GEEK powers test!

[tsudonimh]

Can you tell me what the numbers in each series have in common?

series zro: 0
series one: 1 2 4 8
series two: 3 5 6 9 10 12
series tre: 7 11 13 14
series for: 15

[Update:]
Okay, there's actually a couple series I left out because I'm mean... I've added series zro and for, above.

Comments

[tsudonimh]

I'm sorry Beth. You just aren't geeky enough. You are, as you have already admitted, an ARTEEST. Put on the paint-smeared smock, and the floppy frog warmer on your head, groan through your nose, and be HAPPY. Triangles? I mean, really.

It's interesting that none of the consecutive triplets passes through all three groups...

Actually, 'triangle' is amazingly, tantalizingly related to the explanation.

Work it!

[mircus]

The kingdom of "15".
Let call the rows r1 - r4 and the columns c1 - c6
They are all the ints from 0 to 15.
The sum of each single row and as well the total sum of all the ints are 15 * int.
The following sums gives all 15:
r1c1 + r5c1; r2c1 + r4c4; r2c2 + r4c3; r2c3 + r4c2; r2c4 + r4c1;
r3c1 + r3c6; r3c2 + r3c5; r3c3 + r3c4;
These relationships are 8.
If we substitute the ints with their complement to 15, we get somehow a mirrored structure where r1c1 become r1c5, r2 become r3 but with the row inverted.
The same with r3, where r3c1 become r3c6, r3c2 -> r3c4 and r3c3 -> r3c4.
The relationships are 8 as well.
but if we consider 0 as the complement to 15 of 15, we could consider the relationships r1c1 (complement to 15) + r5c1 (15) as an 1/2 relationships (we need the complement counterpart of our system, the other side of the mirror).
The same with the counterpart of the complements to 15. And so the relationships would be 7 1/2 + 7 1/2 = 15. Another time our golden number!

[tsudonimh]

If I ask how many combinations of 21 can be taken out of 25, I do in effect ask how many combinations of 4 may be taken. For there are just as many ways of taking 21 as there are of leaving 4. -- AUGUSTUS DE MORGAN

This is another clue--a hint of the model you can use to arrive at the answer. The triangle I hint at belongs to Pascal, and can be thought of as either a way of selecting the binomial coefficient of an expansion, or as a way of enumerating ways of selecting r thingies from a collection of n thingies. :)

What are the thingies I'm selecting as I construct a set of series? Here's another:

series 0: 0
series 1: 1 2 4 8 16
series 2: 3 5 6 9 10 12 17 18 20 24
series 3: 7 11 13 14 19 21 22 25 26 28
series 4: 15 23 27 29 30
series 5: 31

[mircus]

Languge problems :-(
Let me understand, thingies means:
little things, the way (ways) of getting the value of the target cell in cellular automata or in the pascal's triangle, the algorithm of iterations in fractals, something explainable in therms of matrices (transpose, convolution)???

[tsudonimh]

In this case, starting from the general and moving to the particular, thingies are whatever is being chosen must have two states, so you could be selecting from n pennies tossed on the table those r among them that landed heads up. The probability of getting r that are heads out of n that are tossed is n!/(r!(n-r)!)/2^n. (in case you aren't familiar with the 'bang' notation, the exclamation point means n!=n*(n-1)*(n-2)*...*1. ) The series for three coins would look like:

0 heads: ttt (probability of 0 heads of three is 1/8)
1 heads: htt tht tth (prob. of 1 head of three is 3/8)
2 heads: hht hth thh (prob. of 2 heads of three is 3/8)
3 heads: hhh (prob. of 3 heads of three is 1/8)

In the case above whatever r is selected from whatever n determines the numbers you see, and I'm asking that you find out what is the 'thingies' being selected, and their states.