## A325902

Fifty is a peculiar integer.
When looking at its neighbors — the largest integer strictly beneath and the smallest strictly above –, more specifically their prime factorization, one finds

$49=\underbrace{7^2<50<3\cdot 17}_{7+7+3=17}=51,$

notably there exists a partition of the neighbor’s factors into two multisets such that both parts’ sums equal another.

Positive integers with the above described property can be found in my most recent addition to the OEIS: sequence A325902.

## Digit sums

Interessant war es auch, drei aufeinanderfolgende Zahlen zu nehmen, von denen die größte durch drei teilbar sein musste, sie zu addieren und aus dem Ergebnis so lange die Quersumme zu bilden, bis eine einstellige Zahl übrig blieb. Diese Zahl war immer sechs.
— Child, Lee: Der Anhalter. München: Blanvalet, 2015; p. 73.

Jack Reacher’s at most tangentially to interpreting the sergeant’s reply related base ten factoid’s formal form is

$\forall n\in\mathbb{N}^+:\mathrm{fds}_{10}\left(\sum\limits_{j=0}^2 3n-j\right) = 6,$

where $\mathrm{fds}_{10}$ represents the final digit sum in base ten.

A proof of the above claim together with the underlying digit sum results is presented in digit_sums.pdf (source: digit_sums.tex).

## Factoids #0

#### I) unit polynomials with non-vanishing degree

$2t+1\in\mathbb{Z}_4[t]$ is its own multiplicative inverse, showing that $R[t]^*=R^*$ does not hold in a general commutative Ring with one.

This phenomenon is uniquely characterized by the following equivalence:

$R[t]^* = R^* \iff \nexists \,0\neq a,b\in R:a\cdot b=0=a+b$

Proof. Negated replication.

Let $R\not\owns f=\sum_{i=0}^n\alpha_it^i\in R[t]^*,\alpha_n\neq0$ be a unit polynomial of non-vanishing degree $n\geq 1$. Let $g=\sum_{j=0}^m\beta_jt^j\in R[t]^*,\beta_m\neq0$ denote its multiplicative inverse, i.e. $f\cdot g=1$.

Claim. The polynomial $g$ has non-vanishing degree $m\geq 1$.
Proof. Suppose $g \in R$. Since $f\cdot g=\sum_{i=0}^n(\alpha_i\cdot g)t^i$, it follows from $\alpha_n\cdot g=0$ that $g$ is a zero divisor. However, at the same time $a_0\cdot g=1$ implies that $g$ is a unit, arriving at a contradiction.

Since both $n,m\geq 1$, one concludes $\exists 1\leq k\leq m$ as well as $\alpha_n\cdot\beta_m=0$.

Existence of the desired ring elements $a,b$ is assured by the following construction.

Let $k=1 \nearrow m$ rise discretely.

If $a:=\alpha_n\beta_{m-k}\neq 0$, implying $b:=\sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}\neq 0$, holds, since the construction arrived at this point, one finds

$a\cdot b=\alpha_n\beta_{m-k}\cdot \sum_{i=1}^k\alpha_{n-i}\beta_{m-k+i}=\sum_{i=1}^k \underbrace{\alpha_n\beta_{m-k+i}}_{=0}\cdot \alpha_{n-i}\beta_{m-k}=0.$

The above condition is met for at least one $1\leq k\leq m$, since otherwise $k=m$ would imply $\alpha_n\beta_{m-m} = 0$, which is impossible since $\alpha_n\neq 0$ and $\beta_0$ is a unit element.

By construction, $0\neq a,b$ as well as $a+b=0$ are given.

Negated implication.

Setting $f:=at+1, g:=bt+1$, one calculates

$f\cdot g=(at+1)\cdot (bt+1)=abt^2+(a+b)\cdot t+1=0t^2+0t+1=1$,

showing $R\not\owns f,g\in R[t]^*$.

q.e.d.