# Finding a squence.

Problem #14 at project euler is straightforward. You need to find an iterative sequence or rather the starting number that results in the most terms. If you are a strict mathematician the problem description isn't very sound but anyway here is the solution:

public void problem14(){

int n0;

long n;

int maxTerms=0, terms=0, nMax=0;

int results[] = new int[1000002];

for(n0=5 ; n0 < 1000000 ; n0++, terms=0)

{

n = n0;

do {

if(n % 2 == 0)

{

n = n/2;

}

else

{

n = 3*n + 1;

}

terms++;

//System.out.print(n +",");

if(n < n0)

{

if(n < 0)

{

System.out.println("N =" + n +" for n0="+n0);

break;

}

if(results[(int)n] != 0)

{

terms += results[(int)n];

break;

}

}

}while(n != 1);

results[n0] = terms;

if(maxTerms < terms)

{

maxTerms = terms;

nMax = n0;

}

//System.out.println("");

}

System.out.println(nMax +" gives the most terms" + maxTerms);

}

At first I used *int*egers instead of *longs*s and ran into a situation where some of the terms exceeded 2^32 -1 which lead to array out of bounds exceptions. Switching to longs solved that one easily.