A242415 Reverse the deltas of indices of distinct primes in the prime factorization of n.

1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 11, 12, 13, 35, 10, 16, 17, 18, 19, 45, 21, 77, 23, 24, 25, 143, 27, 175, 29, 30, 31, 32, 55, 221, 14, 36, 37, 323, 91, 135, 41, 105, 43, 539, 20, 437, 47, 48, 49, 75, 187, 1573, 53, 54, 33, 875, 247, 667, 59, 60, 61, 899, 63, 64, 65

Offset: 1

Author

Antti Karttunen, May 24 2014

Keywords

• nonn

Comments

This self-inverse permutation (involution) of natural numbers preserves both the total number of prime divisors and the (index of) largest prime factor of n, i.e., for all n it holds that A001222(a(n)) = A001222(n) and A006530(a(n)) = A006530(n) [equally: A061395(a(n)) = A061395(n)]. It also preserves the exponent of the largest prime: A053585(a(n)) = A053585(n).

From the above it follows, that this fixes prime powers (A000961), among other numbers.

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 reverses the order of horizontal line segments of the "steps" in Young (or Ferrers) diagram of a partition, but keeps the order of vertical line segments intact. Please see the last example in the example section and compare also to the comments given in A242419.

Examples

			For n = 10 = 2*5 = p_1 * p_3, we get p_(3-1) * p_3 = 3 * 5 = 15, thus a(10) = 15.
For n = 20 = 2*2*5 = p_1^2 * p_3^1, we get p_(3-1)^2 * p_3^1 = 3^2 * 5 = 45, thus a(20) = 45.
For n = 84 = 2*2*3*7 = p_1^2 * p_2 * p_4, when we reverse the deltas of indices, but keep the exponents same, we get p_(4-2)^2 * p_(4-1) * p_4 = p_2^2 * p_3 * p_4 = 3^2 * 5 * 7 = 315, thus a(84) = 315.
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:
   _
  | |
  | |
  | |_ _
  |     |
  |     |_ _
  |_ _ _ _ _|
Reversing the order of horizontal line segment lengths (1,2,2) to (2,2,1), but keeping the order of vertical line segment lengths as (3,2,1), we get a new Young diagram
   _ _
  |   |
  |   |
  |   |_ _
  |       |
  |       |_
  |_ _ _ _ _|
which represents the partition (2,2,2,4,4,5), encoded in A112798 by p_2^3 * p_4^2 * p_5^1 = 3^3 * 7^2 * 11 = 14553, thus a(2200) = 14553.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8192
• Wikipedia, Young diagram
• Index entries for sequences that are permutations of the natural numbers

Crossrefs

Fixed points: A242413.

Cf. A112798, A122111, A153212, A000040, A241919, A067029, A242378, A051119, A225891.

{A000027, A069799, A242415, A242419} form a 4-group.

Formula

If n = p_a^e_a * p_b^e_b * ... * p_h^e_h * p_i^e_i * p_j^e_j * p_k^e_k, where p_a < ... < p_k are distinct primes (sorted into ascending order) in the prime factorization of n, and e_a .. e_k are their nonzero exponents, then a(n) = p_{k-j}^e_a * p_{k-i}^e_b * p_{k-h}^e_c * ... * p_{k-a}^e_j * p_k^e_k.

As a recurrence: a(1) = 1, and for n>1, a(n) = (A000040(A241919(n))^A067029(n)) * A242378(A241919(n), a(A051119(A225891(n)))).

By composing/conjugating related permutations:

a(n) = A069799(A242419(n)) = A242419(A069799(n)).

a(n) = A122111(A069799(A122111(n))) = A153212(A069799(A153212(n))).