A292524 Interpret the values of the Moebius function mu(k) for k = 1 to n as a balanced ternary number.
0, 1, 2, 5, 15, 44, 133, 398, 1194, 3582, 10747, 32240, 96720, 290159, 870478, 2611435, 7834305, 23502914, 70508742, 211526225, 634578675, 1903736026, 5711208079, 17133624236, 51400872708, 154202618124, 462607854373, 1387823563119, 4163470689357
Offset: 0
Examples
mu(1) = 1, so a(1) = 1 * 3^0 = 1. mu(2) = -1, so a(2) = 1 * 3^1 + -1 * 3^0 = 3 - 1 = 2. mu(3) = -1, so a(3) = 1 * 3^2 + -1 * 3^1 + -1 * 3^0 = 9 - 3 - 1 = 5. mu(4) = 0, so a(4) = 1 * 3^3 + -1 * 3^2 + -1 * 3^1 + 0 * 3^0 = 27 - 9 - 3 + 0 = 15.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..2097
- Wikipedia, Balanced ternary
Programs
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Maple
a:= proc(n) option remember; `if`(n=0, 0, a(n-1)*3+numtheory[mobius](n)) end: seq(a(n), n=0..33); # Alois P. Heinz, Oct 13 2017 -
Mathematica
Table[Plus@@(3^Range[n - 1, 0, -1] MoebiusMu[Range[n]]), {n, 50}] -
PARI
a(n) = sum(k=1, n, moebius(k)*3^(n-k)); \\ Michel Marcus, Oct 01 2017
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PARI
my(N=40, x='x+O('x^N)); concat(0, Vec(sum(k=1, N, moebius(k)*x^k)/(1-3*x))) \\ Seiichi Manyama, May 19 2021 -
PARI
a(n) = if(n==0, 0, 3*a(n-1)+moebius(n)); \\ Seiichi Manyama, May 19 2021
Formula
a(n) = Sum_{k = 1..n} mu(k) 3^(n - k).
a(n) = 3 * a(n-1) + mu(n) for n > 0. - Alois P. Heinz, Oct 13 2017
a(n) ~ A238271 * 3^n. - Vaclav Kotesovec, May 19 2021
Extensions
a(0)=0 prepended by Alois P. Heinz, Oct 13 2017
Comments