Solve: 4, 5, 14, 185, ....

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  • #1
karush
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ok somebody sent me this on youtube but I don't think is a workable series

Rimbus Jift
@ray salmon Quick IQ test... Solve: 4, 5, 14, 185, ...

so first the difference between the numbers is 1,9,171


so far just some ideas
The factors of 171 are 1, 3, 9, 19, 57, 171
$9^0 =1$ and $4+1=5$
$9^1 =9$ and $5+9=14$

$9+5=14$
$9+14=19$
$9\cdot 19=171 $
anyway ?

i plugged into W|F but didn't return a series
 
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  • #2
topsquark
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ok somebody sent me this on youtube but I don't think is a workable series

Rimbus Jift
@ray salmon Quick IQ test... Solve: 4, 5, 14, 185, ...

so first the difference between the numbers is 1,9,171


so far just some ideas
The factors of 171 are 1, 3, 9, 19, 57, 171
$9^0 =1$ and $4+1=5$
$9^1 =9$ and $5+9=14$

$9+5=14$
$9+14=19$
$9\cdot 19=171 $
anyway ?

i plugged into W|F but didn't return a series
You should know by now that you can choose any number to be the next one. For example we can use
\(\displaystyle f(x) = -28 x^4 + \dfrac{917}{3} x^3 - 1130 x^2 + \dfrac{5014}{3} x - 815\)

and get f(1) = 4, f(2) = 5, f(3) = 14, f(4) = 185. and the next number in the series will be f(5) = 0.

-Dan
 
  • #3
karush
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ok actually I haven't seen that,
the few series I worked on just plug and played with values of n till you got an eq to generate the series
the imperative "solve" does not insist that it is series generated by an eq but I assume that was the intention
why woufd f(5)=0 or is that just arbitrary

Anyway it does seem slam dung stuff
 
  • #4
topsquark
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ok actually I haven't seen that,
the few series I worked on just plug and played with values of n till you got an eq to generate the series
the imperative "solve" does not insist that it is series generated by an eq but I assume that was the intention
why woufd f(5)=0 or is that just arbitrary

Anyway it does seem slam dung stuff
f(5) is arbitrary. For example:
\(\displaystyle f(x) = -\dfrac{671}{24} x^4 + \dfrac{1221}{4} x^3 - \dfrac{27085}{24} x^2 + \dfrac{6677}{4} x -814\)

gives f(1) = 4, f(2) = 5, f(3) = 14, f(4) = 185, and f(5) = 1.

etc. And you can do other fits aside from polynomials pretty much so long as you have 5 unknowns and the system can be solved. (Polynomials are easy to fit which I why I prefer to use them for demonstrations.)

A problem like this assumes that you can figure out a pattern but unfortunately any more information that you might get could change that answer. So I feel that problems like this are just silly.

-Dan
 
  • #5
karush
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ok I think so too ,...

but curious
what online series calculators are good if you just give a list of 6 numbers which assumes a generator eq
I guess W|F will but haven't tried
 

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