A352928 -id:A352928 - cp's OEIS frontend

A351499 Odd m in A352928.

1, 9, 15, 75, 105, 315, 525, 735, 945, 1155, 1365, 1575, 1995, 3465, 4305, 5775, 6615, 7035, 8085, 8925, 9765, 10395, 11235, 12495, 12705, 13545, 15015, 19635, 26565, 28875, 31185, 33495, 38115, 45045, 255255, 405405, 525525, 765765, 975975, 1036035, 1786785, 2297295

Offset: 1

Michael De Vlieger, May 03 2022

nonn

Cf. A093714, A351498, A352928.

nn = 2^20; c = {1}; j = 1; s = 0; u = 1; {1}~Join~Reap[Do[k = u; While[Nand[FreeQ[c, k], CoprimeQ[j, k], k != j + 1], k++]; j = k; AppendTo[c, k]; If[k == u, If[OddQ[u], Sow[u]]; While[MemberQ[c, u], u++]; c = DeleteCases[c, _?(# < u &)]], {i, 2, nn}]][[-1, -1]]

from math import gcd
from itertools import islice
def agen(): # generator of terms
    an, aset, mink, seen = 1, {1}, 2, {1}
    yield 1
    while True:
        if mink%2 and mink not in seen: yield mink; seen.add(mink)
        k = mink
        while k in aset or gcd(an, k) != 1 or k-an == 1: k += 1
        an = k; aset.add(an)
        while mink in aset: mink += 1
print(list(islice(agen(), 42))) # Michael S. Branicky, May 03 2022

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

Reinhard Zumkeller, Apr 12 2004

nonn

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

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Index entries for sequences that are permutations of the natural numbers

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.

nn = 120; c[] = 0; a[1] = c[1] = 1; u = 2; Do[k = u; While[Nand[c[k] == 0, CoprimeQ[#, k], k != # + 1], k++] &@ a[i - 1]; Set[{a[i], c[k]}, {k, i}]; If[a[i] == u, While[c[u] > 0, u++]], {i, 2, nn}]; Array[a, nn] (* _Michael De Vlieger, May 02 2022 *)

from math import gcd
from itertools import islice
def agen(): # generator of terms
    an, aset, mink = 1, {1}, 2
    while True:
        yield an
        k = mink
        while k in aset or gcd(an, k) != 1 or k-an == 1: k += 1
        an = k
        aset.add(an)
        while mink in aset: mink += 1
print(list(islice(agen(), 72))) # Michael S. Branicky, May 02 2022

A351498 Position k of first term in a run of odd terms with length > 1 in A093714.

Original entry on oeis.org

1, 10, 17, 78, 1787, 15022, 38123, 45052, 1036043

Offset: 1

Michael De Vlieger, May 03 2022

Let S = A093714, known to be a permutation of the natural numbers.
S(k) and S(k+1) are odd.
Theorem: S(n) = n implies S(n) and n have the same parity. Proof: S(n) = n iff S(n) mod n = 0, since n | n. If prime q|n yet (S(n),q) = 1, then S(n) != n, a contradiction.
Corollary: Fixed points in S(n) appear iff S(n) and n have same parity.
Corollary: Consecutive odd terms S(k), S(k+1) imply that the smallest missing number u in S is either odd or (S(k),u) > 1.
Fixed points S(k) appear at k in intervals [78, 1787], [15022, 38123], [45052, 1036043]. These positions result from a dataset of S(1..2^30). Are there any more runs of odd terms?

			S(1) = 1 and S(2) = 3, therefore this sequence begins with 1.
S(10) = 13 and S(11) = 9, therefore 10 is in this sequence.
S(17) = 23 and S(18) = 15, therefore 17 is in this sequence.
For 2 <= k <= 10, S(k) and k have different parity, but for 11 <= k <= 17, S(k) and k have the same parity.

Cf. A093714, A351499, A352928, A352932.

nn = 2^20, c = {1}, j = 1, s = 0; t = {}; u = 2; Reap[Do[k = u; While[Nand[FreeQ[c, k], CoprimeQ[j, k], k != j + 1], k++]; j = k; AppendTo[c, k]; If[# != s, s = #; Sow[i - 1]] &@ Mod[Abs[k - i], 2]; If[k == u, While[MemberQ[c, u], u++]; c = DeleteCases[c, _?(# < u &)]], {i, 2, nn}] ][[-1, -1]]

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A351499 Odd m in A352928.

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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.

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A351498 Position k of first term in a run of odd terms with length > 1 in A093714.

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