A Sidewinder Sequence

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Consider the recursive sequence defined by

a 1 = 1 ; a 2 = 2 ; a n = a n − 1 3 − a n − 2 for n ≥ 3. a_1=1;\; a_2=2;\; a_n=a_{n-1}\sqrt{3}-a_{n-2}\text{ for }n\ge 3.

Then a 2014 a_{2014} can be expressed as m + n p m+n\sqrt{p} , where m , n , and p are integers, and p is positive and square-free. Find m + n + p m+n+p .


The answer is 0.

Matt Enlow by Matt Enlow , 46, USA

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2 solutions

Omar AbdelWanis
Jan 24, 2014

If you list up the first few terms of the sequence you find that a 3 = 2 3 − 1 a_{3} = 2\sqrt{3}-1 , a 4 = − 3 + 4 a_{4} = -\sqrt{3}+4 , a 5 = 2 3 − 2 a_{5} = 2\sqrt{3}-2 , a 6 = − 3 + 2 a_{6} = -\sqrt{3}+2 , a 7 = − 1 a_{7} = -1 , a 8 = − 2 a_{8} = -2 and so on... we find that a 7 = − a 1 a_{7} = - a_{1} and therefore a 13 = − a 7 = a 1 a_{13} = -a_{7} = a_{1} So to find a i a_{i} we can take i m o d 12 i mod 12 (the difference between the two equal terms) a 2014 = a 2014 m o d 12 = a 10 a_{2014} = a_{2014 mod 12} = a_{10} and a 10 = − a 4 = − ( − 3 + 4 ) = − 4 + 3 a_{10} = -a_{4} = -(-\sqrt{3}+4) = -4+\sqrt{3} .

− 4 + 1 + 3 = 0 -4+1+3=0 so the answer is 0 0

1 Helpful 0 Interesting 0 Brilliant 0 Confused

i tried to make a closed form of this recurrence relation but it really became so complicated so i couldn't solve it any way nice thinking !!!!!!

joe ashour - 7 years, 4 months ago

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Look at the sequence that uses the same recurrence relation, but with initial terms a 1 = 0 a_1=0 and a 2 = 1 2 a_2=\frac{1}{2} . That might look a little more familiar...

Matt Enlow - 7 years, 4 months ago

The closed form is a n = x e n α i + y e − n α i a_n = x e^{n\alpha i} + y e^{-n\alpha i} , where α = π 6 \alpha = \frac{\pi}{6} . Without solving for the values of x , y x,y , we can see that a 2014 = − a 4 a_{2014} = -a_4 since 2014 = 6 ⋅ 335 + 4 2014 = 6\cdot 335 + 4 .

George G - 7 years, 4 months ago
Rizky Riman
Jan 25, 2014

How I do it is pretty simple:

  1. find a 3 a_{3} with the formula above
  2. find a 4 a_{4} with the a 3 a_{3} you've got before
  3. find a 5 a_{5} with the a 4 a_{4} you've got before
  4. find a 6 a_{6} with the a 5 a_{5} you've got before . . .
  5. find a 15 a_{15} with the a 14 a_{14} you've got before, and now you'll notice the pattern
  6. 2014 − 3 2014-3 because a 1 , a 2 , a 3 a_{1}, a_{2}, a_{3} is not on the pattern
  7. 2011 2011 divided by 12 12 because there's 12 different pattern
  8. The remainder should be 7 7 and so, we pick the 7th pattern which is − 4 + 3 -4+\sqrt{3}
  9. m + n + p = − 4 + 1 + 3 = 0 m + n + p = -4 + 1 + 3 = 0
  10. So, the answer is 0 \boxed{0}
0 Helpful 0 Interesting 0 Brilliant 0 Confused

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