What would it look like if we used PCA rather than UMAP? PCA is simpler than UMAP, so it's in some sense "less arbitrary". If the image is similar then we know we're seeing something about numbers rather than about our methods.
What would it look like if we used PCA rather than UMAP? PCA is simpler than UMAP, so it's in some sense "less arbitrary". If the image is similar then we know we're seeing something about numbers rather than about our methods.
What does this mean?
Also, I don't think that the (very legitimate) "dimensional" critique of PCA applies here. The units on the coordinates of the representation are the same: the presence or absence of that prime factor.
To the original question: my suspicion is that PCA might pull out the even numbers (first PC) and the divisible by 3 numbers (second PC), because these two factors may explain the most variability in the underlying vector representation. If it did, that would be pretty intuitive, although not as interesting.
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Edited to add: Suspicion turned out to be true. For the first 2000 integers, the top 6 PCs turned out to correspond to the first 6 primes (2, 3, 5, 7, 11, 13).
Plot at: https://imgur.com/a/qi2Sx5u?
function [nums,pcs]=pca_prime(nMax,nPC)
nums = zeros(nMax, nMax);
for k = 2:nMax,
nums(k,factor(k)) = 1; % vector representation of "k"
end;
% 2:end because don't care about 1 as a "prime"
pcs = pca(nums(2:end,:), 'NumComponents', nPC);
-- [nums,pcs]=pca_prime(2000,10); % "svd" would work too
plot(pcs(:,1:6)); % first 6 PCsThose are the same thing. Is your first case just a reader who has no idea what the SVD is?
floor(n / (p[i]*p[j])) / n - floor(n / p[i]) * floor(n/p[j]) / n^2
and the ith diagonal entry will be floor(n / p[i]) / n - ( floor(n / p[i]) / n )^2
for n large, you approximately get a diagonal matrix with diagonal entries / eigenvalues 1/p[i] - 1/p[i]^2.