Find OGF for Recurrence: a_n = 6a_{n-1} + a_{n-2}, a_0=2, a_1=1

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The discussion focuses on finding the ordinary generating function (OGF) for the recurrence relation defined by a_n = 6a_{n-1} + a_{n-2} with initial conditions a_0 = 2 and a_1 = 1. The user initially derives the OGF as A = (6x^2 + x^3 + 2x) / (1 - 6x - x^2) but believes the correct form should be (2 - x) / (1 - 6x - x^2). The conversation reveals discrepancies in defining the generating function and suggests that the OGF may vary based on the starting index of the series.

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Punkyc7
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Find the OGF for the recurrence

a_{n}= 6 * a_{n-1}+ a_{n-2} a_{0}=2, a_{1}=1
So here is what I did

I said let A = \sum_{2>=n}a_{n}x^{n}then I got

A = 6x (A+x) + x^{2}(A +x+2)

which gets me

A= \frac{6x^2+x^{3} +2x}{1-6x - x^2}
ButI should get \frac{2-x}{1-6x - x^2}Can anyone tell me what I am doing wrong ?
 
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Punkyc7 said:
Find the OGF for the recurrence

a_{n}= 6 * a_{n-1}+ a_{n-2} a_{0}=2, a_{1}=1

So here is what I did

I said let A = \sum_{2>=n}a_{n}x^{n}

then I got

A = 6x (A+x) + x^{2}(A +x+2)  = A(6x) + 6x2 + A(x2) + x3 +2x2
This gets you a different A.
which gets me

A= \frac{6x^2+x^{3} +2x}{1-6x - x^2}

ButI should get \frac{2-x}{1-6x - x^2}

Can anyone tell me what I am doing wrong ?

Congrats on your 400th post ! :smile:
 
Thank you on the congratulations. But I do not see how that get me a different A. Did I define what A is right. Usually the generating function start at zero but with this problem that gives you bad subscripts.
 
Punkyc7 said:
Find the OGF for the recurrence

a_{n}= 6 * a_{n-1}+ a_{n-2} a_{0}=2, a_{1}=1



So here is what I did

I said let A = \sum_{2>=n}a_{n}x^{n}


then I got

A = 6x (A+x) + x^{2}(A +x+2)

which gets me

A= \frac{6x^2+x^{3} +2x}{1-6x - x^2}



ButI should get \frac{2-x}{1-6x - x^2}


Can anyone tell me what I am doing wrong ?

I usually find it safer, to write things out a bit more:
A = a_2 x^2 + a_3 x^3 + a_4 x^4 + \cdots = (6a_1 + a_0)x^2 <br /> + (6a_2 + a_1)x^3 + (6a_3 + a_2)x^4 + \cdots\\<br /> = (6+2)x^2 + x^3 + 6x A + x^2 A = x^3 + 8x^2 + (x^2 + 6x)A.
This will give you a different A from what you obtained.

Besides that difference, maybe the answer was for a GF starting at a_0*x^0, or at a_1*x^1. However, I checked that, and could not get the posted answer for any three variants of the problem. Just to be sure, I used Maple to get a solution, and got an answer in agreement with a_0 + a_1 x + A. Here is the Maple command, but I won't give the output, since that would deprive you of the fun of getting it yourself.
> rsolve({a(n)=6*a(n-1)+a(n-2),a(1)=1,a(0)=2},a,'genfunc'(x));

RGV
 

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