A353313 If n is of the form 3k, then a(n) = k, and if n is of the form 3k+r, with r = 1 or 2, then a(n) = 5*k + 3 + r.

0, 4, 5, 1, 9, 10, 2, 14, 15, 3, 19, 20, 4, 24, 25, 5, 29, 30, 6, 34, 35, 7, 39, 40, 8, 44, 45, 9, 49, 50, 10, 54, 55, 11, 59, 60, 12, 64, 65, 13, 69, 70, 14, 74, 75, 15, 79, 80, 16, 84, 85, 17, 89, 90, 18, 94, 95, 19, 99, 100, 20, 104, 105, 21, 109, 110, 22, 114, 115, 23, 119, 120, 24, 124, 125, 25, 129, 130, 26

Offset: 0

Author

Antti Karttunen, Apr 13 2022

Keywords

• nonn

Comments

It is conjectured that all iterations of this sequence starting from any n >= 0 will eventually reach a finite cycle, which by necessity then contains at least one multiple of three. See Drozd links and A349876.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..19683
• Nicholas Drozd, A Busy Beaver Champion Derived from Scratch
• Nicholas Drozd, Feedback to Doron Zeilberger's opinion #155, Jan. 4, 2022

Crossrefs

Cf. A349876, A353310, A353311, A353312, A353314 (variant).

Cf. A353305 (the smallest number reached after the starting point n), A353309 (the largest base-3 digit sum reached after the starting point n).

Programs

Mathematica

Table[With[{c=Mod[n,3]},If[c==0,n/3,(5n-2c+9)/3]],{n,0,80}]  (* Harvey P. Dale, Aug 09 2025 *)

PARI

A353313(n) = { my(r=(n%3)); if(!r,n/3,((5*((n-r)/3)) + r + 3)); };