Sum of digits in base 3

**Ben92** · May 12th 2010, 11:12 AM

Hello!

I conjectured today that $s_{3}(2^{k})=2\left\lfloor \frac{k-1}{2}\right\rfloor$ is true for every $k > 9$ integer, where $s_{3}(2^{k})$ is giving the sum of $2^{k}$ digits in base 3.

Has it already solved or hard to proof?

Ben

**undefined** · May 12th 2010, 04:21 PM

Originally Posted by Ben92

Hello!

I conjectured today that $s_{3}(2^{k})=2\left\lfloor \frac{k-1}{2}\right\rfloor$ is true for every $k > 9$ integer, where $s_{3}(2^{k})$ is giving the sum of $2^{k}$ digits in base 3.

Has it already solved or hard to proof?

Ben

Hi, Ben92,

I must have misinterpreted what you wrote. I wrote a program that determined three sequences:

$2^n$ : 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, ...

$2^n$ (in base 3): 1, 2, 11, 22, 121, 1012, 2101, 11202, 100111, 200222, 1101221, 2210212, 12121201, 102020102, 211110211, 1122221122, ...

$s_3(2^n)$ : 1, 2, 2, 4, 4, 4, 4, 6, 4, 8, 8, 10, 10, 8, 10, 16, ...

The third sequence is the sum of digits of the corresponding member in the second sequence.

Could you clarify the meaning?

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