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Numbers k such that 2^k, 2^(k+1) and 2^(k+2) have the same number of digits.
1

%I #16 Sep 08 2022 08:45:24

%S 1,4,7,10,11,14,17,20,21,24,27,30,31,34,37,40,41,44,47,50,51,54,57,60,

%T 61,64,67,70,71,74,77,80,81,84,87,90,91,94,97,100,103,104,107,110,113,

%U 114,117,120,123,124,127,130,133,134,137,140,143,144,147,150,153,154

%N Numbers k such that 2^k, 2^(k+1) and 2^(k+2) have the same number of digits.

%C The density of this sequence is 1 - 2*log_10(2) = 0.3979400086720376...

%H Robert Israel, <a href="/A115566/b115566.txt">Table of n, a(n) for n = 1..10000</a>

%F floor(log_10(2)*k) = floor(log_10(2)*(k+1)) = floor(log_10(2)*(k+2)).

%e 2^4 = 16, 2^5 = 32, 2^6 = 64: all these numbers have two digits.

%e 2^10 = 1024, 2^11 = 2048, 2^12 = 4096: all these numbers have three digits.

%p select(n -> ilog10(2^n)=ilog10(2^(n+2)), [$1..1000]); # _Robert Israel_, May 19 2019

%t Select[Range[220], Floor[Log[10, 2]*# ] == Floor[Log[10, 2]*(# + 2)] &]

%o (Magma) [k:k in [1..160]|#Intseq(2^k) eq #Intseq(2^(k+2))]; // _Marius A. Burtea_, May 20 2019

%Y Cf. A001682 (same definition with 3 instead of 2).

%Y Cf. A034887 (number of digits in 2^n).

%K nonn,base

%O 1,2

%A _Stefan Steinerberger_, Mar 11 2006