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

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

Offset: 1

Author

Michael De Vlieger, May 03 2022

Keywords

• nonn
• more

Comments

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?

Examples

			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.
		

Crossrefs

Cf. A093714, A351499, A352928, A352932.

Programs

Mathematica

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]]