A100129 Numbers k such that 2^k starts with k.
6, 10, 1542, 77075, 113939, 1122772, 2455891300, 2830138178, 136387767490, 2111259099790, 3456955336468, 4653248164310, 10393297007134, 321249146279171, 972926121017616, 72780032758751764
Offset: 1
Examples
2^6 = 64, which begins with 6; 2^10 = 1024, which begins with 10.
Links
- J. van de Lune, A note on a problem of Erdős, Department of Pure Mathematics, ZW 87/78, Mathematisch Centrum, Amsterdam 1978, 1-3.
Programs
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Mathematica
f[n_] := Floor[ 10^Floor[ Log[10, n]](10^FractionalPart[n*N[ Log[10, 2], 24]])]; Do[ If[ f[n] == n, Print[n]], {n, 125000000}] (* Robert G. Wilson v, Nov 16 2004 *) -
Python
# Caveat: fails for large n due to rounding error. from math import log10 as log frac = lambda x: x - int(x) is_a100129 = lambda n: 0 <= frac(n * log(2)) - frac(log(n)) < log(n + 1) - log(n) # David Radcliffe, Jun 02 2019
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Python
from itertools import count, islice def A100129_gen(startvalue=1): # generator of terms a = 1<<(m:=max(startvalue,1)) for n in count(m): if (s:=str(n))==str(a)[:len(s)]: yield n a <<= 1 A100129_list = list(islice(A100129_gen(),4)) # Chai Wah Wu, Apr 10 2023
Formula
The sequence contains k if and only if 0 <= {k*log_10(2)} - {log_10(k)} < log_10(k+1) - log_10(k), where {x} denotes the fractional part of x. See the van de Lune article. - David Radcliffe, Jun 02 2019
Extensions
a(5) and a(6) from Robert G. Wilson v, Nov 16 2004
More terms from Robert Gerbicz, Aug 22 2006
Comments