A353653 Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 6.

2, 3, 9, 10, 11, 12, 13, 14, 15, 21, 27, 33, 39, 45, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 93, 99, 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165

Offset: 1

Author

Yifan Xie, Jul 15 2022

Keywords

• nonn
• easy

Comments

Numbers m such that the base-6 representation of (5*m-3) starts with 11 or 12 or 13 or 14 or 15 or ends with 0.

First differences give a run of 6^i 1's followed by a run of 6^i 6's, for i >= 0.

Examples

			a(5) = 11 because (2*6^1 + 3)/3 <= 5 < (7*6^1 + 3)/5, hence a(5) = 5 + 6^1 = 11;
a(10) = 21 because (7*6^1 + 3)/5 <= 10 < (12*6^1 + 3)/5, hence a(10) = 6*(10 - 6^1) - 3 = 21.
		

Links

• Yifan Xie, Table of n, a(n) for n = 1..10000
• Index entries for sequences of the a(a(n)) = 2n family

Crossrefs

For other values of k: A080637 (k=2), A003605 (k=3), A353651 (k=4), A353652 (k=5), this sequence (k=6).

Programs

Mathematica

okQ[m_] := With[{id = IntegerDigits[5 m - 3, 6] }, MatchQ[id[[1 ;; 2]], {1, 1}|{1, 2}|{1, 3}|{1, 4}|{1, 5}] || id[[-1]] == 0];
Join[{2}, Select[Range[3, 1000], okQ]] (* Jean-François Alcover, Sep 22 2023 *)

Formula

For n in the range (2*6^i + 3)/5 <= n < (7*6^i + 3)/5, for i >= 0:

a(n) = n + 6^i.

a(n+1) = 1 + a(n).

Otherwise, for n in the range (7*6^i + 3)/5 <= n < (12*6^i + 3)/5, for i >= 0:

a(n) = 6*(n - 6^i) - 3.

a(n+1) = 6 + a(n).