# Tag Info

5

Hint: by definition, $b^c = \exp(c \log b)$. Use all values of $\log b$.

3

What you suggest basically seems to work - I'm going to change the leading value into $n$ rather than $n-1$ for ease of reading: First, $2^1 = 1+1$ and $2^2 = 1+1+2 = 1+2^0+2^1$ Then using induction, assuming that the summation holds for lower $n$, we can use your observation $2^n = 2^{n-1} + 2^{n-1}$ and change the first $2^{n-1}$ into the summation form, ...

3

In binary, this is $$1+1+10+100+...+1\underbrace{00...00}_{n-2\text{ times}}=1+\underbrace{11...11}_{n-1\text{ times}}=1\underbrace{00...00}_{n-1\text{ times}}$$ which in decimal is $2^{n-1}$.

3

I cannot confirm your counterexample. Here are the details so nobody else has to duplicate this effort. Following the method from Brian Moehring's answer, we want to find a value of $n$ such that $\log_5(3^n - 1), \log_7(3^n - 1), \log_{13}(3^n - 1)$ all have fractional parts close to $1$. If their integer parts are $n_5, n_7, n_{13}$, it then suffices to ...

2

It's because it's a geometric progression. Since $$1+p_i+{p_i}^2+··· +{p_i}^{a_i} = \dfrac{{p_i}^{a_i+1}-1}{p_i-1}.$$ you just add an extra one at the start of the progression.

2

Let me suggest 4 ways: $a^{n}-b^{n}=(a-b)(a^{n-1} + ba^{n-2}+ ... + b^{n-1}) \Rightarrow 2^{n} = 2^{n} - 1 + 1 = 1+2+ ... +2^{n-1} +1$ Considering all subsets from set with $n>1$ elements and considering map between subsets and binary numbers Mathematical Induction Newton's binomial theorem

1

This problem seems difficult, see the information provided in OEIS. However, brute force approaches give some results. One may for instance list all numbers satisfying your conditions, until a given maximal value, and then sort them. The maximal value must be large enough to reach $k$ values (it seems from experiments and OEIS that it should be close tp $k^2$...

1

The word you are looking for is probably tetration at least if you are looking for something like $$\underbrace{a^{(a^{(\cdot^{\cdot^{a)\cdots)}}}}}_n$$

1

Let $S:=1+2+4+\dots+2^{n-2}$. Then, since it's a geometric series, we have $S=\dfrac{1-2^{n-1}}{1-2}=2^{n-1}-1\implies S+1=2^{n-1}$.

1

Let $$x = \sum_{m=2}^{n}2^{m-2},$$ and note that $$1 + x = 1 + 1 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{1}.$$ Then, multiplying $(1)$ by $2$, we have $$2+2x = 2 + 2 + 4 + \dotsb + 2^{n-2} + 2^{n-1}\tag{2}.$$ We can rewrite $(1)$ as $$1 + x = 2 + 2 + 4 + \dotsb + 2^{n-3} + 2^{n-2}\tag{3}.$$ Subtracting $(3)$ from $(2)$ gives us $$1 + x = 2^{n-1},$$ which ...

1

Consider: from deMoivre's $e^{i \theta} = \cos \theta + i \sin \theta \tag 1$ with $\theta = 2n \pi + \dfrac{\pi}{2}, \; n \in \Bbb Z, \tag 2$ we have $e^{i(2n\pi + \pi/2)} = i; \tag 3$ thus $i^{\sqrt 3} = e^{i(2n\pi + \pi/2)\sqrt 3}$ $= \cos ((2n\pi + \pi/2)\sqrt 3) + i \sin ((2n\pi + \pi/2)\sqrt 3), \ n \in \Bbb Z. \tag 4$ We may check for consistency: (4) ...

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