(1) The product of exactly two consecutive numbers $k$ and $k+1$ can be settled as follows: $n^6+5n^3+4n+116=k(k+1)$ is equivalent to $4(n^6+5n^3+4n+116)+1=4k^2+4k+1$ is equivalent to $4n^6+20n^3+16n+465=(2k+1)^2$. Hence $4n^6+20n^3+16n+465$ is an odd square, whose root is of the form $2n^3+2a+1$ with integer $a\ge0$. The resulting equation simplifies to $a^2+a(2n^3+1)=4n^3+4n+116$. For $a\ge11$, the left hand side is larger than the right hand side. For $0\le a\le 10$, the only integer solution is $a=4$ and $n=3$. (2) The product of five or more consecutive numbers can be settled as follows: The product of at least five consecutive numbers is a multiple of $5$, whereas $n^6+5n^3+4n+116$ is never a multiple of $5$. (3) The product of exactly three consecutive numbers $k-1$, $k$ and $k+1$ can be settled as follows: $n^6+5n^3+4n+116=(k-1)k(k+1)$ with the substitution $k=n^2+b$ with integer $b$ yields $3n^4b+3n^2b^2+b^3-b=5n^3+n^2+4n+116$. For $b\ge5$, the left hand side is larger than the right hand side. For $b\le0$, the left hand side is non-positive and the right hand side is positive. For $1\le b\le4$, there is no solution. (4) The product of exactly four consecutive numbers $k$, $k+1$, $k+2$ and $k+2$ can be settled as follows: $n^6+5n^3+4n+116=k(k+1)(k+2)(k+3)=(k^2+3k+1)^2-1$ implies that $n^6+5n^3+4n+117$ is the square of an odd integer $n^3+c$ with $c\ge1$. This simplifies to $c^2+(2c-5) n^3=4n+117$. For $c\ge11$, the left hand side is larger than the right hand side. For $1\le c\le 10$, there are no integer solutions.