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May 16

What would a graph of integers vs the percent of the next 1000 integers that are non-tautologically figurate look like?

Counting polygonal numbers, centered polygonal numbers, Platonic solid numbers, centered Platonic solid numbers, regular pyramidal numbers, centered regular pyramidal numbers and sure why not maybe also the bipyramidal and prism analogs of those pyramids (with n copies of the nth k-gonal number stacked). Sagittarian Milky Way (talk) 17:04, 16 May 2024 (UTC)[reply]

100% of all integers are polygonal (in at least in 2 ways that I can think of from the top of my head) and therefore figurate numbers. Perhaps you meant integers that are figurate numbers in non-trivial ways. Dhrm77 (talk) 18:32, 16 May 2024 (UTC)[reply]

Right, it'd have to be ones that are also figurate in non-trivial ways. Sagittarian Milky Way (talk) 20:44, 16 May 2024 (UTC)[reply]

Denoting the function giving the fraction of non-trivial figurates by

f,

we have

f(n)\sim n^{-{\frac {1}{2}}}

as

n

grows, but when

n

gets very large, in the millions, the proportionality breaks down, There will be increasingly often no figurate numbers at all among the next 1000 integers; either

f(n)\geq 0.001

or

f(n)=0.

--Lambiam 05:06, 17 May 2024 (UTC)[reply]

Of course, every positive integer n is both the nth 2-gonal number and the 2nd n-gonal number. GTrang (talk) 14:12, 21 May 2024 (UTC)[reply]

Truncated square tiling and lines through

I was looking at bathroom tile in Truncated square tiling. Am I correct that for a line passing through opposite vertices in one octagon, that it never passes through another vertex in the tiling? Naraht (talk) 21:06, 16 May 2024 (UTC)[reply]

This is one of those problems which is not hard to solve in theory, but it's so easy to make a mistake in calculation that the result shouldn't be trusted on the first attempt. But according to my calculations the line passing through two opposite corners of an octagon will pass through an additional two vertices. You can enumerate the vertices as ((1+√2)k+a,(1+√2)l+b) where (a,b) is one of:

(1/√2, 0), (0, 1/√2), (0, -1/√2), (-1/√2, 0).

The line through (0, 1/√2) and (1+1/√2, 1+√2) is given by y=(1+√2)(x-1/√2). There are four vertices which lie on this line, the (1/√2, 0) and (1+1/√2, 1+√2) we started with, plus (0, -1-1/√2) and (1+√2, 2+3/√2). You can easily verify that the points satisfy the equation of the line, and the values of k, l, a and b for the points are:

(1/√2, 0): k=0, l=0, a=1/√2, b=0

(1+1/√2, 1+√2): k=1, l=1, a=-1/√2, b=0

(0, -1-1/√2): k=0, l=-1, a=0, b=1/√2

(1+√2, 2+3/√2): k=1, l=2, a=0, b=-1/√2

You can prove these are the only four points. For a specific pair (a, b) the equation (1+√2)k+a=(1+√2)((1+√2)l+b-1/√2) is linear in k and l. But since √2 is irrational, you can set coefficients of 1 and √2 equal to each other and obtain two equations in two unknowns. This is a non-degenerate system in this case and so the solution is unique. (For vertical, horizontal and diagonal lines, i.e. m=±1, the system is degenerate and so there will be either no solutions or an infinite number.) --RDBury (talk) 01:25, 17 May 2024 (UTC)[reply]