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Answer 1

\begin{align} 2^{10}10^{2018}+7^4 &= 2^{10}100^{1009}+(50-1)^2 \\ &= 2^{1019}50^{1009} + 50^2 -2\cdot50 + 1 \\ &= 50(2^{1019}50^{1008} + 48)+1 \end{align}

factoring $50(3n+2)+1$ \begin{array}{r} 50( 2)+1 &= & 101 \\ 50( 5)+1 &= & 251 \\ 50( 8)+1 &= & 401 \\ 50(11)+1 &= & 551 &= &19\cdot29 &= &(24-5)(24+5) \\ 50(14)+1 &= & 701 \\ 50(17)+1 &= & 851 &= &23\cdot37 &= &(30-7)(30+7) \\ 50(20)+1 &= &1001 &= &7\cdot 11 \cdot 13 \\ 50(23)+1 &= &1151 \\ 50(26)+1 &= &1301 \\ 50(29)+1 &= &1451 \\ 50(32)+1 &= &1601 \\ 50(35)+1 &= &1751 &= &17\cdot 103 &= &(60-43)(60+43) \\ 50(38)+1 &= &1901 \\ 50(41)+1 &= &2051 &= &7\cdot 293 &= &(150-143)(150+143) \\ 50(44)+1 &= &2201 &= &31\cdot 71 &= &(51-20)(51+20) \\ 50(47)+1 &= &2351 \\ 50(50)+1 &= &2501 &= &41\cdot 61 &= &(51-10)(51+10) \\ \end{array}

We may as well use the points $P_k = (\cos(\theta k), \sin(\theta k))$ where $\theta = \dfrac{2\pi}{5}$ and $k \in \{0,1,2,3,4\}$.

$$\frac{XA+XB}{XC+XD+XE}$$

Let $\theta = \dfrac{2\pi}{5}$. Then we can define \begin{align} A &= (\cos(2\theta),\sin(2\theta)) \\ B &= (\cos(3\theta),\sin(3\theta)) \\ C &= (\cos(4\theta),\sin(4\theta)) \\ D &= (\cos(0\theta),\sin(0\theta)) \\ E &= (\cos(1\theta),\sin(1\theta)) \\ \end{align}

$\begin{array}{ccc} \theta & \cos \theta & \sin \theta \\ 0 \cdot\dfrac{2 \pi}{5} & 1 & 0 \\ 1 \cdot\dfrac{2 \pi}{5} & \dfrac{\sqrt 5 - 1}{4} & \sqrt{\dfrac{5+\sqrt 5}{8}} \\ 2 \cdot\dfrac{2 \pi}{5} & -\dfrac{\sqrt 5 + 1}{4} & \sqrt{\dfrac{5-\sqrt 5}{8}} \\ 3 \cdot\dfrac{2 \pi}{5} & -\dfrac{\sqrt 5 + 1}{4} & -\sqrt{\dfrac{5-\sqrt 5}{8}} \\ 4 \cdot\dfrac{2 \pi}{5} & \dfrac{\sqrt 5 - 1}{4} & -\sqrt{\dfrac{5+\sqrt 5}{8}} \\ \end{array}$

Answer 2

My post needs more work, and I don't have the time for it today. Leaving this to be used by anyone.

Answer 3

a(n)= floor( (9*(2k+1)+3 )/4) = floor( (18k+12)/4) = floor( 4k+3+k/2) = 4k+3+floor(k/2)
b(n)= floor( (3*(2k+1)+1 )/4) = floor( ( 6k+ 4)/4) = floor( k+1+k/2) = k+1+ floor(k/2)
c(n)= floor( (3*(2k+1)+1 )/2) = floor( ( 6k+ 4)/2) = floor( 3k+2)   = 3k+2
s(n) =floor( ((-1)^a(n)+(-1)^b(n) + 3*(-1)^c(n)+7)/4)

a(n)=  4k+3+floor(k/2) =   1 + floor(k/2) % 2
b(n)=  k+1+ floor(k/2) = k+1 + floor(k/2) % 2
c(n)=  3k+2            = k                % 2

Pari/GP-routine so far:

a(n)= floor( (9*n+3 )/4)
b(n)= floor( (3*n+1 )/4)
c(n)= floor( (3*n+1 )/2)
s(n) =floor( ((-1)^a(n)+(-1)^b(n) + 3*(-1)^c(n)+7)/4)

First, we observe that the functions $a(n),b(n),c(n)$ are used as exponents at $-1$ so each $(-1)^e(n)$ gives only periodic values, and we can as well use the modular residue $\pmod 2$ of the floor expressions. Since we have $3$ functions we can, if they are independent, at most generate $8$ different combinations and can thus code the sum $s(n)$ by evaluating the $8$ residue classes of $n \pmod 8$.
So we can look at the functions $f(n)$ when $n=8k+r$ and get

$$ a(8k+r) = \lfloor 18k + {9r+3 \over 4} \rfloor \pmod {2} \\\ b(8k+r) = \lfloor 6k + {3r + 1 \over 4} \rfloor \pmod {2} \\\ c(8k+r) = \lfloor 12k + {3r + 1 \over 2 }\rfloor \pmod {2} $$

We see that the result is independent from the $k$-component because it is integer, can be taken out of the floor and is then congruent to zero by $\pmod 2$.

So indeed we can look at the $8$ residues of $n \pmod 8$ only and the empirical test for the $4$ odd residue classes $ n \pmod 8$ show that your coding of the problem is valid. This answers the last question of your post.

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But there is another scheme to arrive at your $C_8()$ function and which is more obvious to generalize. We establish the following actiontable with the function $\nu_2^*(x)=\min(3,\nu_2(x))$ giving the exponent to which the prime $2$ is factor of $x$ but restricted to a max value of $3$ (because we want look at this modulo $8$ here) . $$ \begin{array} {r|r|r|r|r|r|r} n & n^* & \nu_2^*(n^*) && n & n^* & \nu_2^*(n^*) \\ \hline 0& 0& 3 && 1 & 4 & 2 \\\ 2&2 & 1 && 3 & 10 & 1\\\ 4&4 & 2 && 5 & 16 & 3\\\ 6&6 & 1 && 7 & 22 & 1\\\ \end{array} $$

We can now define for the even and odd residue classes using the $\text{mod}$ symbol for getting the modular residue as in some programming languages reproducing the $\nu^*()$ values by sum-of-floors, let's denote them as $E_\text{even}(n)$ and $E_\text{odd}(n)$ for even and odd $n$ respectively: $$ E_\text{even}(n) = 1 + \left \lfloor 1- {n \text{ mod } 4 \over 4} \right \rfloor + \left \lfloor 1- {n \text{ mod } 8 \over 8} \right \rfloor \\\ E_\text{odd}(n) = 1 + \left \lfloor 1- {n -1 \text{ mod } 4 \over 4} \right \rfloor + \left \lfloor 1- {n -5 \text{ mod } 8 \over 8} \right \rfloor $$

With this one can define the $C_8()$-transformation either $$ C_8(n) = \begin{cases} {n \over 2} & \text { if $n$ is even} \\\ {3n+1 \over 2^{E_\text{odd}(n)}} & \text { if $n$ is odd} \end{cases} $$ or $$ C_8(n) = \begin{cases} {n \over 2^{E_\text{even}(n)}} & \text { if $n$ is even} \\\ {3n+1 \over 2^{E_\text{odd}(n)}} & \text { if $n$ is odd} \end{cases} $$

This way of defining seems meaningful for me especially when one wants to generalize to larger modules.

This slot is as vacant as the expression of someone who's been staring at the same line of a proof for three hours.

Answer 4

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Answer 8

My first ansatz, which gives at best an asymptotic power series (with convergence radius zero!), seemed to be rough and likely much suboptimal. But for completeness I'll give the method here, also because it provides that nice initial guess $a_\text{init}$ which is very near at the true value.

The ansatz employs essentially the Bernoulli-numbers, but not as in the classic Bernoulli-polynomials but in some "transposed" manner.

The individual $\log(1+1/a_k)$ - epressions have a mercator series, such that we can reformulate the original sum $S_d(a,N)$ as a double sum $$ \begin{array}{rll} S_d(a,N) &=& 1/a - 1/2/a^2 + 1/3/ a^3 - 1/4/ a^4 + \cdots - \cdots \\ &+& 1/(a+d) - 1/2/(a+d)^2 + 1/3/ (a+d)^3 - 1/4/ (a+d)^4 + \cdots - \cdots \\ &+& 1/(a+2d) - 1/2/(a+2d)^2 + 1/3/ (a+2d)^3 - 1/4/ (a+2d)^4 + \cdots - \cdots \\ &...& \\ &+& 1/(a+(N-1)d) - 1/2/(a+(N-1)d)^2 + 1/3/ (a+(N-1)d)^3 - 1/4/ (a+(N-1)d)^4 + \cdots - \cdots \\ \end{array}$$ Summing this vertically gives the column-sums $$ \begin{array}{rll} C_{d,1}(a,N) &=& 1/a + 1/(a+d)+ 1/(a+2d) + ... + 1/(a+(N-1)d) \\ C_{d,2}(a,N) &=&\frac12 ( 1/a^2 + 1/(a+d)^2+ 1/(a+2d)^2 + ... + 1/(a+(N-1)d)^2 )\\ C_{d,3}(a,N) &=&\frac13 ( 1/a^3 + 1/(a+d)^3+ 1/(a+2d)^3 + ... + 1/(a+(N-1)d)^3 )\\ &...& \end{array}$$ Such sums are essentially Hurwitz-Zeta sums, and can be expressed as power-series with zeta at negative integers as coefficients.
For easiness of handling and memorizing I've the "ZETA"-matrix in my toolbox, which provides that coefficients for the application to power series.
The ZETA-matrix looks like

and has the composition of Zeta-values
Reading it horizontally one observes the coefficients for the integrals of the Bernoulli-polynomials which are configured to allow the summing-of-like-powers.
However, reading vertically, they give the coeffients for power series to determine the sums-of-reciprocals-of-like-powers. However, that power series have convergence-radius zero, but because of alternating signs we can use them as asymptotic power series, truncated at some meaningful index.

For instance, the power series taken from the second,third, fourth and so on column, and calculated as $$ f_1(a) = a z_{0,1} + a^2 z_{1,1} + a^3 z_{2,1} + ... \to \zeta(2,a) \\ f_2(a) = a z_{0,2} + a^2 z_{1,2} + a^3 z_{2,2} + ... \to \zeta(3,a) \\ ... $$ where $z_{r,c}$ indicates the element of the ZETA-matrix from row $r$ and column $c$ (index based at $0$).
That allows immediately to determine the sums of reciprocals from $a$ to $a+(N-1)$ by $$ s_1(a) = f_1(a) - f_1(a+N) = 1/a^2 + 1/(a+1)^2+... + 1/(a+(N-1))^2 \\ s_2(a) = f_2(a) - f_2(a+N) = 1/a^3 + 1/(a+1)^3+... + 1/(a+(N-1))^3 \\ ... $$

For the first column we need a correction factor which comes out to be $$ f_0(a) = a z_{0,0} + a^2 z_{1,0} + a^3 z_{2,0} + ... \to \\ \log(1/a)+\left(\sum_{k=1}^{a-1} \frac1k \right) -\gamma $$
such that $$ s_0(a) = f_0(a) - f_0(a+N) = 1/a + 1/(a+1)+... + 1/(a+(N-1)) \\ + \log(1/a) - \log(1/(a+(N-1)) $$

Answer 9

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Answer 11

Answers awaiting merge here

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Answer by Felix Marin $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 1}^{n + 1}{n{n \choose k - 1} \over {2n \choose k}} & = n\sum_{k = 0}^{n}{{n \choose k} \over {2n \choose k + 1}} \\[5mm] & = n\sum_{k = 0}^{n}{n \choose k} {1 \over \pars{2n}!/\bracks{\pars{k + 1}!\pars{2n - k - 1}!}} \\[5mm] & = n\pars{2n + 1}\sum_{k = 0}^{n} {n \choose k}{\Gamma\pars{k + 2}\Gamma\pars{2n - k}! \over \Gamma\pars{2n + 2}} \\[5mm] & = n\pars{2n + 1}\sum_{k = 0}^{n} {n \choose k}\int_{0}^{1} t^{k + 1}\pars{1 - t}^{2n - k - 1}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}\pars{1 - t}^{2n}\, {t \over 1 - t}\sum_{k = 0}^{n}{n \choose k} \pars{t \over 1 - t}^{k}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}\pars{1 - t}^{2n - 1}\, t \pars{1 + {t \over 1 - t}}^{n}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}t\, \pars{1 - t}^{n - 1}\,\dd t \\[5mm] & = n\pars{2n + 1}\int_{0}^{1}\pars{1 - t}\,t^{n - 1}\,\dd t \\[5mm] & = n\pars{2n + 1}\pars{{1 \over n} - {1 \over n +1}} = \bbx{2n + 1 \over n + 1} \end{align}

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Answer by Marko Riedel

Starting from

$$n\sum_{k=1}^{n+1} {2n\choose k}^{-1} {n\choose k-1}$$

we find

$$n\sum_{k=1}^{n+1} \frac{n!}{(k-1)! \times (n+1-k)!} \frac{k! \times (2n-k)!}{(2n)!} \\ = \frac{n\times n!}{(2n)!} \sum_{k=1}^{n+1} \frac{k}{(n+1-k)!} (2n-k)! \\ = \frac{n!^2}{(2n)!} \sum_{k=1}^{n+1} k {2n-k\choose n+1-k}.$$

This is

$$\frac{n!^2}{(2n)!} \sum_{k=0}^{n+1} k [z^{n+1-k}] (1+z)^{2n-k} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \sum_{k=0}^{n+1} k z^k (1+z)^{-k}$$

Now when $k\gt n+1$ there is no contribution to the coefficient extractor and we get

$$\frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \sum_{k\ge 0} k z^k (1+z)^{-k} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n} \frac{z/(1+z)}{(1-z/(1+z))^2} \\ = \frac{n!^2}{(2n)!} [z^{n+1}] (1+z)^{2n+2} \frac{z}{1+z} \\ = \frac{n!^2}{(2n)!} [z^{n}] (1+z)^{2n+1} = \frac{n!^2}{(2n)!} {2n+1\choose n} \\ = n!^2 \times (2n+1) \times \frac{1}{n! \times (n+1)!} = \frac{2n+1}{n+1}.$$

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Answer 16

Function $f$ is defined as $$f(x)=\frac{1}{(x-1)^3(x^3+2x^2+x+1)}\tag 1$$ Note that $$x^3+2x^2+x+1=(x+1)(x^2+x+1)\tag 2$$ and so $$f(x)=\frac{1}{(x-1)^3(x+1)(x^2+x+1)}\tag 3$$ and also $$f(x)=\frac{1}{(x-1)}\cdot\frac{1}{(x^2-1)}\cdot\frac{1}{(x^3-1)}=\sum_{i=0}^\infty x^i \cdot\sum_{k=0}^\infty x^{2k} \cdot\sum_{j=0}^\infty x^{3j}$$

$$\frac{1}{(x-1)^3(x+1)(x^2+x+1}=\frac{A_1}{(x-1)}+\frac{A_2}{(x-1)^1}+\frac{A_3}{(x-1)^3}+\frac{B}{(x+1)}+\frac{Cx+D}{(x^2+x+1)}$$ $$1=A_1(x-1)^2(x+1)(x^2+x+1)+A_2(x-1)(x+1)(x^2+x+1)+A_3(x+1)(x^2+x+1)+B(x-1)^3(x^2+x+1)+(Cx+D)(x-1)^3(x+1)$$

Answer 17

It's been a long time since I last did this stuff, but nobody better seems to have turned up, so I'll have a go.

I'll give the box $\square \phi$ its usual meaning here: "when I translate $\phi$ into the formal system using Gödel numbers, then there is a collection of mechanical transformations inside the formal system that together form an internal-system-proof of $\phi$'s quotation". I will also refer to the formal system in question in the first person, with the pronoun "I".

Then the formula $\square (\square \phi \to \phi) \to \square \phi$ should be read as:

If there is a number which encodes a proof that "having a number which encodes a proof of $\phi$ means I can actually prove $\phi$", then there is a number which encodes a proof of $\phi$.

Or, a bit more loosely,

If I, internally, can prove that my proof of $\phi$ respects the logic in which I operate (so my proof of $\phi$ means that in my surrounding logic, $\phi$ is actually true), then in fact there is a proof of $\phi$ that I will be satisfied with.

This is more concretely understood through its universal quantification, Löb's theorem (which is true in any system rich enough to support Peano arithmetic):

If I can prove within myself that my own proof-checking process respects the semantics of the logic in which I operate, then the logic realises that actually I can satisfy myself of any statement at all.

Specifically, $\square (\square \phi \to \phi)$ means "I contain a proof that my proofs are semantically valid in the surrounding logic", while $\square \square \phi \to \square \phi$ means "it's true in the surrounding logic that if I want to prove $\phi$, it's enough to prove provability-of-$\phi$".

The two statements were very unlikely to be the same: one is talking about proof within the system, and one is talking about facts of the surrounding logic.