| Enumeration of Features |
Robert P. Munafo, 2007 Jan 27.
Mu-Atoms
We start by considering the lemniscates as polynomials set equal to zero, and solving them, i.e. "finding the roots". Each root gives the nucleus of some mu-atom. Since the nth lemniscate is of order 2n-1, it follows that for each period n there are 2n-1 roots.
However, a root of period 3 (for example) also shows up as a root of period 6, 9, etc. but we don't count it that way. It only counts as a root of period 3. So, from 2n-1 we must subtract the number of roots of all lower periods that are divisors of n. This gives the following formula for the number of Mu-Atoms of period n:
Na(n) = 2n-1 - Σ(f:(f<n) and (n≡0 mod f))Na(f)
where n≡0 mod f means "f is a divisor of n". The extra condition f < n is there to exclude f equal to n.
For example, when computing N6 we start with 25, and then subtract the values of N1, N2 and N3 because 1, 2 and 3 are divisors of 6.
Here are the first few values of the sequence:
N1=1, N2=1, N3=3, N4=6, N5=15, N6=27, N7=63, N8=120, N9=252, N10=495, N11=1023, N12=2010, N13=4095, N14=8127, N15=16365, N16=32640, N17=65535, N18=130788, N19=262143, N20=523770, N21=1048509, N22=2096127, N23=4194303, N24=8386440, N25=16777200, N26=33550335, N27=67108608, N28=134209530, N29=268435455, N30=536854005, N31=1073741823, ...
(This is Sloane's sequence A000740.)
Secondary Continental Mu-Atoms
There is one secondary continental mu-atom for each rational number between 0 and 1. The secondary continental mu-atoms of period n correspond to the rational numbers with n in the denominator. Because of this, the number of secondary continental mu-atoms is equal to Euler's Totient function:
υ(n) = n - Σ(f:gcd(n,f)=1)1
That is, take n and subtract 1 for every number that is relatively prime to n. This sequence is Sloane's sequence A000010. It starts:
0, 1, 2, 2, 4, 2, 6, 4, 6, 4, 10, 4, 12, 6, 8, 8, 16, 6, 18, 8, 12, 10, 22, 8, 20, 12, 18, 12, 28, 8, 30, ...
Because of period scaling, each mu-atom has children that have the same distribution as the secondary continental mu-atoms, but with scaled-up periods. From this, we can derive the formula for the number of continental mu-atoms of period n:
Nc(1) = 1 (special case)
Nc(n) = Σ(f:n≡0 mod f) [Nc(f) . υ(n/f)]
This sequence starts:
1, 1, 2, 3, 4, 6, 6, 9, 10, 12, 10, 22, 12, 18, 24, 27, 16, 38, 18, 44, 36, 30, 22, 78, 36, 36, 50, 66, 28, 104, 30, ...
The Islands
Next we consider the number of island mu-molecules. So far we have a total number of mu-atoms, and a number of continental mu-atoms, for each period. All mu-atoms that are not continental are part of an island.
However, some of these mu-atoms are descendants of the seeds of the island mu-molecules. So, first we have to look at the relation between descendants in general (which includes continental descendants) and mu-molecule seeds (including the continent seed).
The number of mu-molecules is equal to the number of seeds, since each mu-molecule has one seed. Furthermore, every mu-atom is either a seed or a descendant, but not both. Therefore, we have this relation between mu-molecules and descendants:
Nm(n) = Na(n) - Nd(n)
Descendants
Number of descendants:
Nd(n) = Σ(f:n≡0 mod f) [Na(f).υ(n/f)]
The sequence starts:
0, 1, 2, 3, 4, 7, 6, 12, 12, 23, 10, 51, 12, 75, 50, 144, 16, 324, 18, 561, 156, 1043, 22, 2340, 80, 4119, 540, 8307, 28, 17521, 30, ...
Mu-Molecules
From this we get the values of Nm(n):
1, 0, 1, 3, 11, 20, 57, 108, 240, 472, 1013, 1959, 4083, 8052, 16315, 32496, 65519, 130464, 262125, 523209, 1048353, 2095084, 4194281, 8384100, 16777120, 33546216, 67108068, 134201223, 268435427, 536836484, 1073741793, ...
The functions in maxima
Here are each of the functions implemented in maxima.
See also Exploring, largest mu-atoms, largest islands, Phi(N).
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