# Thread: Recurrence Relations: Constraint Conditions

1. ## Recurrence Relations: Constraint Conditions

Consider the recurrence relation $b_n = 3b_{n-1} + 10b_{n-2}, b_0 = x. b_1 = y$
Under what conditions on x and y are the constants in the general solution to the recurrence integers?

Assume this is correct:

$b_n = C_1 * (5)^n + C_2 * (-2)^n$
then

$b_0 = x = C_1 + C_2$
$b_1 = y = 5C_1 - 2C_2$

I have figured out by trial and error that
x and y can be any integer that when modded with 7 is congruent to 0.
or
If x and y are divided by 7 and they leave no remainder,
then that will produce the constants $C_1$ and $C_2$

Can someone please explain to me why this happens?

2. Originally Posted by Celcius
Consider the recurrence relation $b_n = 3b_{n-1} + 10b_{n-2}, b_0 = x. b_1 = y$
Under what conditions on x and y are the constants in the general solution to the recurrence integers?

Assume this is correct:

$b_n = C_1 * (5)^n + C_2 * (-2)^n$
then

$b_0 = x = C_1 + C_2$
$b_1 = y = 5C_1 - 2C_2$

I have figured out by trial and error that
x and y can be any integer that when modded with 7 is congruent to 0.
or
If x and y are divided by 7 and they leave no remainder,
then that will produce the constants $C_1$ and $C_2$

Can someone please explain to me why this happens?
Have you tried solving the simulate nous equations??

3. Originally Posted by Drexel28
Have you tried solving the simulate nous equations??
Sure, sort of.

$C_2 = (-y + 5x)/7$
and
$C_1 = (y + 2x)/7$

I kind of see something there, but still not 100% sure.