Euler_Problem-065.b93

v $$$    #######################################################################          // numerator

         #######################################################################          // denominator

    v                 -1<
>"F">:9+"0"\0p:9+"0"\2p:|
v"c"p040p2"O1"p0"O0"   $<
          >$1         v@.<v02-"0"p0g03:g2g03-"0"g0:p03:<-1<
  >:1-3%:!|   >:1+3/2*v  +>g*+40g+:55+%"0"+30g2p55+/40p:9-|
>:|       >2-#^_1     >20 p                      "O"v  #  $
| ># #+ #1 #: #- #12# ^#                            ># ^# <
$               >\v  >\:!|
>0"F">:9+2g"0"-:| >:!|1 +<
     ^        -1_ ^#1<

[2, 0]        frac
[3, 0] (calc) pos
[4, 0] (calc) carry


---------------------------------------

Nice algorithm if you see the pattern in the numerators and denominators.

~~~
denom(n+1) = denom(n) + numer(n) * frac(n)
numer(n+1) = denom(n)
~~~

and the fraction at position n is calculated by ([OEIS-A003417](https://oeis.org/A003417)):

~~~
int GetFrac(int idx)
{
	if (idx == 0) return 2;
	if ((idx-1) % 3 == 0) return 1;
	if ((idx-1) % 3 == 1) return ((idx+1)/3)*2;
	if ((idx-1) % 3 == 2) return 1;
	return 2;
}
~~~

The rest is just multiplication and long addition (we exceed the 64bit range) a hundred times ...