A365459 a(n) = n - the largest power of 3 that is less than or equal to n.

0, 1, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 0, 1, 2, 3, 4, 5, 6, 7

Offset: 1

Author

Antti Karttunen, Sep 17 2023

Keywords

• nonn

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..19683 (3^9 = 19683).

Crossrefs

Cf. A000244, A053735 (gives the number of iterations needed to reach 0), A365458.

Programs

Mathematica

Array[# - 3^Floor@ Log[3, #] &, 88] (* Michael De Vlieger, Sep 17 2023 *)

PARI

A365459(n) = if(1==n,0,my(k=0); while((3^k) < n, k++); if((3^k) > n,k--); n-(3^k));

Python

def A365459(n):
    kmin, kmax = 0, 1
    while 3**kmax <= n:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if 3**kmid > n:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return n-3**kmin # Chai Wah Wu, Sep 17 2023

Formula

a(n) = n - A365458(n).

a(n) = n - 3^floor((log n)/(log 3)). - Michael De Vlieger, Sep 17 2023