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Let tau(n) = # of pos odd divisors of n
Let sigma(n) = sum of pos. odd divisors of n
Note that tau and sigma are both multiplicative functions.
1.) Evaluate the following:
tau(12)
tau(16)
sigma(12)
sigma(16)
2.) Let n = p_(1)^(a_1)*p_(2)^(a_2)*p_(3)^(a_3)* ... *p_(k)^(a_k)
a.) Determine a formula for tau(n) in terms of the p's and a's.
b.) Determine a formula for sigma in terms of the p's and a's
I see.
Factorize n>1,
n=p_1^a_1 * p_2^a_2 * ... * p_n^a_n
Say, p_1=2, the even part of the number in factorization.
You can choose 1 choice for a_1 because you want them all odd, hence a_1=0 is only possibility.
You can choose a_2+1 for second and so on.
Thus,
(a_2+1)(a_3+1)...(a_n+1)
Same thing as all but without first factor.