I think I have the perfect solution:
0, 1, 2, infinite, -6, ....
It is easy: 6/(3-n)
Zara
Juan Antonio Zaratiegui Vallecillo escribió:
0, 1, 2, infinite, -6, ....
It is easy: 6/(3-n)
Zara
Juan Antonio Zaratiegui Vallecillo escribió:
> Mark escribió:
>> Like your thinking Zara but...
>> ääääääääääää... <annoying game-show buzzing sound>
>> ...not the answer I'm looking for.
>>
>> 1. The series is in base 10; so not 10, 11, 12...
>>
>> 2. I prepped Steve by mentioning his apparent bent for maths. It would
>> be underhanded of me to then present a series that was
>> alphabetically based. Also using "nul" (or "nil") instead of "zero"
>> is not really playing fair; so not "nul, one, two, six, ten" (you
>> forgot that one :-)
>>
>> 3. The fibonacci series extends to the left of zero as well as to the
>> right, thus - ...5 -3 2 -1 1 0 1 1 2 3 5... For the '0' of my
>> series to be generated in this manner requires a previous term
>> of '0' and prior to that -1 and so on (see example below).
>>
>> "Mine": ...4 1 -2 0 -1 0 0 1 2 4 7 12...
>> Fibon.: ...5 -3 2 -1 1 0 1 1 2 3 5 8...
>>
>> My series begins at '0', there are no previous terms that can
>> exist. However, it is infinite and the fourth term is a little
>> unexpected.
>>
>>
>> Have another go.
>>
>> Mark
>>
>
>
> Another go:
>
> Series of primer numbers:
> 1,2,3,5,7,11,13,17...
> Mark series, 1 less:
> 0,1,2,4,6,10,12,16...
>
> Do I get the cookie?
>
> Zara
>
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