3 7 4 2 4 6 8 5 9 375 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

This is a really wonderful problem, another great example of dynamic programming.**(continued)**

This problem got me thinking about turning numerals into words, and I actually spent quite a while solving this problem much more fully than I had to. I’m not going to explain my full solution: that’s for a different post.

Suffice to say, I wrote a script that will print, in British English short-scale words, any integer from to ( is one *vigintillion,* after which Wikipedia runs out of names). It took quite a lot of effort and made me realise just how subtle our numbering rules actually are. Here’s the code:

# D Falck.

Easy. Define a quick digits function to return a tuple of the digits of `N`:

def digits(N): return tuple(int(i) for i in str(N))

Now just sum the digits of . Done.

def digits(N): return tuple(int(i) for i in str(N)) print(sum(digits(2**1000)))

**(continued)**

This looks confusing at first, and it’s easy to get lost combinatorially. However, this type of problem is a typical example of where we can use basic dynamic programming.

Dynamic programming is where, instead of trying to do the whole thing at once, you build up gradually piece by piece. In this case, we want to consider very carefully what it means to go from one corner to the next.

Say we label the vertical lines as and the horizontal lines as for an grid. Any route from corner to corner can only go downwards and rightwards, and so there are only two options for which corner came before corner : either corner or corner .**(continued)**

This is a really interesting problem. Before even thinking about what we’re going to have to do later on, let’s just make a quick function to determine the next number in a Collatz sequence:

def nextCollatz(n): if n % 2 == 0: return int(n/2) else: return 3*n+1

This quite literally is nothing more than the rule given in the question. Next, we want to generate the whole sequence given a particular starting number:

def CollatzSequence(n): sequence = [n] while True: n = nextCollatz(n) sequence.append(n) if n == 1: break return sequence

We’re just appending the next term (using the function `nextCollatz(n)`

we just defined) endlessly until we reach 1, and then we return that whole list.**(continued)**