OFFSET
1,2
COMMENTS
Start with a point marked on an axis at unit distance from the origin. How many distinct points can be marked on the axis, if the offset from the origin of each additional point is a positive integer multiple or submultiple of that of a point already marked, and no less than unit distance nor greater than a given maximum, d? The possible finite answers, which occur only if d < 27/5, are the numbers in this sequence.
If we think of an interval with lower bound 1 as growing in time, equivalence classes of cardinality a(n) appear when the interval reaches [1, r_min(n)] for a rational r_min(n) whose numerator and denominator are 5-smooth, and disappear when the interval reaches [1, r_max(n)) where r_max(n) = (r_min(n+1))^2/r_min(n).
The initially known sequence terms (n <= 20) have the following property: a(n) = 2 * a(n-1) or, for some i < n, a(n) = 2 * a(n-1) - a(i); the former case occurring if r_min(n) equals a prime number, the latter occurring if and only if r_min(n) = r_max(i). Having identified underlying reasons that look like the basis of a proof, the author conjectures this is true for all n > 1.
Infinite equivalence classes occur if the interval is a scaled image of an interval containing (1, 27/5). Intervals that are scaled images of an interval containing [1, 50/9) or (1, 50/9] have no finite equivalence classes; for other intervals, the smallest equivalence class has cardinality 1, 2, 4 or 12 (cf. A215795).
The 5-smooth rationals in the critical semi-open interval [1, 50/9) are therefore partitioned into infinite equivalence classes. I believe these classes can be numbered, ordered by the power(s), k, of 5, for which 5^k*2^j is in the class for at least one integer j. 8/5, 1/1 and 5/1 are in class 0; 25/8 and 125/64 are in class 1; 32/25 and 128/125 are in class -1. If we close the critical interval, the inclusion of 50/9 unites classes 0 and 1 as defined above. If an interval, I, (or any scaled image of I) is a proper superset of the closed interval [1, 50/9], I am confident all 5-smooth rationals in I are in the same equivalence class. - Peter Munn, Jun 29 2022
From Peter Munn, Sep 16 2022: (Start)
The linked figure, "The disappearance of finite classes", shows various interval parts of the interval [1, 27/5] as rectangles. The arrowed lines between the rectangles represent the relation R. It can be seen as summarizing a graph between uncountably many points.
There are 12 rectangles around the circumference annotated as representing points that are on a finite path. These correspond to the points (equivalently real numbers) that are in equivalence classes of cardinality 12 -- the only finite equivalence classes when the interval is large enough to have infinite equivalence classes.
If we animate this figure to show the upper bound of the interval increasing from 27/5, the 12 outer rectangles would shrink. Consider, in particular, the one on the right marked with bounds 27/25 and 10/9. 10/9 derives from (((1 * 5) / 3) * 2) / 3, a progression that can followed around the inner edge of the circuit in the figure. 10/9 stays fixed. 27/25 grows as 1/5 of the upper bound, and the finite classes/paths disappear when the upper bound reaches 50/9 = 5 * 10/9.
(End)
LINKS
Peter Munn, The disappearance of finite classes
Eric Weisstein's World of Mathematics, Closure
Eric Weisstein's World of Mathematics, Equivalence Class
Wikipedia, Transversal (combinatorics)
FORMULA
a(n) = |S_n| where S_n = S_(n-1) U {min( {2*x, 3*x, 5*x : x in S_(n-1)} \ S_(n-1)) / y : y in S_(n-1)} n >= 2, S_1 = {1.0}.
EXAMPLE
The real interval [1.0, 2.0] has equivalence classes including {1.5} and {1.0, 2.0}, so |{1.5}| = 1 and |{1.0, 2.0}| = 2 are in the sequence.
To establish whether 3 is in the sequence, let x be the smallest member (by absolute value) of an equivalence class E on an interval I. E has cardinality greater than 1 only if 2*x is in I. If 3*x is not in I, there are no other multiples or submultiples of {x, 2*x} in I, so E has cardinality less than 3. Otherwise 3*x is in I, 3/2*x is in I, so {x, 2*x, 3*x, 3/2*x} is a subset of E and E has cardinality greater than 3. So 3 is not in the sequence.
Equivalence classes of cardinality 7 take the form {x, 2*x, 3*x, 3/2*x, 4*x, 4/3*x, 8/3*x} for real x <> 0.
Equivalence classes of cardinality 12 take the form {x, 2*x, 3*x, 3/2*x, 4*x, 4/3*x, 8/3*x, 9/2*x, 9/4*x, 9/8*x, 27/8*x, 27/16*x} for real x <> 0.
In the table below the columns are as follows. |E|: cardinality of equivalence class; I_min: smallest interval with lower bound 1 that has such an equivalence class; I_max: largest interval with lower bound 1 that has such an equivalence class; u_max: upper bound of I_max in decimal notation.
|E| I_min I_max u_max
1 [1, 1] (1, 4) 4.0
2 [1, 2] (1, 9/2) 4.5
4 [1, 3] (1, 16/3) 5.3333333333...
7 [1, 4] (1, 81/16) 5.0625
12 [1, 9/2] (1, 50/9) 5.5555555555...
24 [1, 5] (1, 6561/1280) 5.12578125
41 [1, 81/16] (1, 531441/102400) 5.189853515625
58 [1, 6561/1280] (1, 3^16/(2^16*5^3)) 5.2547266845...
...
214 [1, 16/3] (1, 2^52*5^8/3^43) 5.3592727015...
...
oo (1, 27/5) not defined not defined
PROG
(PARI)
{ my (points = Set([1]), newmax);
print (1, " ", matsize(points)[2]);
for (n=2, 25,
newmax = vecmin(setminus(setbinop((x, y)->x*y, [2, 3, 5], points), points));
points = setunion(setbinop((x, y)->x/y, [newmax], points), points);
print (n, " ", matsize(points)[2]); )} \\ Peter Munn, Jun 29 2022
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Peter Munn, Oct 04 2019
EXTENSIONS
a(21)-a(25) from Peter Munn, Jun 29 2022
STATUS
approved