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Problem 561


Problem 561


Divisor Pairs

Let S(n) be the number of pairs (a,b) of distinct divisors of n such that a divides b.
For n=6 we get the following pairs: (1,2), (1,3), (1,6), (2,6) and (3,6). So S(6)=5.
Let pm# be the product of the first m prime numbers, so p2# = 2*3 = 6.
Let E(m, n) be the highest integer k such that 2k divides S((pm#)n).
E(2,1) = 0 since 20 is the highest power of 2 that divides S(6)=5.
Let Q(n)=i=1nE(904961,i).
Q(8)=2714886.

Evaluate Q(1012).


约数对

记S(n)为n的不同约数组成的约数对(a,b)的数目,其中a整除b。
对于n=6,我们有如下约数对:(1,2),(1,3),(1,6),(2,6)和(3,6)。因此S(6)=5。
pm#是前m个素数的乘积,因此p2# = 2*3 = 6。
记E(m, n)为使得2k整除S((pm#)n)的最大整数k。
已知E(2,1) = 0,因为20是能够整除S(6)=5的2的幂中最大的数。
Q(n)=i=1nE(904961,i)
已知Q(8)=2714886。

Q(1012)


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