
Re: Recurrence Relations
Posted:
Mar 29, 2012 4:47 AM



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.

