# Talk:Sequence: nth number with exactly n divisors

## Handy hints[edit]

Handy optimization hints: Terms in position n that is a prime number are **always** the nth prime raised to the (n-1)th power. E.G.

# 1 2 3 4 5 6 7 8 9 10 11 primes 2 3 5 7 11 13 17 19 23 29 31

Term 7 is 17^6 == 24137569. Term 11 is 31^10 == 819628286980801.

and so on.

Non prime odd terms are **always** a square number. --Thundergnat (talk) 18:53, 11 April 2019 (UTC)

*All*odd terms are**always**a square number. -- Gerard Schildberger (talk) 01:02, 12 April 2019 (UTC)

- You are 100% correct, but these were meant to be optimization hints, not statements of general fact. Sort of an either / or kind of thing. --Thundergnat (talk) 09:59, 12 April 2019 (UTC)

- That is one
~~hel~~heck of some dandy hints!! -- Gerard Schildberger (talk) 18:56, 11 April 2019 (UTC)

May this is a naive question: Why 2nd term of sequence is not 2 but 3.

2 has exactly 2 divisors: 1 and 2 (CalmoSoft)

- 2 is the
**first**number with exactly 2 divisors but 3 is the**second**such number. So, 3 is the second term of the sequence. --PureFox (talk) 13:48, 8 March 2021 (UTC)

- Thanks PureFox, now I understand it (CalmoSoft)