Here's a small example. We'll set $k=8$ so that we can hash messages of up to $2^8-1=255$ bits. We note that the eight smallest primes are $p_1=2, p_2=3$, $p_3=5,\ldots p_8=19$. We'll also pick a "hard to factor" number $n=314159265358979323$ (note that it is not actually necessary to know the factors of $n$).
To hash the 60-bit number $x$ 100110000111101011110010110011111110010111101110000110010100 we divide it into blocks of eight bits: 10011000 01111010 11110010 11001111 11100101 11101110 00011001 0100; pad the last block with zeros: 10011000 01111010 11110010 11001111 11100101 11101110 00011001 01000000 and add a final block representing the length: 10011000 01111010 11110010 11001111 11100101 11101110 00011001 01000000 00111100.
The VSH of $x$ is $p_1^{e_1}\cdots p_8^{e_8}\mod n$ where the $e_i$ is constructed by taking the $i$th least significant bit of each block and concatenating. Thus $e_1$ is the leftmost bits of each block $000110100=52$, similarly for $e_2$ we have $011101000=232$, $e_3=57$,
$e_4=429$, $e_5=453$, $e_6=217$, $e_7=250$ and $e_8=376$ and so
$$VSH(x)=2^{52}3^{232}5^{57}7^{429}11^{453}13^{217}17^{250}19^{376}\mod n$$
$$= 200591979525226314$$
Computation of the hash can be made ore efficient and pipelined by what is effectively a use of Shamir's trick. To do this we start with an initial value 1 mod $n$, then use each block to determine a subset of small primes to multiply our value but, followed by a square of our value. Thus we start $x_0=1$, our first block is 10011000 which selects the primes $p_8, p_5, p_4=19,11,7$ and compute
$$x_1=x_0^2\times(7\times 11\times 19)\mod n=1463$$
the next block is 01111010 which selects the primes 3,7,11,13,17 so that
$$x_2=x_1^2\times(3\times7\times11\times13\times 17)\mod n=109267977819$$
followed by
$$x_3=x_2^2\times(3\times11\times13\times17\times19)\mod n=25314401756605599$$
and
$$x_4=251897391984383465$$
$$x_5=259626239061055217$$
$$x_6=190834175947594433$$
$$x_7=249101195160685902$$
$$x_8=78295427833839690$$
$$x_9=200591979525226314$$
and so on, eventually getting the same answer as above for $x_9$.
