On certain series of numbers - correction

[Doug Kerr]

I have discussed many times in recent threads that there can be a series of numbers that obeys both these rules:

• If we start with any value and go forward in the list by 3 values, the 2nd value is exactly 2 times the first.

• If we start with any value and go forward in the list by 10 values, the 2nd value is exactly 10 times the first.

I had said that the longest list following those rules would be 30 values in length. That was in error. We can have a list following those rules with up to 59 values.

It is true that if we start with some value and build the list only "upwards" from that, following both rules, the longest valid list will be of length 30. (For the next number, both rules would apply, giving different values.)

But if we start with some value and build the list both upwards and downwards from that, following both rules, the longest valid list will be of length 59.

Note that in general valid values in these lists will not necessarily all be integers. If however we start with certain integer values (e.g., 80) and build the list only upwards, all values in the list will be integers.

Sorry for the bum steer before.

Best regards,

Doug

 

[Doug Kerr]

Here is an explanation of the length limits for such a series of values.

Suppose we start building our list with a certain value, and that we number that value as "Value number 1". As out "root value", it is by definition a member of two sets:

• Every 3rd value
• Every 10th value

We assign values to all the values in the first set and all the values of the second set.

For quite a while there would be no value that was a member of both sets. (We note that as we do this almost 2/3 of the "value positions" would have no value assigned by this process. We can take care of that later, but for now don't worry about it.)

But when we get to position 31, we find it to be a member of both sets. It thus gets a value from the rule "every 3rd value is 2 times the value" and the rule "every 10th value is 10 times the value". But those two values are (slightly) different. So value 31 cannot fit in with this story. Only 30 consecutive values can fully fit in.

However, if we also build our list downward from the root, given values to every 3rd position and every 10th position, we find that the 30th position of that batch is a member of both sets. So it would get a value from both rules. Again, those two values are (slightly) different. So that 30th downward value cannot fit in with this story. Only 29 downward positions will fit in.

Thus the longest list of numbers that would obey both rules is of length 59.

Best regards,

Doug