A242420
Self-inverse permutation of positive integers: a(n) = (A006530(n)^(A071178(n)-1)) * A243057(n).
Original entry on oeis.org
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
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:
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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:
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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:
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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.
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
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.
Apart from 1 also a subsequence of
A102750.
A242413
Numbers in whose prime factorization the first differences of indices of distinct primes form a palindrome; fixed points of A242415.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 21, 23, 24, 25, 27, 29, 30, 31, 32, 36, 37, 41, 43, 47, 48, 49, 53, 54, 59, 60, 61, 63, 64, 65, 67, 70, 71, 72, 73, 79, 81, 83, 89, 90, 96, 97, 101, 103, 107, 108, 109, 113, 120, 121, 125, 127, 128, 131, 133, 137, 139, 140, 144
Offset: 1
1 is present because it has an empty factorization, so both the sequence of the prime indices and their first differences are empty, and empty sequences are palindromes as well.
12 = 2*2*3 = p_1^2 * p_2 is present, as the first differences (deltas) of prime indices (1-0, 2-1) = (1,1) form a palindrome.
60 = 2*2*3*5 = p_1^2 * p_2 * p_3 is present, as the deltas of prime indices (1-0, 2-1, 3-2) = (1,1,1) form a palindrome.
61 = p_18 is present, as the deltas of prime indices, (18), form a palindrome.
144 = 2^4 * 3^2 = p_1^4 * p_2^2 is present, as the deltas of prime indices (1-0, 2-1) = (1,1) form a palindrome.
Also, any of the cases mentioned in the Example section of A242417 as being present there, are also present in this sequence.
Differs from
A243068 for the first time at n=36, where a(36)=60, while
A243068(36)=61.
A242417
Numbers in whose prime factorization both the first differences of indices of distinct primes and their exponents form a palindrome; fixed points of A242419.
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 16, 17, 19, 21, 23, 25, 27, 29, 30, 31, 32, 36, 37, 41, 43, 47, 49, 53, 59, 61, 64, 65, 67, 70, 71, 73, 79, 81, 83, 89, 90, 97, 101, 103, 107, 109, 113, 121, 125, 127, 128, 131, 133, 137, 139, 149, 151, 154, 157, 163, 165, 167, 169
Offset: 1
1 is present because it has an empty factorization, so both sequences are empty, thus palindromes.
3 = p_2^1 is present, as both the sequence of the first differences (deltas) of prime indices (2-0) = (2) and the exponents (1) are palindromes.
6 = p_1^1 * p_2^1 is present, as both the deltas of prime indices (1-0, 2-1) = (1,1) and the exponents (1,1) form a palindrome.
8 = p_1^3 is present, as both the deltas of prime indices (1) and the exponents (3) form a palindrome.
300 = 4*3*25 = p_1^2 * p_2^1 * p_3^2 is present, as both the deltas of prime indices (1-0, 2-1, 3-2) = (1,1,1) 1, 2 and the exponents (2,1,2), form a palindrome.
144 = 2^4 * 3^2 = p_1^4 * p_2^2 is NOT present, as although the deltas of prime indices (1-0, 2-1) = (1,1) are palindrome, the sequence of exponents (4,2) do NOT form a palindrome.
441 = 9*49 = p_2^2 * p_4^2 is present, as both the deltas of prime indices (2-0, 4-2) = (2,2) and the exponents (2,2) form a palindrome.
30030 = 2*3*5*7*11*13 = p_1 * p_2 * p_3 * p_4 * p_5 * p_6 is present, as the exponents are all ones, and the deltas of indices, (6-5,5-4,4-3,3-2,2-1,1-0) = (1,1,1,1,1,1) likewise are all ones, thus both sequences form a palindrome. This is true for all primorial numbers, A002110.
47775 = 3*5*5*7*7*13 = p_2^1 * p_3^2 * p_4^2 * p_6^1 is present, as the deltas of indices (6-4,4-3,3-2,2-0) = (2,1,1,2) and the exponents (1,2,2,1) both form a palindrome.
90000 = 2*2*2*2*3*3*5*5*5*5 = p_1^4 * p_2^2 * p_3^4 is present, as the deltas of indices (3-2,2-1,1-0) = (1,1,1) and the exponents (4,2,4) both form a palindrome.
Showing 1-4 of 4 results.
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