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.

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A093714 a(n) = smallest number coprime to a(n-1), not equal to a(n-1)+1, and not occurring earlier; a(1)=1.

Original entry on oeis.org

1, 3, 2, 5, 4, 7, 6, 11, 8, 13, 9, 14, 17, 10, 19, 12, 23, 15, 22, 21, 16, 25, 18, 29, 20, 27, 26, 31, 24, 35, 32, 37, 28, 33, 38, 41, 30, 43, 34, 39, 44, 47, 36, 49, 40, 51, 46, 45, 52, 55, 42, 53, 48, 59, 50, 57, 56, 61, 54, 65, 58, 63, 62, 67, 60, 71, 64, 69, 68, 73, 66, 79
Offset: 1

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Author

Reinhard Zumkeller, Apr 12 2004

Keywords

Comments

Lexicographically earliest infinite sequence of distinct positive numbers such that gcd(a(n-1), a(n)) = 1, a(n) != a(n-1) + 1. - N. J. A. Sloane, May 02 2022
Permutation of natural numbers with inverse A093715: a(A093715(n))=A093715(a(n))=n.
Comments from N. J. A. Sloane, May 02 2022: (Start)
Proof that this is a permutation of the natural numbers.
1. As usual for a "lexicographically earliest sequence" of this class, there is a function B(k) such that a(n) > k for all n > B(k).
2. For any prime p, p divides a(n) for some n. [Suppose not. Using 1, find n_0 such that a(n) > p^2 for all n >= n_0. But if a(n) > p^2, then p is a smaller choice for a(n+1), contradiction.]
3. For any prime p, p divides infinitely many terms. [Suppose not. Let p^i be greater than any multiple of p in the sequence. Go out a long way, and find a term greater than p^i. Then p^i is a smaller candidate for the next term. Contradiction.]
4. Every prime p appears naked. [If not, using 3, find a large multiple of p, G*p, say. But then p would have been a smaller candidate than G*p. Contradiction.]
5. The next term after a prime p is the smallest number not yet in the sequence which is relatively prime to p. Suppose k is missing from the sequence, and find a large prime p that does not divide k. Then the term after p will be k. So every number appears.
This completes the proof.
Conjecture 1: If p is a prime >= 3, p-1 appears after p.
Conjecture 2: If p is a prime, the first term divisible by p is p itself.
Conjecture 3: If a(n) = p is a prime >= 5, then n < p.
(End)
Coincides with A352588 for n >= 17. - Scott R. Shannon, May 02 2022

Crossrefs

Cf. A085229, A347113, A352588, A352928 (smallest missing number).
A352929 gives indices of prime terms, A352930 = first differences, A352931 = a(n)-n. See also A352932.
See Comments in A109812 for a set-theory analog.

Programs

A085230 Inverse of A085229.

Original entry on oeis.org

1, 3, 2, 5, 4, 7, 6, 9, 10, 11, 8, 13, 12, 15, 16, 17, 14, 19, 18, 21, 22, 23, 20, 25, 24, 27, 28, 29, 26, 31, 30, 33, 34, 35, 36, 37, 32, 39, 40, 41, 38, 43, 42, 45, 46, 47, 44, 49, 48, 51, 52, 53, 50, 55, 54, 57, 58, 59, 56, 61, 60, 63, 64, 65, 66, 67, 62, 69, 70
Offset: 1

Views

Author

Reinhard Zumkeller, Jun 22 2003

Keywords

Comments

Permutation of natural numbers: a(A085229(n)) = A085229(a(n)) = n.
Odd numbers appear in order, at a(2k) for k > 0. - Michael De Vlieger, Apr 13 2022

Crossrefs

Programs

A353706 n appears in A093714 a(n) steps earlier than expected.

Original entry on oeis.org

0, -1, 1, -1, 1, -1, 1, -1, -2, -4, 3, -4, 3, 2, -3, -5, 4, -5, 4, -5, 1, 3, 6, -5, 3, -1, 1, -5, 5, -7, 3, 1, -1, -5, 5, -7, 5, 3, -1, -5, 5, -9, 5, 3, -3, -1, 5, -5, 5, -5, 5, 3, 1, -5, 5, -1, 1, -3, 5, -5, 3, -1, 1, -3, 5, -5, 3, -1, 1, -3, 5, -5
Offset: 1

Views

Author

N. J. A. Sloane, May 05 2022

Keywords

Comments

It appears that a positive number n appears in A093714 at about the n-th term; if in fact A093714(k) = n, then a(n) = n - k.
An equivalent definition uses the inverse to A093714: a(n) = n - A093715(n).

Examples

			A093714(8) = 11, so 11 has appeared 3 steps early, therefore a(11) = 11 - 8 = 3.
		

Crossrefs

Programs

  • PARI
    See Links section.
Showing 1-3 of 3 results.