Mind blown: The article made me realize that you can generate one of those trippy "continued fraction" equalities any time you have x on both sides of the equation, one side by itself.

Golden ratio: x = 1 + 1/x -> keep re-plugging the right-hand side into the x on the RHS:

x = 1 + 1/(1 + 1/(1 + 1/...

Generating Functions and their applications may be of interest to you.

https://en.wikipedia.org/wiki/Generating_function

A PDF on the topic: https://www.math.upenn.edu/~wilf/gfology2.pdf

I was introduced to them in Concrete Mathematics by Graham, Knuth, and Patashnik, but then got to see their application in later statistics courses.

It occurs to me that a continued surd also works here:

  x = sqrt(1 + sqrt(1 + sqrt(1 + sqrt(1 + ...

You might also have your mind blown by this esoteric result: https://news.ycombinator.com/item?id=11976554

... actually, just mired in incomprehension.

You can generate many of them, not all of them converges though.

Isn't the golden ratio x = 1/(x-1) ?

Never mind, they both work out to the same value.

I guess it is the same equation (x-1 = 1/x)

Its great to see Ramunujan getting a fair bit of attention these days. His story is really fascinating. I can recommend:

https://www.amazon.com/Man-Who-Knew-Infinity-Ramanujan/dp/06...

This was also made into a movie a couple of months ago, I haven't seen the movie but it should be on netflix/itunes by now since I don't think it had a wide theatrical release.

http://www.imdb.com/title/tt0787524/

Netflix currently says they're getting it in August 2016 (DVD; I don't know about streaming)

The house is #204 out of 288 houses. Assuming the house itself isn't counted in either sum, which wasn't too clear by the question.

I got this by the definition of triangle numbers and solving n^2 = m(m+1)/2 for integer solutions. n is the house and m is the total of all houses. As usual, I have no idea how Ramanujan did this with cont. fractions.

An exposition of how Ramanujan did it: http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-...

Thanks for the spoiler.

Sorry about that. It isn't the full answer though.

   The sum of 1..n is n(n+1)/2. A.
   The sum of 1..N is N(N+1)/2. B.
   The sum of n..N = B-A.
   We want solutions to A = B-A. Or, 1 = B-A/A

I leave the rest as an exercise for the reader. :)

I really enjoy Quanta Magazine's artwork featured on the articles!