At Loggerheads With Log-recursions!

Number Theory Level 5

$\log_n (n T_n) = \log_{n+1} T_{n+1}$

Consider a sequence of positive integers given by $1 = T_2, T_3, T_4, \ldots$ that satisfy the above recurrence relation. What is the number of factors of $T_{100}$ ?

The answer is 38809.

1 solution

Anirudh Chandramouli
Sep 21, 2015

$log_{n} (nT_{n}) = log_{n+1}(T_{n+1})$

Therefore,

$1+log_{n}(T_{n}) = log_{n+1}(T_{n+1})$

Since $\log_2 T_2 = \log_2 1 = 0$ , it follows that $\log_n T_n = n - 2$ .

Thus, $T_{n} = n^{n-2}$ .

This gives us $T_{100} = 100^{98} = 2^{196} * 5^{196}$

Therefore the number of divisors is $(196+1)(196+1)=\boxed{38809}$

In problem it is given that $T_{2} = 1$ so $a_{2} = 0$ and hence $a_{n}$ cannot equal n

Also you wrote that $\sum_{I =2}^n(1) = n+1$ which I think should be $n-1$ Hence $a_{n+1} = n-1$ and hence $a_{100} = 98$

So $T_{100} = 100^ {98}$

so answer should be $(196+1)(196+1) = 38809$ Correct me if I'm wrong

Hargun Preet Singh - 5 years, 3 months ago

I completely agree since I got same answer as that of @Hargun Singh . Please look into it @Rohith M.Athreya

neelesh vij - 5 years, 3 months ago

see the reports section of the problem

ur doubts have been addressed.

Rohith M.Athreya - 5 years, 3 months ago

@Rohith M.Athreya – I don't think you've addressed the doubts in the reports section.

Judging by this solution, since $\log_2 T_2 = 0$ , it is not valid to claim that $\log_n T_n = n$ .

I've flagged this problem. Please clarify what it is you are asking for, otherwise, I believe that the answer is indeed 38809.

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – well yeah!!

u seem to be right. i must have made a mistake along the way.

sorry for the mistake

i got the same answer too.

Rohith M.Athreya - 5 years, 3 months ago

@Rohith M.Athreya – Thanks. I have updated the answer to 38809.

Calvin Lin Staff - 5 years, 3 months ago

At Loggerheads With Log-recursions!

The answer is 38809.

1 solution

0 pending reports