Masking in Japan, and Dec 09
Dec. 9th, 2025 09:17 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
mindstalkSo, you know that Japanese people mask more than Americans or Europeans. But how much more? Some numbers from today: ( Read more... )
So, 40-50% generically outside, and 50-75% on trains. On the "masks and exercise" front, I'd note that many bicyclists have been masked, too.
Further, almost but not entirely all of customer-facing employees have been masked. Train, bus drivers, retail shop employees, the few waiters I've seen. I'd say at least 80% conservatively, 90% likely, maybe not much higher (it takes few outliers to push a ratio away from extremes.) I think Seki said that waiters often aren't, but I dunno.
Now, is this the New Normal after covid? Not necessarily. Japan has been having a bad flu season, huge spike in cases, and a major strain (coming soon to a school or hospital near you) wasn't in the vaccine this year, so I think the government has been urging people to mask again. Also it's winter-ish and some people here may have noticed "masks are like a scarf but better."
( Read more... )
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distinct real numbers, one can find a subsequence of length 
which is either increasing or decreasing; and that one cannot improve the 
, by considering for instance a sequence of 
blocks of length 
can be decomposed into the union of 
be a sequence of distinct numbers, determine 
“.
, this is an explicit piecewise linear function of the variables 
. The day the problem was posted, Desmond Weisenberg proposed studying the quantity 
, defined as the largest constant such that 
coins for some large 
. She then passes the piles to Bob, who is allowed to select a monotone subsequence of the piles (in the weak sense) and keep all the coins in those piles. What is the largest fraction 
the problems can easily be seen to be equivalent.)
, clearly 
: Alice has to put all the coins into one pile, which Bob simply takes. Similarly 
: regardless of how Alice divides the coins into two piles, the piles will either be increasing or decreasing, so in either case Bob can take both. The first interesting case is 
. Bob can again always take the two largest piles, guaranteeing himself 
of the coins. On the other hand, if Alice almost divides the coins evenly, for instance into piles 
for some small 
, then Bob cannot take all three piles as they are non-monotone, and so can only take two of them, allowing Alice to limit the payout fraction to be arbitrarily close to 
.
. It is also easy to see that 
. Less than an hour after that, 
, and posed the problem of determining the asymptotic limit of 
as 
, given that this was now known to range between 
and 
. This version was 
, which would also imply that 
as 
from the original Erdős-Szekeres theorem by a standard “blow-up” argument which we can give here in the Alice-Bob formulation. Take a large 
, and replace each pile of 
new piles, each of size 
, chosen so that the longest monotone subsequence in this collection is 
. Among all the new piles, the longest monotone subsequence has length 
. Applying Erdős-Szekeres, one concludes the bound 
, and applying Cauchy-Schwarz, one obtains 
for all 
) with a small a value of 
as possible. After an hour of run time, AlphaEvolve produced the following upper bounds on 
, with some intriguingly structured potential extremizing solutions: 
and 
. After a bit of effort, he also produced an explicit upper bound construction:
. 
of numbers clustered around the “red number” 
and “blue number” 
, consisting of 
“blue” numbers, followed by 
be the least number such that, whenever one packs 
into a square of sidelength 
, with all sides parallel to the coordinate axes, one has 
, let 
be the maximal sum over all increasing subsequences ending in 
be the maximal sum over all decreasing subsequences ending in 
, we have either 
(if 
) or 
(if 
). In particular, the squares 
and 
are disjoint. These squares pack into the square ![{[0, S(x_1,\dots,x_n)]^2}](https://s0.wp.com/latex.php?latex=%7B%5B0%2C+S%28x_1%2C%5Cdots%2Cx_n%29%5D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)
, so by definition of 
, we have 
with 
, one has 
. It then suffices to show that if we pack the length 
by 
. Then we have a 
grid 
square, when restricted to this grid, becomes a discrete rectangle 
for some finite sets 
with 
we conclude that 
is at most 
, giving the claim. Similarly if it is the other sum. 
with 
. We can rescale so that the square one is packing into is ![{[0,k]^2}](https://s0.wp.com/latex.php?latex=%7B%5B0%2Ck%5D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)
. Thus, we pack 
squares of sidelength 
into 
![\displaystyle [0,k]^2 \subset [0,N]^2 \subset [0,N + |c| \frac{N}{N-\varepsilon}]^2.](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5B0%2Ck%5D%5E2+%5Csubset+%5B0%2CN%5D%5E2+%5Csubset+%5B0%2CN+%2B+%7Cc%7C+%5Cfrac%7BN%7D%7BN-%5Cvarepsilon%7D%5D%5E2.&bg=ffffff&fg=000000&s=0&c=20201002)
![{[0,N]^2 \backslash [0,k]^2}](https://s0.wp.com/latex.php?latex=%7B%5B0%2CN%5D%5E2+%5Cbackslash+%5B0%2Ck%5D%5E2%7D&bg=ffffff&fg=000000&s=0&c=20201002)
by 
unit squares. We can similarly pack ![\displaystyle [0,N + |c| \frac{N}{N-\varepsilon}]^2 \backslash [0,N]^2](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5B0%2CN+%2B+%7Cc%7C+%5Cfrac%7BN%7D%7BN-%5Cvarepsilon%7D%5D%5E2+%5Cbackslash+%5B0%2CN%5D%5E2&bg=ffffff&fg=000000&s=0&c=20201002)
![\displaystyle [0, \frac{N}{N-\varepsilon} (N+|c|-\varepsilon)]^2 \backslash [0, \frac{N}{N-\varepsilon} (N-\varepsilon)]^2](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%5B0%2C+%5Cfrac%7BN%7D%7BN-%5Cvarepsilon%7D+%28N%2B%7Cc%7C-%5Cvarepsilon%29%5D%5E2+%5Cbackslash+%5B0%2C+%5Cfrac%7BN%7D%7BN-%5Cvarepsilon%7D+%28N-%5Cvarepsilon%29%5D%5E2&bg=ffffff&fg=000000&s=0&c=20201002)
squares of sidelength 
. All in all, this produces 
, 
should behave like a gaussian random variable with mean and variance both close to 
; but the question of the joint distribution of consecutive values such as 
is still only partially understood. Aside from some lower order correlations at small primes (arising from such observations as the fact that precisely one of 
will be divisible by 
), the expectation is that such consecutive values behave like independent random variables. As an indication of the state of the art, it was recently 
, 
will be asymptotically decorrelated in the limit 
of the standard deviation, but does not resolve finer-grained information, such as precisely estimating the probability of the event 
.
, such a bound is already to be expected (though not completely universal) from the Hardy–Ramanujan law; the main difficulty is thus with the short shifts 
. If one only had to demonstrate this type of bound for a bounded number of 
“almost prime” in the sense that 
becomes bounded. Thus the problem is that the “sieve dimension” 
grows (slowly) with 
for all 
for a large 
), and then performing a moment calculation analogous to the standard proof (due to Turán) of the Hardy–Ramanujan law, but weighted by the Maynard sieve. (In order to get good enough convergence, one needs to control fourth moments as well as second moments, but these are standard, if somewhat tedious, calculations).
, but in view of the Hardy–Ramanujan law, the 
digit of this number is influenced by about 
nearby values of 
, thus 
, we obtain some relations between shifts 
: 
: 
, and arrive at an identity of the form 
is a parameter (we eventually take 
) and 
are various shifts that we will not write out explicitly here. This looks like quite a messy expression; however, one can adapt proofs of the Erdős–Kac law and show that, as long as one ignores the contribution of really large prime factors (of order 
, say) to the 
, that this sort of sum behaves like a gaussian, and in particular once one can show a suitable local limit theorem, one can contradict 
, which is an error that overwhelms the information provided by 
. Here it is important that the error estimates of Pilatte are quite strong (of order 
); previous correlation estimates of this type (such as those used in 
and 
). Heuristic arguments led 
. They were able to establish an upper bound of 
, while Hildebrand obtained a lower bound of 
, due to 
(the limitation here is the standard one, which is that the current technology on pairwise correlation estimates either requires logarithmic averaging, or is restricted to almost all scales rather than all scales). Roughly speaking, the idea is to use the circle method to rewrite the above density in terms of expressions 
, use the estimates of Pilatte to handle the minor arc 
and 
of increasing natural numbers can grow if one constrains its translates 
to interact with the set 
of squarefree numbers in various ways. For instance, Erdős defined a sequence 
to have “Property 
” if each of its translates 
in finitely many points. Erdős believed this to be quite a restrictive condition on 
for all sufficiently large 
and any specified function 
that tends to infinity as 
. For instance, one can find a sequence that grows like 
. The density zero claim can be proven by a version of the Maier matrix method, and also follows from known moment estimates on the gaps between squarefree numbers; the latter claim is proven by a greedy construction in which one slowly imposes more and more congruence conditions on the sequence to ensure that various translates of the sequence stop being squarefree after a certain point.
, which asserts that for infinitely many 
of 
are square-free. Since the squarefree numbers themselves have density 
, it is easy to see that a sequence with property 
for each 
). This requires a recursive sieve construction, in which one starts with an initial scale 
such that 
is squarefree for most of the squarefree numbers 
(and all of the squarefree numbers 
, but need not obey this property if it only grows like 
. This is achieved through further application of sieve methods.
is squarefree for all 
. Erdős writes, “In fact one can find a sequence which grows exponentially. Must such a sequence really increase so fast? I do not expect that there is such a sequence of polynomial growth.” Here our results are relatively weak: we can construct such a sequence that grows like 
, but do not know if this is optimal; the best lower bound we can produce on the growth, coming from the large sieve, is 
. (Somewhat annoyingly, the precise form of the large sieve inequality we needed was not in the literature, so we have an appendix supplying it.) We suspect that further progress on this problem requires advances in inverse sieve theory.
, asserts that there are infinitely many 
of elements of an admissible set ![{|{\mathcal SF} \cap [x]|}](https://s0.wp.com/latex.php?latex=%7B%7C%7B%5Cmathcal+SF%7D+%5Ccap+%5Bx%5D%7C%7D&bg=ffffff&fg=000000&s=0&c=20201002)
of squarefree elements up to 
![\displaystyle |{\mathcal SF} \cap [x]| - \frac{6}{\pi^2} x \ll x^{1/2} \exp(-c \log^{3/5} x / (\log\log x)^{1/5}).](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle++%7C%7B%5Cmathcal+SF%7D+%5Ccap+%5Bx%5D%7C+-+%5Cfrac%7B6%7D%7B%5Cpi%5E2%7D+x+%5Cll+x%5E%7B1%2F2%7D+%5Cexp%28-c+%5Clog%5E%7B3%2F5%7D+x+%2F+%28%5Clog%5Clog+x%29%5E%7B1%2F5%7D%29.&bg=ffffff&fg=000000&s=0&c=20201002)
![{A(x) > |{\mathcal SF} \cap [x]|}](https://s0.wp.com/latex.php?latex=%7BA%28x%29+%3E+%7C%7B%5Cmathcal+SF%7D+%5Ccap+%5Bx%5D%7C%7D&bg=ffffff&fg=000000&s=0&c=20201002)
for all sufficiently large 
:
