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Math Help - Sum of digits in base 3

  1. #1
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    Sum of digits in base 3

    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
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  2. #2
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    Quote Originally Posted by Ben92 View Post
    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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