A243057 -id:A243057 - cp's OEIS frontend

A242420 Self-inverse permutation of positive integers: a(n) = (A006530(n)^(A071178(n)-1)) * A243057(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, 90, 61, 899, 63, 64, 65

Offset: 1

Antti Karttunen, May 31 2014

nonn

This self-inverse permutation (involution) of positive integers 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 factor (A071178), from which follows that the sequence A102750 is closed with respect to this permutation, i.e., for all n in A102750, a(n) is either same n or some other term of A102750.
Considered as an operation on partitions encoded by the indices of primes in the prime factorization of n (as in table A112798), this implements a self-inverse bijection which is a composition of the effects of A242419 and A225891. (Or equally: A105119 and A242419). For details, please see the respective Comments sections and/or Example section of this entry.

			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:
   _
  | |
  | |
  | |_ _
  |     |
  |     |_ _
  |_ _ _ _ _|
First we apply A242419, which reverses the order of "steps", so that each horizontal and vertical line segment centered around a "convex corner" moves as a whole, so that the first stair from the top (one unit wide and three units high) is moved to the last position, the second one (two units wide and two units high) stays in the middle, and the original bottom step (two units wide and one unit high) will be the new topmost step, thus we get the following Young diagram:
   _ _
  |   |_ _
  |       |
  |       |_
  |         |
  |         |
  |_ _ _ _ _|
which represents the partition (2,4,4,5,5,5), encoded in A112798 by p_2 * p_4^2 * p_5^3 = 3 * 7^2 * 11^3 = 195657.
Then we apply A225891, which rotates the exponents of distinct primes in the factorization of n one left, in this context the vertical line segments one step up, with the top-one going to the bottomost, and so we get:
   _ _
  |   |
  |   |_ _
  |       |
  |       |
  |       |_
  |_ _ _ _ _|
which represents the partition (2,2,4,4,4,5), encoded in A112798 by p_2^2 * p_4^3 * p_5 = 3^2 * 7^3 * 11 = 33957, thus a(2200) = 33957.

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

Fixed points: A243068.

Cf. A112798, A242419, A105119, A225891, A006530, A071178, A102750, A243057, A243059.

(define (A242420 n) (* (expt (A006530 n) (- (A071178 n) 1)) (A243057 n)))

a(n) = (A006530(n)^(A071178(n)-1)) * A243057(n).
For all k in A102750, a(k) = A243057(k) = A243059(k).
By composing related permutations:
a(n) = A225891(A242419(n)) = A242419(A105119(n)).

A243058 Fixed points of A243057 and A243059.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 11, 12, 13, 17, 19, 21, 23, 24, 29, 30, 31, 37, 41, 43, 47, 48, 53, 59, 61, 63, 65, 67, 70, 71, 73, 79, 83, 89, 96, 97, 101, 103, 107, 109, 113, 127, 131, 133, 137, 139, 149, 151, 154, 157, 163, 165, 167, 173, 179, 180, 181, 189, 191, 192, 193, 197, 199, 210

Offset: 1

Antti Karttunen, May 31 2014

nonn

Number n is present if its prime factorization n = p_a * p_b * p_c * ... * p_i * p_j * p_k (where a <= b <= c <= ... <= i <= j <= k are the indices of prime factors, not necessarily all distinct; sorted into nondescending order) satisfies the condition that the first differences of those prime indices (a-0, b-a, c-b, ..., j-i, k-j) form a palindrome.

The above condition implies that none of the terms of A070003 are present, as then at least the difference k-j would be zero, but on the other hand, a-0 is at least 1. Cf. also A243068.

			12 = 2*2*3 = p_1 * p_1 * p_2 is present, as the first differences (deltas) of the indices of its nondistinct prime factors (1-0, 1-1, 2-1) = (1,0,1) form a palindrome.
18 = 2*3*3 = p_1 * p_2 * p_2 is NOT present, as the deltas of the indices of its nondistinct prime factors (1-0, 2-1, 2-2) = (1,1,0) do NOT form a palindrome.
65 = 5*13 = p_3 * p_6 is present, as the deltas of the indices of its nondistinct prime factors (3-0, 6-3) = (3,3) form a palindrome.

Antti Karttunen, Table of n, a(n) for n = 1..2048

A subsequence of A243068.
Apart from 1 also a subsequence of A102750.
A000040 is a subsequence.
Cf. A242413, A242417, A243057, A243059, A242417.

A243286 Self-inverse permutation of natural numbers induced by the restriction of A243057 (or A243059 or A242420) to the union of {1} and A102750.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 12, 8, 9, 10, 27, 7, 13, 14, 36, 16, 62, 18, 19, 121, 148, 22, 23, 24, 43, 191, 11, 28, 283, 75, 113, 32, 87, 34, 481, 15, 388, 38, 39, 160, 1456, 42, 25, 795, 213, 602, 47, 74, 49, 818, 51, 52, 339, 54, 2699, 345, 57, 58, 59, 1053, 5219, 17, 914, 64

Offset: 1

Antti Karttunen, Jun 02 2014

nonn

			With n=7, the seventh number in the complement of A070003 (i.e. in the union of {1} and A102750) is A102750(6) = 10. When A243057 (or A243059) is applied to it, the result is another number that is a member of A102750, in this case 15, which occurs there as A102750(11). Thus a(7) = 11+1 = 12.

Antti Karttunen, Table of n, a(n) for n = 1..126
Index entries for sequences that are permutations of the natural numbers

Cf. A243057, A243059, A102750, A243285, A242420.

(define (A243286 n) (if (<= n 1) n (+ 1 (A243285 (A243057 (A102750 (- n 1)))))))

a(1) = 1, and for n>1, a(n) = 1 + A243285(A243057(A102750(n-1))). [Note: instead of A243057 one can also use A243059 or A242420.]

A243056 If n is the i-th prime, p_i = A000040(i), then a(n) = i, otherwise the difference between the indices of the smallest and the largest prime dividing n: for n = p_i * ... * p_k, where p_i <= ... <= p_k, a(n) = (k-i); a(1) = 0 by convention.

Original entry on oeis.org

0, 1, 2, 0, 3, 1, 4, 0, 0, 2, 5, 1, 6, 3, 1, 0, 7, 1, 8, 2, 2, 4, 9, 1, 0, 5, 0, 3, 10, 2, 11, 0, 3, 6, 1, 1, 12, 7, 4, 2, 13, 3, 14, 4, 1, 8, 15, 1, 0, 2, 5, 5, 16, 1, 2, 3, 6, 9, 17, 2, 18, 10, 2, 0, 3, 4, 19, 6, 7, 3, 20, 1, 21, 11, 1, 7, 1, 5, 22, 2, 0, 12, 23, 3, 4, 13, 8, 4, 24, 2

Offset: 1

Antti Karttunen, May 31 2014

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Useful when computing A243057 or A243059.
A025475 (prime powers that are not primes) gives the positions of zeros.
Differs from A241917 for the first time at n=18.
Cf. A243055, A241919, A242411, A049084, A027748, A055396, A061395.

(define (A243056 n) (if (prime? n) (A049084 n) (A243055 n)))

a(1) = 0, for n>1, if n = A000040(i), a(n) = i, otherwise a(n) = A061395(n) - A055396(n) = A243055(n).

A242419 Reverse both the exponents and the deltas of the indices of distinct primes present in the prime factorization of n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 15, 11, 18, 13, 35, 10, 16, 17, 12, 19, 75, 21, 77, 23, 54, 25, 143, 27, 245, 29, 30, 31, 32, 55, 221, 14, 36, 37, 323, 91, 375, 41, 105, 43, 847, 50, 437, 47, 162, 49, 45, 187, 1859, 53, 24, 33, 1715, 247, 667, 59, 150, 61, 899, 147, 64, 65

Offset: 1

Antti Karttunen, May 17 2014

nonn

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)].
Considered as an operation on partitions encoded by the indices of primes in the prime factorization of n (as in table A112798), this implements a bijection which reverses the order of "steps" in Young (or Ferrers) diagram of a partition (but keeps the horizontal line segment of each step horizontal and the vertical line segment vertical). Please see the last example in the example section.
To understand the given recursive formula, it helps to see that in the above context (Young diagrams drawn with French notation), the sequences employed effect the following operations:
  A001222: gives the height of whole diagram,
  A051119: removes the bottommost step from the diagram,
  A241919: gives the length of the horizontal line segment of the bottom step, i.e. its width,
  A071178: gives the length of the vertical line segment of the bottom step, i.e. its height,
  A242378(k,n): increases the width of whole Young diagram encoded by n by adding a rectangular area A001222(n) squares high and k squares wide to its left,
  and finally, multiplying by A000040(a)^b adds a new topmost step whose width is a and height is b. Particularly, multiplying by (A000040(A241919(n))^A071178(n)) transfers the bottommost step to the top.

			For n = 10 = 2*5 = p_1^1 * p_3^1, we get p_(3-1)^1 * p_3^1 = 3 * 5 = 15, thus a(10) = 15.
For n = 20 = 2*2*5 = p_1^2 * p_3^1, we get p_(3-1)^1 * p_3^2 = 3^1 * 5^2 = 3*25 = 75, thus a(20) = 75.
For n = 84 = 2*2*3*7 = p_1^2 * p_2 * p_4, when we reverse the deltas of indices, and reverse also the order of exponents, we get p_(4-2) * p_(4-1) * p_4^2 = 3 * 5 * 7^2 = 735, thus a(84) = 735.
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 "steps", so that each horizontal and vertical line segment centered around a "convex corner" moves as a whole, means that the first stair from the top (one unit wide and three units high) is moved to the last position, the second one (two units wide and two units high) stays in the middle, and the original bottom step (two units wide and one unit high) will be the new topmost step, thus we get the following Young diagram:
   _ _
  |   |_ _
  |       |
  |       |_
  |         |
  |         |
  |_ _ _ _ _|
which represents the partition (2,4,4,5,5,5), encoded in A112798 by p_2 * p_4^2 * p_5^3 = 3 * 7^2 * 11^3 = 195657, thus a(2200) = 195657.

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

Fixed points: A242417.
Cf. A112798, A122111, A153212, A000040, A241919, A071178, A242378, A051119, A243057, A105119, A242420.
{A000027, A122111, A153212, A242419} form a 4-group.
{A000027, A069799, A242415, A242419} form also a 4-group.

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_k * p_{k-i}^e_j  * p_{k-h}^e_i * ... * p_{k-a}^e_b * p_k^e_a.
As a recurrence:
a(1) = 1, and for n>1, a(n) = (A000040(A241919(n))^A071178(n)) * A242378(A241919(n), a(A051119(n))).
By composing related permutations:
a(n) = A122111(A153212(n)) = A153212(A122111(n)).
a(n) = A069799(A242415(n)) = A242415(A069799(n)).
a(n) = A105119(A242420(n)).

A243059 If n = p_a * p_b * ... * p_h * p_i * p_j * p_k, with p_a <= p_b <= ... <= p_k, then a(n) = p_{k-j} * p_{k-i} * p_{k-h} * ... * p_{k-a} * p_k, where p_{0} = 0 and for k>=1, p_{k} = A000040(k). a(1)=1 by convention.

Original entry on oeis.org

1, 2, 3, 0, 5, 6, 7, 0, 0, 15, 11, 12, 13, 35, 10, 0, 17, 0, 19, 45, 21, 77, 23, 24, 0, 143, 0, 175, 29, 30, 31, 0, 55, 221, 14, 0, 37, 323, 91, 135, 41, 105, 43, 539, 20, 437, 47, 48, 0, 0, 187, 1573, 53, 0, 33, 875, 247, 667, 59, 90, 61, 899, 63, 0, 65, 385, 67, 2873, 391, 70, 71, 0

Offset: 1

Antti Karttunen, May 31 2014

nonn

A243058 gives all n such that a(n) = n (the fixed points of this sequence, which include primes).

Differs from A243057 in that the "degenerate cases" A070003 are here zeros, but is otherwise equal to it (at the points given by A102750), i.e. for all n, a(A102750(n)) = A243057(A102750(n)) = A242420(A102750(n)).

			For n = 9 = 3*3 = p_2 * p_2, we have a(n) = p_{3-3} * p_3 = 0*3 = 0. [Like all terms in A070003 this is an example of "degenerate case", where some p's in the product get index 0, and thus are set to 0 by the convention used here.]
For n = 10 = 2*5 = p_1 * p_3, we have a(n) = p_{3-1} * p_3 = 3*5 = 15.
For n = 12 = 2*2*3 = p_1 * p_1 * p_2, we have a(n) = p_{2-1} * p{2-1} * p_2 = p_1^2 * p_2 = 12.
For n = 15 = 3*5 = p_2 * p_3, we have a(n) = p_{3-2} * p_3 = 2*5 = 10.
For n = 2200 = 2*2*2*5*5*11 = p_1 * p_1 * p_1 * p_3 * p_3 * p_5, we have a(n) = p_{5-3} * p_{5-3} * p_{5-1} * p_{5-1} * p_{5-1} * p_5 = 3*3*7*7*7*11 = 33957.
For n = 33957 = 3*3*7*7*7*11 = p_2 * p_2 * p_4 * p_4 * p_4 * p_5, we have a(n) = p_{5-4} * p_{5-4} * p_{5-4} * p_{5-2} * p_{5-2} * p_5 = 2*2*2*5*5*11 = 2200.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Fixed points: A243058 (includes primes).
Positions of zeros: A070003.
Cf. A243057, A000040, A032742, A243056, A242419, A242420, A243286.

a(1)=1, and for n>1, a(n) = q_{A243056(n)} * a(A032742(n)). Here q_{k} stands for 0 when k=0, and otherwise for the k-th prime, A000040(k).

If n = p_a * p_b * ... * p_h * p_i * p_j * p_k, where p_a <= p_b <= ... <= p_k are (not necessarily distinct) primes (sorted into nondescending order) in the prime factorization of n, then a(n) = p_{k-j} * p_{k-i} * p_{k-h} * ... * p_{k-a} * p_k, where p_{0} = 0 and for k>=1, p_{k} = A000040(k).

A243074 a(1) = 1, a(n) = n/p^(k-1), where p = largest prime dividing n and p^k = highest power of p dividing n.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 2, 3, 10, 11, 12, 13, 14, 15, 2, 17, 6, 19, 20, 21, 22, 23, 24, 5, 26, 3, 28, 29, 30, 31, 2, 33, 34, 35, 12, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 7, 10, 51, 52, 53, 6, 55, 56, 57, 58, 59, 60, 61, 62, 63, 2, 65, 66, 67, 68, 69, 70, 71, 24, 73, 74, 15, 76, 77, 78, 79, 80, 3

Offset: 1

Antti Karttunen, May 31 2014

nonn

After 1, A102750 gives such k that a(k) = k, which are also the positions of the records as for all n, a(n) <= n. After 1, only terms of A102750 occur, each an infinite number of times.

			For n = 18 = 2*3*3, we discard all instances of the highest prime factor 3 except one, thus a(18) = 2*3 = 6.
For n = 54 = 2*3*3*3, we discard two copies of 3, and thus also the value of a(54) is 2*3 = 6.
For n = 20 = 2*5, the highest prime factor 5 occurs only once, so nothing is cast off, and a(20) = 20.

Antti Karttunen, Table of n, a(n) for n = 1..10001

Differs from A052410 for the first time at n=18.

Cf. A006530, A051119, A102750, A070003, A243057.

(define (A243074 n) (* (A051119 n) (A006530 n)))

a(n) = A006530(n) * A051119(n).

cp's OEIS Frontend

A242420 Self-inverse permutation of positive integers: a(n) = (A006530(n)^(A071178(n)-1)) * A243057(n).

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A243058 Fixed points of A243057 and A243059.

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A243286 Self-inverse permutation of natural numbers induced by the restriction of A243057 (or A243059 or A242420) to the union of {1} and A102750.

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A243056 If n is the i-th prime, p_i = A000040(i), then a(n) = i, otherwise the difference between the indices of the smallest and the largest prime dividing n: for n = p_i * ... * p_k, where p_i <= ... <= p_k, a(n) = (k-i); a(1) = 0 by convention.

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A242419 Reverse both the exponents and the deltas of the indices of distinct primes present in the prime factorization of n.

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A243059 If n = p_a * p_b * ... * p_h * p_i * p_j * p_k, with p_a <= p_b <= ... <= p_k, then a(n) = p_{k-j} * p_{k-i} * p_{k-h} * ... * p_{k-a} * p_k, where p_{0} = 0 and for k>=1, p_{k} = A000040(k). a(1)=1 by convention.

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A243074 a(1) = 1, a(n) = n/p^(k-1), where p = largest prime dividing n and p^k = highest power of p dividing n.

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