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
Keywords
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..19683 (3^9 = 19683).
Programs
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Mathematica
Array[# - 3^Floor@ Log[3, #] &, 88] (* Michael De Vlieger, Sep 17 2023 *)
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PARI
A365459(n) = if(1==n,0,my(k=0); while((3^k) < n, k++); if((3^k) > n,k--); n-(3^k));
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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
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