algorithm 4

Solve the recurrence an=3an−1−3an−2+an−3for n>2 with a0=a1=0 and a2=1. Solve the same recurrence with the initial condition on a1 changed to a1=1.

Note by Pryhant Kielh
7 years, 8 months ago

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Comments

look easy but actually difficult. hehehe :)

Pryhant Kielh - 7 years, 8 months ago

Hi;

If this is the recurrence

a(n)=3a(n−1)−3a(n−2)+a(n−3)a(n) = 3 a(n - 1)- 3 a(n - 2) + a(n - 3)

then there are many methods. perhaps the simplest is by using the techniques of experimental math.

Generate the first couple of terms

0, 0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66,...

if you do not recognize that sequence then take it over to the the OEIS. This is a tremendous online resource available to everyone so why "Rediscover America"

[a_n=\frac{1}{2} \left(n^2-n\right)/]

bobbym none - 7 years, 8 months ago

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Sorry, bad latexing: an=12(n2−n)a_n=\frac{1}{2} \left(n^2-n\right)

bobbym none - 7 years, 8 months ago

First recurrence:

a(n)=12(n2−n)a(n)=\frac{1}{2} \left(n^2-n\right)

Second recurrence:

a(n)=12(3n−n2)a(n)=\frac{1}{2} \left(3 n-n^2\right)

Louie Tan Yi Jie - 7 years, 8 months ago
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