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Numbers n such that 2^n in base 3 has same number of 2's as 2^(n+1) in base 3 and 2^n and 2^(n+1) have the same number of digits in base 3.
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%I #5 Aug 27 2012 16:19:43

%S 5,16,27,40,65,92,124,138,143,265,368,457,476,501,634,707,839,842,848,

%T 929,1013,1086,1289,1303,1587,1685,1812,1926,1994,2213,2308,2522,2565,

%U 2950,3286,3674,3774,3942,4034,4318,4381,4438,4719,4728,4909,4971

%N Numbers n such that 2^n in base 3 has same number of 2's as 2^(n+1) in base 3 and 2^n and 2^(n+1) have the same number of digits in base 3.

%C Using empirical data for 1 <= n <= 10000, it has been found that the distribution of these terms correlates well (R^2 = 0.9936) to h(n) = c*n^(1/2) with 'c' a constant approximately 0.64. In addition, h'(n) approximates the probability that any particular n has this property. Any terms in sequence A056154 must also satisfy this sequence.

%e First term: 2^5 = 1012, 2^6 = 2101, both with 1 two and both of length 4. Second term: 2^16 = 10022220021, 2^17 = 20122210112, both with 5 twos and both of length 11.

%t sn2Q[n_]:=Module[{a=2^n,b=2^(n+1)},DigitCount[a,3,2]==DigitCount[b,3,2] && IntegerLength[a,3]==IntegerLength[b,3]]; Select[Range[5000],sn2Q] (* _Harvey P. Dale_, Aug 27 2012 *)

%Y A056154.

%K easy,nonn,base

%O 1,1

%A Russell Harper (rharper(AT)intouchsurvey.com), Aug 13 2000