# Thread: recurrence involving floors and square roots

1. ## recurrence involving floors and square roots

Hey,

Would appreciate any help on this one.

Solve the recurrence:

a(0) = 1;

a(n) = a(n-1) + floor(sqrt[a(n-1)]) ; for n >0.

This is exercise number 3.28 from Concrete mathematics. The solution at the back kinda makes a bit of sense....but i still cant grasp entirely how they've reached the solution.

Thanks

2. Originally Posted by schumi316
Hey,

Would appreciate any help on this one.

Solve the recurrence:

a(0) = 1;

a(n) = a(n-1) + floor(sqrt[a(n-1)]) ; for n >0.

This is exercise number 3.28 from Concrete mathematics. The solution at the back kinda makes a bit of sense....but i still cant grasp entirely how they've reached the solution.

Thanks
In general the formula is,
a_n=(1+2+...+n)+([\sqrt(1)]+[sqrt(2)]+...+[\sqrt(n)])
With some work I think this can be simplified.