A242415
Reverse the deltas of indices of distinct primes in the prime factorization of 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, 60, 61, 899, 63, 64, 65
Offset: 1
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:
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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
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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.
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.
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.
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, 61, 63, 64, 65, 67, 70, 71, 72, 73, 79, 81, 83, 89, 96, 97, 101, 103, 107, 108, 109, 113, 121, 125, 127, 128, 131, 133, 137, 139, 144
Offset: 1
4 = p_1^2 is present, as the first differences (deltas) of the prime indices (excluding the extra copies of the largest prime factor 2), form a palindrome: (1-0) = (1).
18 = 2*3*3 = p_1 * p_2 * p_2 is present, as the deltas of the indices of its nondistinct prime factors, (excluding the extra copies of the largest prime factor 3) form a palindrome: (1-0, 2-1) = (1,1).
60 = 2*2*3*5 = p_1 * p_1 * p_2 * p_3 is NOT present, as the deltas of prime indices (1-0, 1-1, 2-1, 3-2) = (1,0,1,1) do NOT form a palindrome.
Also, any of the cases mentioned in the Example section of A243058 as being present there, are also present in this sequence.
Differs from
A242413 for the first time at n=36, where a(36)=61, while
A242413(36)=60.
Showing 1-4 of 4 results.
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