Divisors

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

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Quote:

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Originally Posted by ThePerfectHacker

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I believe this is the formula for reg tau(n) and reg sigma(n). I need to find it for the ODD divisors of n (number/sum, respectively for tau/sigma)

Quote:

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Originally Posted by flash101

I believe this is the formula for reg tau(n) and reg sigma(n). I need to find it for the ODD divisors of n (number/sum, respectively for tau/sigma)

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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.