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a(n+1)=(15-21a(n))/(20a(n)+34)

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Subject: a(n+1)=(15-21a(n))/(20a(n)+34)
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by: carlo berlingn - Sun, 11 Jun 2023 17:41 UTC

Re: a(n+1)=(15-21a(n))/(20a(n)+34)

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From: inva...@invalid.invalid (Learning math)
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Subject: Re: a(n+1)=(15-21a(n))/(20a(n)+34)
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by: Learning math - Mon, 12 Jun 2023 18:51 UTC

"carlo berlingn" <carlo.numerik@t-online.de> wrote in message
news:39ce12ed-f5d0-46d0-84b8-fb2e090a0693n@googlegroups.com...
> How can I find the explicit form for this recursion
> Carlo

No idea, but I'd start here just to get an idea of where I might be going.
https://www.wolframalpha.com/input?i=a%28n%2B1%29%3D%2815-21a%28n%29%29%2F%2820a%28n%29%2B34%29
And I'd probably stop there depending on my enthusiasm for learning a method
able to find a solution of that complexity.

Otherwise I'd go here
https://www.google.com/search?q=Recurrence+equation
Or here
https://www.google.com/search?q=Difference+equation
to see if I can find either an example of a means of solving the problem or
a reference to a book which explains how to solve it.

Re: a(n+1)=(15-21a(n))/(20a(n)+34)

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From: chris.m....@gmail.com (Chris M. Thomasson)
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Subject: Re: a(n+1)=(15-21a(n))/(20a(n)+34)
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by: Chris M. Thomasson - Mon, 12 Jun 2023 21:19 UTC

On 6/11/2023 10:41 AM, carlo berlingn wrote:
> How can I find the explicit form for this recursion
> Carlo

a(n+1)=(15-21a(n))/(20a(n)+34)

i[0] = 666 /* init the recursion, so to speak... */
i[1] = (15 - 21 * i[0]) / (20 * i[0] + 34)
i[2] = (15 - 21 * i[1]) / (20 * i[1] + 34)
i[3] = (15 - 21 * i[2]) / (20 * i[2] + 34)
....

Two wrights don't make a rong, they make an airplane. Or bicycles.

tech / sci.math / a(n+1)=(15-21a(n))/(20a(n)+34)

Subject	Author
a(n+1)=(15-21a(n))/(20a(n)+34)	carlo berlingn
Re: a(n+1)=(15-21a(n))/(20a(n)+34)	Learning math
Re: a(n+1)=(15-21a(n))/(20a(n)+34)	Chris M. Thomasson