Talk:Numbers whose count of divisors is prime: Difference between revisions
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m (Peak moved page Talk:Numbers which count of divisors is prime to Talk:Numbers whose count of divisors is prime: "which" is ungrammatical here, "whose" is acceptable and yields a similar title.) |
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That is only possible if all a..z are even.<BR> |
That is only possible if all a..z are even.<BR> |
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That means all n has to be m*m.<BR> |
That means all n has to be m*m.<BR> |
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Therefor only square numbers need to be tested. |
Therefor only square numbers need to be tested.[[User:Horsth|Horsth]] |
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:Brilliant, thanks! --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 15:42, 11 July 2021 (UTC) |
:Brilliant, thanks! --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 15:42, 11 July 2021 (UTC) |
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::I think, Nigel enjoys weekend :-) [[User:Horsth|Horsth]] 15:47, 11 July 2021 (UTC) |
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:::Yes, he's usually onto these like a flash :) --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 16:17, 11 July 2021 (UTC) |
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:Thanks, I was enjoying a weekend until 11pm on Sunday when a bunch of Italians spoiled it. Taking Horsth's analysis a step further, considering n with two Prime factors p<sub>1</sub><sup>a</sup>p<sub>2</sub><sup>b</sup> then (a+1) and (b+1) are factors of CoD(n) which is therefore not prime. This makes the solution set the Cartesian product of n and g mapped to n<sup>g-1</sup> where n is the set of primes and g is the set of of off odd primes. No need to factor, no need to determine primality.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 14:04, 13 July 2021 (UTC) |
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Latest revision as of 07:15, 7 September 2021
The question is: Which numbers got odd count of divisors
If you look at prime decompostion of a number like n = p1^a*p2^b*...pn^z,
than the count of divisors is CoD = (a+1)*(b+1)*(c+1)*....(z+1).
That is only possible if all a..z are even.
That means all n has to be m*m.
Therefor only square numbers need to be tested.Horsth
- Brilliant, thanks! --PureFox (talk) 15:42, 11 July 2021 (UTC)
- Thanks, I was enjoying a weekend until 11pm on Sunday when a bunch of Italians spoiled it. Taking Horsth's analysis a step further, considering n with two Prime factors p1ap2b then (a+1) and (b+1) are factors of CoD(n) which is therefore not prime. This makes the solution set the Cartesian product of n and g mapped to ng-1 where n is the set of primes and g is the set of of off odd primes. No need to factor, no need to determine primality.--Nigel Galloway (talk) 14:04, 13 July 2021 (UTC)