Hey good people,
it has been a while now since I posted the small enigmas. I have been asked by 
Solution for Number Sequence #1
Enigma:
What is the missing number and how to compute?
[1; 3; 7; 15; 13; 9; 19; 21; *; 15]
Answer:
If we take a closer look at the first few entries we see that from the first to the second, from the second to the third ... The value almost double. It we look even closer we see that the sequnce follows the rule:
First assumption:
Multiply the previous value by 2 and add 1.
or
F(n) = 2*F(n-1)+1
But this fomrula fails for all number after the 15 (4th entry) and we need one formula for the whole sequence. Ok, now we take a look at the follwoing numbers to get an idea. The first false number is 13.
- 13 would be correct (according to assumption) if the number before would be 6, but it is 15.
- 9 would be correct if the number before would be 4, but it is 13.
- 19 would be correct if the number before would be 9, and it is 9
- 21 would be corect if the number before would be 10 but is is 19
Do you see the realtion between:
6-<>15
4-<>13
9<->9
10-<->19
The first number is the sum of digits of the second! So with this we have the final solution.
Multiply the sum of digits of the previous value by 2 and add 1.
or
F(n) = 2*SumOfDigits(F(n-1))+1
And of with this...the missing number was 7
Solution for Number Sequence #2
Enigma:
What is the missing number and how to compute?
[0; -2; -3; 8; 95; *]
Answer:
We see that the growth of this sequnce is very very fast. There are not many possibilties if the squence is increasing this fast. The faculty of n is one example.
What is n! ?
n=1 -> n! = 1
n=2 -> n! = 2
n=3 -> n! = 6
n=4 -> n! = 24
n=5 -> n! = 120
As we see the growth of n! is very simular to the our sequence. But we are not yet there. What do we need to change for each n to get to our sequence?
n =1 -> we need to do -1
n=2 -> we need to do -4
n= 3 ->we need to do -9
n=4 -> we need to do -16
n=5 -> we need to do -25
And I think one can see now that we need to subtract the squares of n to get to our sequence.
So the solution is F(n) = n!-n^2
I think this is quite challenging if one doesn't know that there are not many way to obtain a number sequence that increases so fast. Nevertheless, in a retrosperspective this enigma was too difficult.
And that were the solutions.. Feel free to comment and give me your opinion on the difficulty. Should I decrease it?
