A376566 Lexicographically earliest sequence of positive integers such that for any n > 0, there is an even number of k's such that 1 <= k < n and a(n) divides a(k).
1, 2, 1, 3, 1, 4, 1, 2, 1, 5, 1, 6, 1, 2, 1, 3, 1, 7, 1, 8, 1, 2, 1, 4, 1, 2, 1, 9, 1, 3, 1, 10, 1, 2, 1, 5, 1, 11, 1, 12, 1, 2, 1, 3, 1, 4, 1, 2, 1, 6, 1, 2, 1, 3, 1, 13, 1, 14, 1, 2, 1, 7, 1, 15, 1, 3, 1, 5, 1, 16, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1
Offset: 1
Keywords
Examples
The first terms, alongside the corresponding k's, are: n a(n) k's -- ---- --------------------------------------------- 1 1 None 2 2 None 3 1 1, 2 4 3 None 5 1 1, 2, 3, 4 6 4 None 7 1 1, 2, 3, 4, 5, 6 8 2 2, 6 9 1 1, 2, 3, 4, 5, 6, 7, 8 10 5 None 11 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 12 6 None 13 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 14 2 2, 6, 8, 12 15 1 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 16 3 4, 12
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A025523.
Programs
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PARI
{ nb = vector(16); for (n = 1, 86, for (v = 1, oo, if (nb[v]%2==0, print1 (v ", "); fordiv (v, d, nb[d]++;); break;););); }
Formula
a(2*k-1) = 1 for any k > 0.
a(2*A025523(n-1)) = n for any n > 1 (and this is the first occurrence of n in the sequence). - Hugo Pfoertner, Oct 01 2024
Comments