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A100129
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Numbers k such that 2^k starts with k.
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11
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6, 10, 1542, 77075, 113939, 1122772, 2455891300, 2830138178, 136387767490, 2111259099790, 3456955336468, 4653248164310, 10393297007134, 321249146279171, 972926121017616, 72780032758751764
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OFFSET
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1,1
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COMMENTS
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According to van de Lune, Erdős observed that 2^6 = 64 and 2^10 = 1024 were two examples where the decimal expansion of 2^k starts with that of k. At that time no other examples were known. Jan van de Lune computed the first 6 terms in 1978. - Juan Arias-de-Reyna, Feb 12 2016
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LINKS
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FORMULA
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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
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EXAMPLE
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2^6 = 64, which begins with 6;
2^10 = 1024, which begins with 10.
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MATHEMATICA
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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 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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Mark Hudson (mrmarkhudson(AT)hotmail.com), Nov 15 2004
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EXTENSIONS
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STATUS
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approved
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