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.

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

Author

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

Keywords

• easy
• nonn
• base

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.

Examples

			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.
		

Crossrefs

A056154.

Programs

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 *)