A392107 a(n) = position of first occurrence of the n-th binary word (using letters 1 and 2 as in A392106) in the sequence of residues modulo 3 of the primes.

4, 1, 11, 4, 3, 9, 36, 11, 4, 8, 10, 3, 9, 15, 271, 36, 11, 21, 19, 4, 8, 14, 35, 10, 3, 7, 9, 23, 15, 54, 271, 273, 46, 36, 18, 11, 21, 37, 34, 19, 4, 6, 8, 22, 14, 53, 270, 35, 10, 20, 33, 3, 7, 13, 71, 9, 23, 51, 15, 39, 54, 290, 271, 272, 273, 437, 110

Offset

1,1

Comments

Conjecture: every positive integer occurs infinitely many times. In particular, the words that start in position 1 (i.e., the positions of 4 in the sequence), are indexed as 1, 4, 9, 20, 41, 84, 170, 341, 683, 1368, 2737, ...

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..2000

Example

The primes mod 3 are (2, 0, 2, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 2,...), and the first 14 binary words are 1, 2, 11, 12, 21, 22, 111, 112, 121, 122, 211, 212, 221, 222, so a(1) = 4 because the first word, 1, occurs in position 4; a(6) = 9 because the 6th word, 22, first occurs in position 9.

Mathematica

t=Mod[Prime[Range[10^4]], 3];
w=Flatten[Map[Tuples[{0, 1}, #]&, Range[10]]+1, 1];
Quiet[TakeWhile[Map[SequencePosition[t, #, 1][[1]][[1]]&, w], IntegerQ]]
(* Peter J. C. Moses, Jan 01 2026 *)

Prog

(Magma) t := [ NthPrime(i) mod 3 : i in [1..1000] ]; res := []; for k in [1..10], v in CartesianPower({1, 2}, k) do pos := 0; for i in [1..#t-k+1] do if t[i..i+k-1] eq [v[j] : j in [1..k]] then pos := i; break; end if; end for; if pos eq 0 then break k; end if; Append(~res, pos); end for; res; // Vincenzo Librandi, Jan 21 2026

Crossrefs

Cf. A039701, A392106.

Sequence in context: A094503 A113897 A158753 * A183884 A135552 A181690

Adjacent sequences: A392104 A392105 A392106 * A392108 A392109 A392110

Keyword

nonn

Author

Clark Kimberling, Jan 09 2026

Status

approved