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Topic: Weird leap-year rule (fwd)
Replies: 5   Last Post: Jan 4, 2013 1:17 PM

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Posts: 6
Registered: 11/10/12
Re: Weird leap-year rule (fwd)
Posted: Jan 4, 2013 1:17 PM
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On Fri, 04 Jan 2013 01:44:02 -0500, quasi wrote:

> Keith F. Lynch wrote:
>>quasi wrote:
>>>quasi wrote:
>>>>Keith F. Lynch wrote:
>>>>> Is there any (integer number of days) length, positive,
>>>>> negative, or zero, that a year can never have?

>>>> No, every integer length is possible.
>>Correct. Does every length occur infinitely many times?

> Yes, with two exceptions.
> The year y has 364 days iff y = 4
> The year y has 366 days iff y = 1
> All other year lengths occur infinitely often.
> quasi

That was my own conclusion as well.


Consider p^n, p an odd prime. It has n + 1 factors, all odd: 1, p, p^2,
..., p^n, or p^k as k ranges from 0 to n inclusive. There are infinitely
many distinct such p, so infinitely many distinct p^n for each n > 0 and so
infinitely many distinct y for which (odd factors - even factors) equals n
+ 1. So, infinitely many y for which (odd factors - even factors) is any
given integer greater than or equal to 2. That gives us infinitely many
occurrences of 367 and above.

Now, take 2p^n: doubling the number adds a second set of factors,
equinumerous to the first and all even resulting in infinitely many
distinct y for which odd minus even is zero. That gives us infinitely many
occurrences of 365.

Finally, consider 4p^n, which adds a third set of factors over and above
2^n, also equinumerous to the first and also all even. That gives us odd
minus even equal to any integer less than or equal to -2 and so every y
from 364 on down occurring infinitely often.


However, given a number with r odd factors and s even factors:

* Multiplying a number by an odd prime p not already a factor of it gives
a number with r doubled and s doubled, as each previous factor is joined
by a factor that is p times larger and has the same odd/even status and
is distinct from the previous factors and other new factors. Thus, this
move doubles r - s, either moving the length farther from 365 or leaving
it exactly on 365.
* More generally, the number np^k, p an odd prime not dividing n, has k + 1
times as many odd factors and k + 1 times as many even factors as n and
therefore gives either 365 or a length farther from 365 than n gives.
* n*2^k, n odd, has the (odd) factors of n, plus k times
as many *even* factors, so:
- If k = 1, the result is r = s.
- If k = 2, the result is s = 2r.
- More generally, s = k*r.
This means that the result again either gives a length of 365 or gives a
length at least as far away from 365 as n gave.

Consequently, multiplying numbers that give lengths outside {364,366} by
any additional relatively-prime factors gives lengths outside {364,366}.
So, the lengths we can get infinitely often depend entirely on what we get
with 1 and prime powers.

1 gives 366 due to having a lone, odd factor.
All odd prime numbers give 367 due to having just two factors, both odd.
Odd prime powers p^k, as worked out previously, give 366 + k (k > 0), so
367 or more.

The latter two are, therefore, unable to give us 364 or 366, themselves or
no matter what additional factors they're multiplied by. NO NUMBER WITH ANY

That leaves powers of 2, and we've already seen that 2^n gives 366-n, which
gets us 366 and 364 exactly once each. Thus, those occur only once each.
All the other lengths occur infinitely often.

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