There is no set rule for sequences. Also they are NOT unique. You could have gotten a pattern your way with differences. 32 3 2 3 2 41 35 5 -1 5 43 40 4 4 48 44 8 52 52 60 112 One way would be to put two more 41's under that 41 and work your way backward: 32 3 2 3 2 41 35 5 -1 5 43 41 40 4 4 48 84 41 44 8 52 132 125 52 60 184 257 112 244 441 356 685 1041 That's one way to get two more terms 356 and 1041. But you could have put 42 and 43 under that 41 and gotten a different sequence. Or 43 and 45 and gotten a different one still, etc., etc. ---------------------------------------------------------- Here's prabably the sequence that the person who made that had in mind: Each number in the sequence is a representation of the number 32 in different bases, starting with base 10 and going down to 2. Here's how we do that. To get get the nth term we start out dividing 32 by k, where k=10,9,8,...2. Then we divide each quotient by k, noting the remainder each time. Sooner or later the quotient will be become 0. Then the nth term of the sequence will consist of the remainders taken as digits in the reverse order. R will stand for remainder. Remember, we stop dividing each time when the quotient gets to 0. Here we go: 32÷10 = 3 R=2 3÷10 = 0 R=3 Take remainders as digits in the reverse order. 1st term = 32, which is 32 in base ten. Now do the same thing with 32 and 9 32÷9 = 3 R=5 3÷9 = 0 R=3 Take remainders as digits in the reverse order. 2nd term = 35, which is 32 in base 9 Now do the same thing with 32 and 8 32÷8 = 4 R=0 4÷8 = 0 R=4 Take remainders as digits in the reverse order. 3rd term = 40, which is 32 in (octal) base 8 Now do the same thing with 32 and 7 32÷7 = 4 R=4 4÷7 = 0 R=4 Take remainders as digits in the reverse order. 4th term = 44, which is 32 in base 7 Now do the same thing with 32 and 6 32÷6 = 5 R=2 5÷6 = 0 R=5 Take remainders as digits in the reverse order. 5th term = 52, which is 32 in base 6 Now do the same thing with 32 and 5 32÷5 = 6 R=2 6÷5 = 1 R=1 1÷5 = 0 R=1 Take remainders as digits in the reverse order. 6th term = 112, which is 32 in base 5 Now do the same thing with 32 and 4 32÷4 = 8 R=0 8÷4 = 2 R=0 2÷4 = 0 R=2 Take remainders as digits in the reverse order. 7th term = 200, which is 32 in base 4 Now do the same thing with 32 and 3 32÷3 = 10 R=2 10÷3 = 3 R=1 3÷3 = 1 R=0 1÷3 = 0 R=3 Take remainders as digits in the reverse order. 8th term = 3012, which is 32 in base 3 Now do the same thing with 32 and 2 32÷2 = 16 R=0 16÷2 = 8 R=0 8÷2 = 4 R=0 4÷2 = 2 R=0 2÷2 = 1 R=0 1÷2 = 0 R=1 Take remainders as digits in the reverse order. 9th term = 100000, which is 32 in (binary) base 2. So the next two terms you wanted are 3012 and 10000 There are no more terms. If you tried it with 1 it would never end. Edwin