A354538 a(n) is the least k such that A322523(k) = n.
1, 3, 8, 17, 44, 125, 368, 1097, 3284, 9845, 29528, 88577, 265724, 797165, 2391488, 7174457, 21523364, 64570085, 193710248, 581130737, 1743392204, 5230176605, 15690529808, 47071589417, 141214768244, 423644304725, 1270932914168, 3812798742497, 11438396227484
Offset: 0
Examples
1, 2 go in box 0 {1,2};
3=1+2 so goes in box 1 {3};
4 goes in box 0 {1,2,4};
5=1+4 so goes in box 1 {3,5};
6=2+4 so goes in box 1 {3,5,6};
7 goes in box 0 {1,2,4,7};
8=7+1=3+5 so goes in box 2 {8};
...
Box 0: {1,2,4,7,10,13,16,19,22,...} = {1} mod 3 U misfit entry 2;
Box 1: {3,5,6,12,14,21,23,...} = {3,5} mod 9 U misfit entry 6;
Box 2: {{8,9,11,15} mod 27} U {18};
Box 3: {{17,20,24,26,27,29,33,45} mod 81} U {54};
...
Links
- Paolo Xausa, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (4,-3).
Crossrefs
Cf. A322523.
Programs
-
Mathematica
Join[{1, 3}, (3^Range[2, 50] + 7)/2] (* or *) LinearRecurrence[{4, -3}, {1, 3, 8, 17}, 50] (* Paolo Xausa, Jun 10 2024 *) -
Python
from itertools import count def A354538(n): for k in count(1): c, m = 0, k while not (a:=divmod(m,3))[1]: c += 1 m = a[0] if not (m==2 or m%3==1): c += 1 m = (m+1)//3-2 while (a:=divmod(m,3))[1]: c += 1 m = a[0] if c == n: return k # Chai Wah Wu, Oct 15 2022 def A354538(n): return (n<<1)+1 if n<2 else 3**n+7>>1 # based on formula, Chai Wah Wu, Oct 15 2022
Formula
For n >= 2, a(n) = (3^n + 7)/2.
For n >= 0, the contents of the n-th box are (conjecture):
1) All the positive integers == b(i) mod 3^(n+1) where 1 <= i <= 2^n and {b(i)} is some strictly increasing sequence of 2^n integers such that b(1)=a(n), b(2^n)= 5*3^(n-1).
2) The additional 'misfit' entry 2*3^n.
[For the correctness and the formula and the conjecture see A322523. - Jianing Song, Oct 17 2022]
Extensions
a(7)-a(15) from Michel Marcus, Sep 23 2022
a(16)-a(20) from Chai Wah Wu, Oct 15 2022
a(21)-a(28) from Paolo Xausa, Jun 10 2024
Comments