Usbrth

The wikipedia entry on Ramanujan contains the following passage:

One of his remarkable capabilities was the rapid solution for
problems. He was sharing a room with P. C. Mahalanobis who had a
problem, "Imagine that you are on a street with houses marked 1
through n. There is a house in between $(x)$ such that the sum of the
house numbers to left of it equals the sum of the house numbers to its
right. If $n$ is between $50$ and $500$, what are $n$ and $x$?" This is a
bivariate problem with multiple solutions. Ramanujan thought about it
and gave the answer with a twist: He gave a continued fraction. The
unusual part was that it was the solution to the whole class of
problems. Mahalanobis was astounded and asked how he did it. "It is
simple. The minute I heard the problem, I knew that the answer was a
continued fraction. Which continued fraction, I asked myself. Then the
answer came to my mind," Ramanujan replied.

What was the continued fraction, and how did it give all solutions to the problem? Most importantly, how could someone derive such a solution? Do similar problems also have continued fractions that describe all solutions?

This seems like an interesting and powerful method, and I would like to learn more about it.

An expansion of Andre's comment into a detailed exposition can be found at http://www.johnderbyshire.com/Opinions/Diaries/Puzzles/2009-06.html. It's also at http://www.angelfire.com/ak/ashoksandhya/winners2.html#PUZZANS4.

We will use

Theorem 1: If $rgt0$ is irrational and $left|,frac pq-r,right|lefrac12q^2$ where $(p,q)=1$, then $frac pq$ is a continued fraction convergent for $r$.

and

Theorem 2: If $leftfracp_kq_kright$ is the sequence of continued fraction convergents for an irrational $rgt0$, then $r$ is between $fracp_kq_k$ and $fracp_k+1q_k+1$ and $fracp_k+1q_k+1-fracp_kq_k=frac(-1)^kq_k+1q_k$.

The problem can be written as
$$
overbrace fracx^2-x2 ^substacktextsum of $1$\textto $x-1$=overbracefracn^2+n2-fracx^2+x2^textsum of $x+1$ to $n$tag1
$$
which is equivalent to
$$
8x^2=(2n+1)^2-1tag2
$$
From $(2)$, we get $frac2n+1x=sqrt8+frac1x^2$; thus, we want an over-estimate for $sqrt8$ that has an odd numerator. Furthermore, $(2)$ implies
$$
beginalign
frac2n+1x-sqrt8
&=sqrt8+tfrac1x^2-sqrt8\[9pt]
&=frac1/x^2sqrt8+frac1x^2+sqrt8\
&lefrac12sqrt8,x^2tag3
endalign
$$
Theorem 1 and inequality $(3)$ say that $frac2n+1x$ needs to be a continued fraction approximant for $sqrt8$ to satisfy $(2)$.

The continued fraction for $sqrt8$ is $(2;1,4,1,4,1,dots)$. The convergents are
$$
beginarrayc
k&0&color#C001&2&color#C003&4&color#C005&dots\hline
fracp_kq_k&frac21&color#C00frac31&frac145&color#C00frac176&frac8229&color#C00frac9935&dots
endarraytag4
$$
The repeating continued fraction gives the following numerator/denominator recurrence
$$
beginalign
a_2n&=4a_2n-1+a_2n-2tag5a\
a_2n+1&=a_2n+a_2n-1tag5b
endalign
$$
Theorem 2 says that the red terms, the ones with odd indices, are over-estimates. Furthermore, the recurrences $text(5a)$ and $text(5b)$ guarantee that the red terms have odd numerators.

Combining $text(5a)$ and $text(5b)$, we get that the recurrence for the odd numbered numerators and denominators is
$$
a_2n+1=6a_2n-1-a_2n-3tag6
$$
Thus, we can get the solutions from the odd-indexed terms of $(4)$:
$$
beginarrayc
k&1&3&5&7&9&dots\hline
fracp_kq_k&frac31&frac176&frac9935&frac577204&frac33631189&dots\
fracn_kx_k&frac11&frac86&frac4935&frac288204&frac16811189&dots
endarraytag7
$$
The pair that has $50le nle500$, is $(x,n)=(204,288)$.

Checking:
$$
sum_k=1^203k=20706=sum_k=205^288ktag8
$$

See Pi Mu Epsilon Journal Vol. 14 # 6 pp 389-97, 2017.

"House Number Puzzle: Solution Through Bifurcation"

A summary from the above Pi Mu Epsilon publication is given below:

Continued Fraction of √2 is: 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169 ... The above (x/y) also form solutions to the Pell's equation: x ² −2 y ² = ±1. The house numbers are products of x and y (the solutions to the above Pell's equation): 1*1; 3*2; 7*5; 17*12; 41*29 ... and the total number of houses are provided by x*x and 2*y*y alternatively. Total number of houses: 1*1; 2*2*2; 7*7; 2*12*12; 41*41 ...

For details see the actual publication.