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Topic: Recurrence Relations
Replies: 4   Last Post: Apr 3, 2012 2:42 AM

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Angela Richardson

Posts: 42
From: UK
Registered: 6/22/11
Re: Recurrence Relations
Posted: Mar 29, 2012 4:47 AM
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att1.html (1.4 K)

f(n+2)mod7=f(n+1)mod7 +f(n)mod7
so the sequence begins 1,2,5,6,6,0,6,1,0,1 and after that is 1,2 so the sequence must repeat in cycles of 10 from there. 2009=9mod10 so f(2009)mod7=f(9)mod7=1


________________________________
From: Toni <discussions@mathforum.org>
To: discretemath@mathforum.org
Sent: Thursday, 29 March 2012, 8:04
Subject: Re: Recurrence Relations

Thank you very much Wouter.

Do not worry about my teacher: she just loves me.

I wonder if the result could be deduced only from the recurrence relation without the aid of a computer or a calculator.



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