The figure below illustrates how a torus (donut shape) can be cut in 
pieces with only 3 cuts:
pieces with only 3 cuts:
In fact, is the maximum number of pieces a torus can be sliced with 3 cuts.
is the maximum number of pieces a torus can be sliced with 3 cuts.Wikipedia's article on Torus, gives an amazing formula for calculating the maximum number of pieces using n number of cuts on a single Torus:
Therefore for 
: 
.
: .Given a cutting limit x, write a function to find all positive integers 
, such that 
is a semi-prime. A semi-prime is a positive integer that has only 2 and exactly 2 prime factors. A few example of semi-primes are: 
.
, such that is a semi-prime. A semi-prime is a positive integer that has only 2 and exactly 2 prime factors. A few example of semi-primes are: .As an example for 
, the only semi-prime 's are 
and 
, therefore the function output should be 
.
, the only semi-prime 's are and , therefore the function output should be .Solution Stats
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For n>330289
something weird happens in this series...
Many more numbers pop up as part of the solution.
For example:
330290 ## 463 * 12970537638763 = 6005358926747269
330321 ## 165161 * 36370874563 = 6007050013699643
330326 ## 2678131 * 2243102671 = 6007322799387901
These 3 examples are semi-primes.
You can check in MATLAB command window:
factor( C(330290) )
factor( C(330321) )
factor( C(330326) )
where C(n) is the equation for the cuts-pieces on a torus.
For this reason, the test 6 of this problem may be wrong and should updated for x=3e5.
Actually, factor(C(330290)) would return:
[ 2 5 23 29 33029 27259489]
without rounding errors.
Flintmax ~ 9e15, so if you calculated ~6e15 as n/6, n was clearly above flintmax.
A key point to solve this quickly for large inputs is to notice that any prime factors of n other than 2 or 3 will necessarily be factors of C(n) because the polynomial has a zero constant term.
@GeeTwo
thank you for pointing out a crucial detail that I was missing!