[Guest]

All the following sequences follow the same pattern, try to find what it is (all starting numbers must be 2 digits, all other numbers must be 1 or 2 digits).

10,10,10,10,10,10,10,10,10

11,9,81,79,11,9,81,79,11

12,6,36,6,36,6,36,6,36

13,1,1,1,1,1,1,1,1,1

Now some random starting numbers:

33,21,19,71,69,21,19,71

67,11,9,81,79,119,81

That should be enough, using the same rule, what are the first few terms when 99 is the starting number?

If you do this, what starting number means that there are the most terms before a number is repeated?

Edited May 9, 2011 by James22

[curr3nt]

Did you mean 67,11,9,81,79,11,9,81 instead of 67,11,9,81,79,119,81?

[Guest]

Yes, sorry, last one should be:

67,11,9,81,79,11,9,81

[Guest]

For 99, I think the sequence is one of these

99,11,9,81,79,11,9,81

99,-11,-9,81,79,11,9,81

99,9,81,79,11,9,81

99,-9,81,79,11,9,81

And for the starting number with the most, I have to guess 67.

I am thinking the sequence is determined by a set of rules based on what number is in the one's place in two digit numbers, and for single digit numbers that it is the number squared.

[Guest]

There seem to be different rules for different types of numbers. For example, if a number is less than 10, square it. If the number ends in a 1, subtract 2 & include it in the sequence of numbers if it’s the first number. If it ends in a 3, subtract twelve. If it ends in a 9, square 9 and subtract the tens number, for example 59 would be 9 squared (81) minus 50, which would equal 31.

Use absolute values of negative numbers (not sure if this is a rule or not).

Using these few rules, we can figure out the sequence for 99.

99, 9, 81, 79, 11, 9, 81, 79, 11, etc.

As for which starting number has the most terms before repeating, I have no idea, but I'm guessing it doesn't end in a 1, 3, 9, or any even number, so it must end in a 5 or 7. That's all I have, don't know if it's right or not.

[Guest]

There seem to be different rules for different types of numbers. For example, if a number is less than 10, square it. If the number ends in a 1, subtract 2 & include it in the sequence of numbers if it’s the first number. If it ends in a 3, subtract twelve. If it ends in a 9, square 9 and subtract the tens number, for example 59 would be 9 squared (81) minus 50, which would equal 31.

Use absolute values of negative numbers (not sure if this is a rule or not).

Using these few rules, we can figure out the sequence for 99.

99, 9, 81, 79, 11, 9, 81, 79, 11, etc.

As for which starting number has the most terms before repeating, I have no idea, but I'm guessing it doesn't end in a 1, 3, 9, or any even number, so it must end in a 5 or 7. That's all I have, don't know if it's right or not.

You have the correct sequence, and the rules you listed certainly work at least in this case, but there is a more general rule that is the same no matter what the number is. You and Nana7 are both on the right lines though.

Edited May 9, 2011 by James22

[Molly Mae]

Is to the 6th digit.

37 19 71 69 21 19

47 9 81 79 11 9

67 29 61 59 31 29

All numbers that end in a 7 terminate at the 6th number.

EDIT: Disregard my inability to accomplish basic mathematics. =P

Edited May 9, 2011 by Molly Mae

[k-man]

Expressing every number x between 0 and 99 as x=10*a+b the formula for the next number in the sequence is

abs(b²-10*a)

or

abs(b(b+1)-x)

It's the same thing expressed differently.

[Guest]

Thanks to K-Man, I found the most terms before repeating

84,64,44,24,4,16,26,16

92,86,44,24,4,16,26,16

94,74,54,34,14,6,36,6

Edited May 10, 2011 by Nana7

[Guest]

Let each number be represented in the format xy (e.g., 6 = 06: x=0, y=6).

Each subsequent number is given by |10x - y²|.

When 99 is the starting number, where the numbers within the braces are the repeated sequence, the terms are:

99, {9, 81, 79, 11}, ...

The numbers 8, 84, 92 and 94 each have the most number of terms (seven) before a number is repeated.

8, 64, 44, 24, 4, {16, 26}, ...

84, 64, 44, 24, 4, {16, 26}, ...

92, 86, 44, 24, 4, {16, 26}, ...

94, 74, 54, 34, 14, { 6, 36}, ...

Edited May 10, 2011 by Dej Mar

[Guest]

K-Man and Dej mar got it right. Out of interest, what method did you use to find the longest chain of no repeats? I just used excel and I'm wondering if there is a more elegent way of doing it.

[Guest]

I did it by myself and I need to say:

Given two digits (a and b) number "ab" the formula that generate the serie is:

ABS(10a+b^2)

[Guest]

Let Xn be the nth number of a sequence of that set, while An and Bn be the 2 digits on ten place and integer place of Xn respectively, for n>/=2,

Xn = |(An-1) *10 - (Bn-1)^2|