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A way of drawing natural numbers

Definition

Define a function I(p) from the prime numbers to the integers so that p is the I(p)-th prime. Thus I(2)=1, I(3)=2, and I(7)=4. Define the drawing of any (positive) natural number n recursively as follows:

The drawing of n=1 is just a blank space.
For any other n with s distinct prime factors, the drawing of n is arranged around a horizontal bar divided by s-1 scores into s segments. The scores and the bar divide the space above and below the bar into 2s compartments. Arrange the s distinct prime factors of n in ascending order, and denote the k-th of these factors by B(k). In the k-th compartment above the bar (reading from left to right), place the drawing of the number of times B(k) appears in the prime factorization of n. In the k-th compartment below the bar, place the drawing of I(B(k)) - I(B(k-1)). (For the 1st compartment, pretend that I(B(0))=0).

Thus no two numbers have identical drawings, and also anything which looks like a drawing (a configuration as defined below) corresponds to a unique number n.

A simple C++ program can generate such drawings. The output for 163 and 989 appears at the top of this page. Here are some more examples:

Configurations

Call any set S of nonintersecting, possibly scored, horizontal bars in the plane a configuration if

S is empty, or
S contains a longest bar B such that:
- B extends farther to the left and right than any other bar in S.
- Each score in B is such that a vertical line drawn through the score would intersect no other bar in S. Thus B and its scores cleanly separate the other bars into different compartments.
- The subset of S in each of these compartments is a configuration.

Then the drawing of any natural number is a configuration, and also any configuration is the drawing of a unique number n.

Numerical discussion

Let L(n) denote the number of bar segments in the drawing of n. Then L(1)=0, L(2)=1, L(3)=2, and so on. Let P(n) denote the number of positive integers k for which L(k)=n. Then P(0)=1, P(1)=1, and so on. Note that L(65536)=4 even though P(4) is only 55. To calculate P(n) for n > 1, imagine using n segments to construct a configuration. How many ways can we combine the segments? Take any three configurations A, B, and C where L(A) + L(B) + L(C) = n-1. Draw a single segment in the plane. Place A and B in the compartments above and below this segment, respectively. Attach C to the right end of this segment so that C's longest bar meets it. The result is a configuration with n segments. Clearly any nonempty configuration can be constructed in this way, and so P(n) satisfies this recurrence relation:

P(n) = sigma ( P(a) * P(b) * P(c) )

where "sigma" stands for a summation taken over all nonnegative integers a, b, and c for which a+b+c = n-1. And thus, as Simon Plouffe pointed out to me,

P(n) = binomial(3n,n)/(2n+1)

where "binomial" stands for the binomial coefficient.

Let S(n) denote the smallest k such that L(k)=n. Then S(0)=1, S(1)=2, and so on. I don't know a simple formula for S(n).

This page is kept by me.