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

Author

Mark Hudson (mrmarkhudson(AT)hotmail.com), Nov 15 2004

Keywords

• nonn
• base

Comments

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

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.

Crossrefs

Cf. A064541 (2^k ending with k), A032740 (k a substring of 2^k), A131494.

Programs

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

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