hi all,
Is there a way to find a formula for the nth element of a linear recurrence
? Instead of going linearly one by one
c1, c2, c3 are the three constants.
a(n+3) = c1*a(n+2) - c2*a(n+1) + c3*a(n)
thanks.
shady
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yes,u can use matrix exponentiation
On Sun, Jan 9, 2011 at 3:49 PM, shady sinv...@gmail.com wrote:
hi all,
Is there a way to find a formula for the nth element of a linear recurrence
? Instead of going linearly one by one
c1, c2, c3 are the three constants.
a(n+3) = c1*a(n+2) -
