#This code produces Pascal's triangle up to row nrows.
#Note that it is a list of lists
nrows=10
pascal=[[1],[1,1]]
for r in [2..nrows]:
    nextrow=[1]
    prevrow=pascal[-1]
    lenPre=len(prevrow)
    for k in [0..lenPre-2]:
        nextrow.append(prevrow[k]+prevrow[k+1])
        
    nextrow.append(1)
    pascal.append(nextrow)

#here is a printout to test it
for r in pascal:
    print r

#Now we do the same thing, but mod m
m=3
nrows=10
pascal=[[1],[1,1]]
for r in [2..nrows]:
    nextrow=[1]
    prevrow=pascal[-1]
    lenPre=len(prevrow)
    for k in [0..lenPre-2]:
        nextrow.append(mod(prevrow[k]+prevrow[k+1],m))
        
    nextrow.append(1)
    pascal.append(nextrow)

#test it
for r in pascal:
    print r

#the next lines of code count the number of odd numbers in each row.  You cannot just sum the elements in each row, because 
#the elements are NOT integers but instead are integers modulo 2, so they only sum to 0 (even) or 1 (odd)
#So we use a boolean expression. 

rowNum =0 #keep track of the row for output
for r in pascal: #here "r" is the actual row, with numerous elements if rowNum is large
    nOdd=0 #begin a sum
    for a in r:
        nOdd += a==1  #increment the value only if a equals 1, i.e. a is odd
    print rowNum, nOdd
    rowNum += 1   #increment the row number

#incidentally, you may want to use binary, etc.
n=2018
n.digits()
n.digits(base=2)

rowNum.digits(2)