I'm not quite clear on how you defined p and q. Is p the remainder when n is divided by 3, and q = n - p?
a_n = (1/3)n^3 + n^2 - (1/3)n ... I am going to assume that this part is correct because it probably is. We are then left with proving that every term will be an integer. ...
where floor(n) = integer part of nWrite a function which, for an integer n, gives the number of trailing zeroes in n!.
int(n/5^1)+int(n/5^2)+int(n/5^3)+int(n/5^4)... I think.
bound = 4200000000;
s = 0;
While[Prime[Prime[2^(s + 1)]] < bound, s++];
x = 2^s;
For[t = s - 1, t ≥ 0, t--, If[Prime[Prime[x + 2^t]] < bound, x = x + 2^t]];
Print[Prime[x]]
Print[Prime[Prime[x]]]
Easy one: Deduce the (minimum) number of moves required to solve the Tower of Hanoi puzzle with 3 pegs and n discs. Harder one: By their nature, factorials tend to accumulate trailing zeroes. For example, 5! (120) has one trailing zero, 10! (3628800) has two trailing zeroes, 25! (15511210043330985984000000) has six trailing zeroes, etc. Write a function which, for an integer n, gives the number of trailing zeroes in n!.
There are only 5 trailing zeroes in 25! because 5 goes into it 5 times.
By their nature, factorials tend to accumulate trailing zeroes. For example, 5! (120) has one trailing zero, 10! (3628800) has two trailing zeroes, 25! (15511210043330985984000000) has six trailing zeroes, etc. Write a function which, for an integer n, gives the number of trailing zeroes in n!.
There are only 5 trailing zeroes in 25! because 5 goes into it 5 times. Not sure where your math went wrong as I don't know where to figure out the product without doing it longhand, and that's unnecessary.
f(0) = 0 f(n) = floor(n/5) + f(floor(n/5))
