Talk:Sequence: nth number with exactly n divisors
From Rosetta Code
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
helheck 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)