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A056736
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.
1
5, 16, 27, 40, 65, 92, 124, 138, 143, 265, 368, 457, 476, 501, 634, 707, 839, 842, 848, 929, 1013, 1086, 1289, 1303, 1587, 1685, 1812, 1926, 1994, 2213, 2308, 2522, 2565, 2950, 3286, 3674, 3774, 3942, 4034, 4318, 4381, 4438, 4719, 4728, 4909, 4971
OFFSET
1,1
COMMENTS
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.
EXAMPLE
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.
MATHEMATICA
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 *)
CROSSREFS
KEYWORD
easy,nonn,base
AUTHOR
Russell Harper (rharper(AT)intouchsurvey.com), Aug 13 2000
STATUS
approved