cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A243057 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} = 1 and for k>=1, p_{k} = A000040(k).

Original entry on oeis.org

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

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Author

Antti Karttunen, May 31 2014

Keywords

Comments

A243058 gives all n such that a(n) = n (the fixed points of this sequence, which include primes).
A102750 gives such n that a(a(n)) = n. A243286 is the self-inverse permutation induced when the domain is restricted to A102750. Cf. also A242420.
A070003 gives all n such that a(a(n)) <> n. Another variant, A243059, is defined to be zero when n is one of the terms of A070003.

Examples

			For n = 9 = 3*3 = p_2 * p_2, we have a(n) = p_{3-3} * p_3 = 1*3 = 3. [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 1 by 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.
		

Crossrefs

Fixed points: A243058 (includes primes).

Formula

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} = 1 and for k>=1, p_{k} = A000040(k).
a(1)=1, and for n>1, a(n) = p_{A243056(n)} * a(A032742(n)). Here p_{k} stands for 1 when k=0, and otherwise for the k-th prime, A000040(k).
For all n, a(n) = a(A243074(n)).
For all k in A102750, a(k) = A242420(k).

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

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Author

Antti Karttunen, May 31 2014

Keywords

Comments

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)).

Examples

			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.
		

Crossrefs

Fixed points: A243058 (includes primes).
Positions of zeros: A070003.

Formula

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).

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

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Author

Antti Karttunen, May 31 2014

Keywords

Comments

Number n is present, if its prime factorization n = p_a^e_a * p_b^e_b * p_c^e_c * ... * p_i^e_i * p_j^e_j * p_k^e_k where a < b < c < ... < i < j < k, satisfies the condition that the first differences of prime indices (a-0, b-a, c-b, ..., j-i, k-j) form a palindrome.
More formally, numbers n whose prime factorization is either of the form p^e (p prime, e >= 0), i.e., one of the terms of A000961, or of the form p_i1^e_i1 * p_i2^e_i2 * p_i3^e_i3 * ... * p_i_{k-1}^e_{i_{k-1}} * p_{i_k}^e_{i_k}, where p_i1 < p_i2 < ... < p_i_{k-1} < p_k are distinct primes (sorted into ascending order) in the prime factorization of n, and e_i1 .. e_{i_k} are their nonzero exponents (here k = A001221(n) and i_k = A061395(n), the index of the largest prime present), and the indices of primes satisfy the relation that for each index i_j < i_k present, the index i_{k-j} is also present.

Examples

			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.
		

Crossrefs

Fixed points of A242415.
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

Views

Author

Antti Karttunen, May 20 2014

Keywords

Comments

Numbers that are fixed by the permutation A242419, i.e., for which A242419(n) = n. Also, numbers that are fixed by both A069799 and A242415.
Number n is present if its prime factorization n = p_a^e_a * p_b^e_b * p_c^e_c * ... * p_i^e_i * p_j^e_j * p_k^e_k where a < b < c < ... < i < j < k, satisfies the condition, that both the first differences of prime indices (a-0, b-a, c-b, ..., j-i, k-j) and the respective exponents (e_a, e_b, e_c, ... , e_i, e_j, e_k) form a palindrome.
More formally, numbers n whose prime factorization is either of the form p^e (p prime, e >= 0), i.e., one of the terms of A000961, or of the form p_i1^e_i1 * p_i2^e_i2 * p_i3^e_i3 * ... * p_i_{k-1}^e_{i_{k-1}} * p_{i_k}^e_{i_k}, where p_i1 < p_i2 < ... < p_i_{k-1} < p_k are distinct primes (sorted into ascending order) in the prime factorization of n, and e_i1 .. e_{i_k} are their nonzero exponents (here k = A001221(n) and i_k = A061395(n), the index of the largest prime present), and the indices of primes satisfy the relation that for each index i_1 <= i_j < i_k present, the index i_{k-j} is also present, and the exponents e_{i_j} and e_{i_{(k-j)+1}} are equal.

Examples

			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.
		

Crossrefs

Fixed points of A242419. Intersection of A242413 and A242414.
Subsequences: A000961, A002110.

A243068 Fixed points of A242420.

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

Views

Author

Antti Karttunen, Jun 01 2014

Keywords

Comments

A number n is present if its prime factorization n = p_a * p_b * p_c * ... * p_i * p_j * p_k^e_k, where a <= b <= c <= ... <= i <= j < k are the indices of prime factors, not necessarily all distinct, except that j < k, and the greatest prime divisor p_k [with k = A061395(n)] may occur multiple times, satisfies the condition that the first differences of those prime indices (a-0, b-a, c-b, ..., j-i, k-j) form a palindrome.

Examples

			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.
		

Crossrefs

Fixed points of A242420.
Differs from A242413 for the first time at n=36, where a(36)=61, while A242413(36)=60.
A000040 and A243058 are subsequences.
Showing 1-5 of 5 results.