I'm trying to encode a number. Pulling a prank on a friend who brags is really good at math.  Need an complex equation where the answer will work out to 1346 in some context.  Any help would be appreciated.      
 A: Take a random function with no real roots, say $5^x((x+1)^2+1)$. Then multiply with $x-1346$. Using this example, we get: 
Find all real numbers such that:
$$5^xx^3 -1344 \cdot 5^xx^2-2690 \cdot 5^x x-2692 \cdot 5^x=0$$
Of course, one cold take a much harder starting equation, for example $2^{2^x}\log(x+1)+(x+1)^4+4546$, as long as it has no roots. 
A: Should integration be accepted too (I just noticed the ENT tag...) try the neat $\,a=13\,$ in :
$$\int_0^\infty \frac{x^8-3a}{\cosh(x\frac{\pi}2)}\,dx$$
Concerning number theory let's observe (source) :


*

*$1346=2\cdot(672+1)\quad$ where $\;N:=672=2^5\cdot 3\cdot 7\;$ is the second triperfect number (i.e. the sum of the positive divisors of $N$ is equal to $3N$).

*the smallest triplet $\;(n+29,n+30,n+31)\;$ of positive integers all divisible by a cube different of $1$.

A: Give the prime factorization of $1743388617272249143997555461487119439669521095365209$.
A: $n=1346$ is the $270$th integer solution of:
$$ \phi(\phi(\phi(n)))=2^6$$
with $\phi$ being the Euler totient function, and also the number we get by swapping the last two digits of the $15$th Lucas number. Moreover, $1346$ is a semiprime sandwiched between semiprimes and $1346$ is the number of distinct products of the form $ijk$ for $1\leq i<j<k\leq 27$, or:
$$ 1346 = \sum_{k=7^2}^{8^2-1}\sigma(k) $$
where $\sigma$ is the sum-of-divisors function.
A: Simple, but if you haven't been introduced to geometric series, the following might appear rather strange:
$$1346 = \sum_{n=0}^{\infty} \left(\frac{1345}{1346}\right)^n$$
A: \begin{align}
f(0)&=6\\
f'(1)&=13\\
f''(2)&=18\\
f'''(3)&=6\\
f^{(n)}(n)&=0\quad\text{for }n>3\\
f(10)&=\boxed?
\end{align}
Not sure how hard this is, though.
