 16384, 1875, 45, 13122, 32768, 77, 81, 50, 972, 9375, 65536, 14, 72, 1715, 2916, 39366, 131072, 105, 262144, 118098, 225, 13
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OFFSET

1,2


COMMENTS

In order for the "index difference" to make sense, we consider the factorization to be sorted with respect to the primes but not the powers to which they are raised; that is, first comes the smallest prime and each subsequent prime is larger than the previous disregarding their powers.
For every n it is true that a(a(n)) = n.
From Antti Karttunen, May 29 2014: (Start)
In other words, this is a selfinverse permutation (involution) of natural numbers.
This permutation maps primes (A000040) to the powers of two larger than one (A000079(n>=1)) and vice versa.
The term a(1) = 1 was added on the grounds that as 1 has an empty prime factorization, there is nothing to swap, thus it stays same. It is also needed as a base case for the given recurrence.
Considered as a function on partitions encoded by the indices of primes in the prime factorization of n (as in table A112798), this implements an operation which exchanges the horizontal and vertical line segment of each "step" in Young (or Ferrers) diagram of a partition. Please see the last example in the example section.
(End)


LINKS

Antti Karttunen, Table of n, a(n) for n = 1..1024
Wikipedia, Young diagram
Index entries for sequences that are permutations of the natural numbers


FORMULA

Denote the ith prime with p(i): p(1)=2, p(2)=3, p(3)=5, p(4)=7, etc. Let n = p(a1)^b1 * p(a2)^b2 * ... * p(ak)^bk is the factorization of n where p(i)^j is the ith prime raised to power j. As mentioned above, we assume that the primes are sorted, i.e., a1 < a2 < a3 ... < ak. Then a(n) = p(c1)^d1 * p(c2)^d2 * ... * p(ck)^dk where c1 = b1 and c(i) = b(i) + c(i1) for i > 1 d1 = a1 and d(i) = a(i)  a(i1) for i > 1.
From Antti Karttunen, May 16 2014: (Start)
a(1) = 1 and for n>1, let r = a(A051119(n)). Then a(n) = r * (A000040(A061395(r)+A071178(n)) ^ A241919(n)).
a(n) = A122111(A242419(n)) = A242419(A122111(n)).
(End)


EXAMPLE

For n = 10 we have 10 = 2^1 * 5^1 = p(1)^1 * p(3)^1 then a(10) = p(1)^1 * p(2)^2 = 2^1 * 3^2 = 18.
For n = 18 we have 18 = 2^1 * 3^2 = p(1)^1 * p(2)^2 then a(18) = p(1)^1 * p(3)^1 = 2^1 * 5^1 = 10.
For n = 19 we have 19 = 19^1 = p(8)^1 then a(19) = p(1)^8 = 2^8 = 256.
For n = 2200, we see that it encodes the partition (1,1,1,3,3,5) in A112798 as 2200 = p_1 * p_1 * p_1 * p_3 * p_3 * p_5 = 2^3 * 5^2 * 11. This in turn corresponds to the following Young diagram in French notation:
_
 
 
 _ _
 
 _ _
_ _ _ _ _
Exchanging the order of the horizontal and vertical line segment of each "step", results the following Young diagram:
_ _ _
 _ _
 
 _
 
_ _ _ _ _ _
which represents the partition (3,5,5,6,6), encoded in A112798 by p_3 * p_5^2 * p_6^2 = 5 * 11^2 * 13^2 = 102245, thus a(2200) = 102245.


PROG

(Scheme, with Antti Karttunen's IntSeqlibrary)
(definec (A153212 n) (if (<= n 1) n (let ((r (A153212 (A051119 n)))) (* r (expt (A000040 (+ (A061395 r) (A071178 n))) (A241919 n))))))
(PARI) a(n) = {my(f = factor(n)); my(g = f); for (i=1, #f~, if (i==1, g[i, 1] = prime(f[i, 2]), g[i, 1] = prime(f[i, 2]+ primepi(g[i1, 1]))); if (i==1, g[i, 2] = primepi(f[i, 1]), g[i, 2] = primepi(f[i, 1])  primepi(f[i1, 1])); ); factorback(g); } \\ Michel Marcus, Dec 16 2014


CROSSREFS

Cf. A000040, A051119, A061395, A071178.
Fixed points: A242421.
{A000027, A122111, A153212, A242419} form a 4group.
Cf. also A112798, A069799, A242415, A241909, A241916.
Sequence in context: A122111 A243053 A243052 * A244981 A284571 A124833
Adjacent sequences: A153209 A153210 A153211 * A153213 A153214 A153215


KEYWORD

nonn


AUTHOR

Luchezar Belev (l_belev(AT)yahoo.com), Dec 20 2008


EXTENSIONS

Term a(1)=1 prepended, and also more terms computed by Antti Karttunen, May 16 2014


STATUS

approved



