Ruzzle/Riddle Thread
- Thread starter ProCop
- Start date
*****ANSWER BELOW*****
The answer to this is 65511.
It's complicated to detail the method for achieving the answer, so bear with me please...
Let n be the position in the sequence, and R(n) be the answer - so in this case R(1)=1, R(2)=3, R(3)=5 etc.
Also, let X(n) be the sequence of prime numbers: X(n)=1, 2, 3, 5, 7, 11, 13, 17 etc
Okay - let Y(n) be X(n+1) - X(n).
So Y(n) = 1, 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6 etc.
And thus Y(n)-2 = -1, -1, 0, 0, 2, 0 , 2, 0 , 2, 4 etc
Now let D(n) be the SUM of ALL {Y(n)-2} from n=0 to n
i.e. the sequence D(n) = -1, -2, -2, -2, 0, 0, 2, 2, 4, 8 etc.
The solution is thus:
R(n) = 2^(n-1) - D(n) - 1.
So, when n=6, 2^(6-1) = 32; D(n) = 0; so R(6) = 32 - 0 - 1 = 31.
When n=12, 2^(12-11) = 2048; D(12) = 12; so R(12) = 2048 - 12 - 1 = 2035.
Likewise, when n=17 (i.e. the next in the sequence):
2(n-1) = 2^(17-1) = 65536; D(17) = 24; so R(17) = 65536 - 24 - 1 = 65511.
The only difficulty is in working out D(n).
But I'll leave that to you.
So there you have it.
It's not an error.
Very tricky, mind you.
Wow,
1,1,2,3, 5,8,13,20, 34
hhopefully is this one not too easy for you
My reading
1 1 2 3 5 8 13 20 34 (52 89)
make two rows of it
1 2 5 13 34 =89
a b c d
1 3 8 20 =52
a b c d
a+b + c+ c and so on
But I got a post about this elsewhere that it is 'Fibonacci-mice' problem, where mouses get born after 8 years or so.
But do not worry if it's too much you've done a great job today...
1,1,2,3, 5,8,13,20, 34
hhopefully is this one not too easy for you
My reading
1 1 2 3 5 8 13 20 34 (52 89)
make two rows of it
1 2 5 13 34 =89
a b c d
1 3 8 20 =52
a b c d
a+b + c+ c and so on
But I got a post about this elsewhere that it is 'Fibonacci-mice' problem, where mouses get born after 8 years or so.
But do not worry if it's too much you've done a great job today...
ProCop said:
I think that site is wrong - as it is too close to the Fibonacci series (i.e. one number out on only one of the first 9 numbers in the sequence) that there is not really enough info to make any other assessment.
As for 3,1,4,1,5,9... I guess this goes 2 then 6 then 5?
Okay, back to the 1,1,2,3,5,8,13,20,34
If you split this into two inter-twined series, as you had done:
Series 1: 1, 2, 5, 13, 34, x
Series 2: 1, 3, 8, 20, y
Okay - these become much easier.
Series 1: Each number of the sequence is twice the preceeding number PLUS the sum of all numbers before that preceeding one.
So 13 = 2x5 + 2 + 1
So 34 = 2x13 + 5 + 2 + 1
The next is thus: 34x2 + 13 + 5 + 2 + 1 = 89
Series 2: Each number of the sequence is twice the number before plus 2^(n-2).
So n=2: 1x2 + 2^(2-2) = 3
n=3: 3x2 + 2^1 = 8
n=4: 8X2 + 2^2 = 20
n=5: 20x2 + 2^3 = 48
So the next two numbers are 48 and 89.
However, splitting it into two sequences makes it much easier to come up with workable series.
Alternatively it could be the Fibonacci series with every 8th number in the sequence reduced by 1.
That's the ultimate problem with sequences - if you don't provide enough info then any number of possible answers arise.
If you split this into two inter-twined series, as you had done:
Series 1: 1, 2, 5, 13, 34, x
Series 2: 1, 3, 8, 20, y
Okay - these become much easier.
Series 1: Each number of the sequence is twice the preceeding number PLUS the sum of all numbers before that preceeding one.
So 13 = 2x5 + 2 + 1
So 34 = 2x13 + 5 + 2 + 1
The next is thus: 34x2 + 13 + 5 + 2 + 1 = 89
Series 2: Each number of the sequence is twice the number before plus 2^(n-2).
So n=2: 1x2 + 2^(2-2) = 3
n=3: 3x2 + 2^1 = 8
n=4: 8X2 + 2^2 = 20
n=5: 20x2 + 2^3 = 48
So the next two numbers are 48 and 89.
However, splitting it into two sequences makes it much easier to come up with workable series.
Alternatively it could be the Fibonacci series with every 8th number in the sequence reduced by 1.
That's the ultimate problem with sequences - if you don't provide enough info then any number of possible answers arise.
geistkiesel
Valued Senior Member
You win a gold coin. Did you know it or did you figure it out? Unfair I know,geodesic said:
Geistkiesel
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