# aliquot

## < a quantity that can be divided into another a whole number of time />

I recently subscribed to the n-Category Café RSS feed, and I really enjoyed the last posts on random permutations. For instance, I learned that the distribution of the number of cycles of length k in a a random permutation of an n-element set, thought of as a random variable, follow a Poisson distribution with mean 1/k as $n\to \infty$. Go read the rest of the post if you want to learn more interesting facts on random permutations.

That being said, there’s a lot more to see regarding permutations in Knuth’s artwork, of course, especially in the TAOCP, vol. 4 (ex pre-fascicle 2B), for which there exist some efficient Python solutions (or even the Steinhaus-Johnson-Trotter algorithm). However, I recently stumbled upon the following problem on the Euler project. The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime, and, (ii) each of the 4-digit numbers are permutations of one another. There exists a 4-digit increasing sequence, and the question is: What 12-digit number do you form by concatenating the three terms in this sequence?

Here is my solution, which is heavily inspired by some of the solutions I found online in Haskell, Python or Lisp dialects:

from sympy import isprime

def isperm(a, b):
return sorted(str(a)) == sorted(str(b))

a, f = 1487, 1
d = 3330
while True:
a += 3-f
f = -f
b, c = a+d, a+2*d
if all(elt is True for elt in map(isprime, [a, b, c])) and \
isperm(a, b) and isperm(b, c):
break

print(str(a)+str(b)+str(c))


I had a great time at writing the above if statement, because of the way Python can be used to write concise iterators while allowing functional geekery using map. I didn’t implement a Sieve for finding prime numbers since I believe the sympy package provides a very efficient ones. There are very clever implementations on Stack Overflow, though. Note that the last expression could be rewritten as ''.join(map(str, [a, b, c])). The logic behind the above code is quite simple: we generate prime candidates, a, b, and c, starting with a=1487, and using the same increment, 3330. Put it simply, this is brute force method, but it happens to work in a reasonable amount of time.