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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.

A366908 Lexicographically earliest infinite sequence of distinct positive integers such that, for n > 1, a(n) shares a factor with n but does not equal n, while not sharing a factor with a(n-1).

Original entry on oeis.org

1, 4, 9, 2, 15, 8, 21, 10, 3, 5, 22, 27, 26, 7, 25, 6, 85, 12, 95, 14, 33, 16, 69, 28, 45, 13, 18, 35, 58, 39, 155, 24, 11, 17, 20, 51, 74, 19, 36, 55, 82, 49, 86, 77, 30, 23, 94, 57, 56, 65, 34, 91, 106, 63, 40, 119, 38, 29, 118, 75, 122, 31, 81, 32, 105, 44, 201, 46, 87, 50, 213, 52, 219, 37
Offset: 1

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Author

Scott R. Shannon, Oct 27 2023

Keywords

Comments

To ensure the sequence is infinite a(n) must be chosen so that it does not have as prime factors all the distinct prime factors of n+1. The first time this rule is required is when determining a(15); see the examples below.
For the terms studied the primes appear in their natural order except for 11 and 13 which are reversed. The sequence is conjectured to be a permutation of the positive integers.
Observation: apart from a(4) = 2, a(9) = 3, and a(33) = 11, prime a(n) is such that n is congruent to +- 2 (mod 12). - Michael De Vlieger, Oct 29 2023

Examples

			a(4) = 2 as 2 does not equal 4, shares the factor 2 with 4 while not sharing a factor with a(3) = 9.
a(15) = 25 as 25 does not equal 15, shares the factor 5 with 15 while not sharing a factor with a(14) = 7. Note that 6 is unused and satisfies these requirements but as 15 + 1 = 16 = 2^4 only contains 2 as a distinct prime factor, a(15) cannot also contain 2 as a factor else a(16) would not exist.
		

Crossrefs

Programs

  • Mathematica
    nn = 1000;
      c[] := False; m[] := 1;
      f[x_] := f[x] = Times @@ FactorInteger[x][[All, 1]];
      a[1] = 1; j = a[2] = 4; c[1] = c[4] = True; u = 2;
      Do[k = u;
        If[PrimePowerQ[n], p = FactorInteger[n][[1, 1]]; k = m[p];
         While[
          Or[c[#], ! CoprimeQ[j, #], Divisible[#, f[n + 1]], # == n] &[k p],
           k++]; k *= p; While[c[p m[p]], m[p]++],
        While[
          Or[c[k], ! CoprimeQ[j, k], CoprimeQ[k, n], Divisible[k, f[n + 1]],
            k == n], k++] ];
        Set[{a[n], c[k], j}, {k, True, k}];
        If[k == u, While[c[u], u++]], {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Oct 29 2023 *)