## Re: Simple question on "decades"

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• ... Thank you Phil, enlightening. Also to earlier contributors.
Message 1 of 6 , Dec 31, 2010
--- In primenumbers@yahoogroups.com, Phil Carmody <thefatphil@...> wrote:
>
> --- On Wed, 12/29/10, rupert.wood@... <rupert.weather@...> wrote:
> > I suspect this has been addressed
> > somewhere. But here goes, anyway.
> >
> > In asymptotic (ratio) terms, are the decades 10-19, 40-49,
> > 70-79, ... (which of course have no 3-divisors in the odd
> > numbers except "at 5") any more likely to contain more
> > primes than the other decades? I would guess that this is
> > not so.
>
> That depends on exactly what you're trying to ask. So I'll provide 2 answers, one for each interpretation.
>
> Firstly, assuming that you're isolating the 10s column in the base-10 representation, so looking at ...XY where you look at X=0..9 and compare.
>
> Let's say you have a number ending ...XY. You know nothing about its divisibility by 3, 7, 11, ... because you know nothing about the ... part. (However, you know everything about its divisibilty by 2 and 5 by looking at Y.) Taken over the set of all natural numbers ending ...XY, 1/3 of them will be divisible by 3, and 1/p of them will be divisible by prime p>5. That's independent of X. So there's nothing special at all about the 10 different sets.
>
> Secondly, assuming you're just clumping all naturals into intervals of size 10, and you're interested in how dense each of those intervals is likely to be.
>
> Given an arbitrary interval {10k+0 .. 10k+9}, and knowing nothing about it (i.e. knowing nothing about k), there's very little to say except that the expected density of primes is proportional to 1/log(10k) (with a constant of proportionality equal to 1). We know that half of them will be even, and a fifth of them will be divisible by 5, but the remaining 4 we still know very little about. We do know that they're not divisible by 2 or 5, so the expected density within those 4 will be (5/4)*(2/1)/log(10k). That's the baseline.
>
> However, in the decades where {10k+2, 10k+5, and 10k+8} are divisible by 3, the 4 candidates we're left with are known not to be divisible by 3, and therefore will have an expected density (3/2) times higher than the baseline.
>
> In the decade where {10k+1, 10k+4, 10k+7} are divisible by 3, two of our remaining 4 are removed. The surviving 2 will also have an expected density (3/2) times higher, but there's only 2 of them.
>
> In the decade where {10k+0, 10k+3, 10k+6 and 10k+9} are divisible by 3, again, 2 of the remaining 4 are removed.
>
> So the expected density in a third of the decades, the ones having 4 remaining candidates, is twice that of the other two thirds of the decades, namely the ones with only 2 candidates remaining.
>
> Side note - 4+2+2 = 8. 8=phi(2*5*3).
>
> A quick numerical verification:
>
> ? v=[0,0,0];forprime(p=0,50000,d=p\10;d3=d%3;v[d3+1]++);v
> [1281, 2562, 1290]
>
> As you proposed, but then rejected, the decade with ...5 divisible also by 3 (15, 45, 75, 105, ...) does indeed have a higher density of primes.
>
> low high low
> 0-9 10-19 20-29
> 30-39 40-49 50-59
> 60-69 70-79 80-89
> 90-99 100-109 110-120
> ...
>
> The same pattern appears if you arrange the decades in 7 columns, and ignore divisibility by 3. Some of the 7 (containing 70k+14, 70k+35, or 70k+56) will have 4 remaining candidates, and others (containing 70k+07, 70k+21, 70k+49, or 70k+63) will have 3 remaining. Quick verification:
>
> ? v=[0,0,0,0,0,0,0];forprime(p=0,100000,d=p\10;d7=d%7;v[d7+1]++);v
> %1 = [1197, 1603, 1198, 1605, 1191, 1599, 1199]
>
> 3 denser, 4 sparser with a ratio of 4:3 between them, exactly as predicted.
>
> 2nd side note - 3*4+4*3 = 24 = phi(2*5*7)
>
> If you want to take both 3 and 7 into account, then you'll be looking at 21 different columns, and as the 3 and 7 interact you'll find 4 different densities from those 21. You can verify that yourself in GP.
>
> A general lesson is that if you're looking at prime density, look at the residues modulo small primes ("small" I leave undefined here, it gets a little complicated sometimes), and that tells you pretty much everything you need to know.
>
> Phil
>

Thank you Phil, enlightening. Also to earlier contributors.
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