Small dots indicate that this series will continue. So here is your solution:
Let's look at the index based on 1. You noticed that 1 occurs at index 1, (1 + 2) = 3, (1 + 2 + 3) = 6, (1 + 2 + 3 + 4) = 10, etc. . We have a formula for this. Its n * (n + 1) / 2.
, ( 0, java 0):
index = index + 1; // now it is 1 based index and our formula would fit in nicely.
index = index * 2;
sqroot = integer part of square root of index;
if( sqroot * (sqroot+1) == index)
print 1;
else
print 0;
, O (1) ( )