A336018 a(n) = floor(frac(log_2(n))*n), where frac denotes the fractional part.
0, 0, 1, 0, 1, 3, 5, 0, 1, 3, 5, 7, 9, 11, 13, 0, 1, 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 27, 29, 0, 1, 2, 4, 6, 7, 9, 11, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 45, 47, 49, 52, 54, 56, 59, 61, 0, 1, 2, 4, 5, 7, 9, 10, 12, 13
Offset: 1
Keywords
Links
- Chai Wah Wu, Table of n, a(n) for n = 1..10000
Programs
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Maple
a:= n-> floor(n*log[2](n))-n*ilog2(n): seq(a(n), n=1..80); # Alois P. Heinz, Jan 04 2021
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Mathematica
a[n_]:=Floor[FractionalPart[Log[2, n]]*n]; Table[a[n], {n, 1, 100}]
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PARI
a(n) = floor(n*frac(log(n)/log(2))); \\ Michel Marcus, Jul 07 2020
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Python
def A336018(n): return len(bin(n**n//(2**((len(bin(n))-3)*n))))-3 # Chai Wah Wu, Jul 09 2020
Formula
a(n) = floor((log_2(n) - floor(log_2(n)))*n).
From Alois P. Heinz, Jan 04 2021: (Start)
a(n) = 0 <=> n in { A000079 }. (End)