Growth rate of geodesic balls ... AKA how many numbers can one reach after t steps?

Given that we allow multiplicands up to s, what is the number of numbers reached before t steps?

The following asks how to make numbers from things smaller than them:

In[]:=
MultiplicativeParts[n_]:=Flatten[Table[##]&@@@FactorInteger[n]]
In[]:=
AllMultiplierDecompositions[n_]:=Union[With[{m=MultiplicativeParts[n]},Catenate[Function[u,(Times@@@TakeList[u,#])&/@Catenate[Permutations/@IntegerPartitions[Length[m]]]]/@Permutations[m]]]]
In[]:=
AllMultiplierDecompositions[12]
Out[]=
{{12},{2,6},{3,4},{4,3},{6,2},{2,2,3},{2,3,2},{3,2,2}}
In[]:=
With[{max=40},Grid[Prepend[Transpose[Normal[SparseArray[Flatten[Table[Normal[KeyMap[{n,#}&,KeySort[Counts[Length/@AllMultiplierDecompositions[n]]]]],{n,max}]],Automatic,""]]],Range[max]],Frame->All,Background->{Automatic,{LightBlue,{None}}}]]
Out[]=
1
2
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5
6
7
8
9
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11
12
13
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35
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39
40
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
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2
1
2
4
2
2
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4
4
2
2
6
1
2
2
4
6
4
2
2
2
7
2
2
6
1
3
3
3
3
9
1
3
6
6
12
9
1
4
4
6
4
1
What this shows is how frequently the number will show up at a given step number (i.e. its path weight at that step number)

Prime will only show up after 1 step
Prime powers p^m will show up with a 1 at step m Binomial[m,t]

2nd row: is it number of times the number shows up in a 2D grid smaller than itself

E.g. for 20:
In[]:=
Count[Flatten[Table[ij,{i,19},{j,19}]],20]
Out[]=
4
In[]:=
Table[Count[Flatten[Table[ij,{i,n-1},{j,n-1}]],n],{n,30}]
Out[]=
{0,0,0,1,0,2,0,2,1,2,0,4,0,2,2,3,0,4,0,4,2,2,0,6,1,2,2,4,0,6}
Number of ways to form the number out of pairs of numbers.....
In[]:=
Grid[Table[ij,{i,11},{j,11}]/.12->Item[12,Background->LightRed]]
Out[]=
1
2
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6
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11
2
4
6
8
10
12
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16
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20
22
3
6
9
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15
18
21
24
27
30
33
4
8
12
16
20
24
28
32
36
40
44
5
10
15
20
25
30
35
40
45
50
55
6
12
18
24
30
36
42
48
54
60
66
7
14
21
28
35
42
49
56
63
70
77
8
16
24
32
40
48
56
64
72
80
88
9
18
27
36
45
54
63
72
81
90
99
10
20
30
40
50
60
70
80
90
100
110
11
22
33
44
55
66
77
88
99
110
121
Doesn’t work for cube....
In[]:=
Count[Flatten[Table[ijk,{i,19},{j,19},{k,19}]],20]
Out[]=
15

Total number of numbers reached after 1, 2, 3 ... steps given

In[]:=
gg=Table[Last[ResourceFunction["GraphNeighborhoodVolumes"][With[{m=k},NestGraph[n|->Table[an,{a,m}],1,10]],{1}]],{k,2,13}]
Out[]=
{{1,2,3,4,5,6,7,8,9,10,11},{1,3,6,10,15,21,28,36,45,55,66},{1,4,9,16,25,36,49,64,81,100,121},{1,5,14,30,55,91,140,204,285,385,506},{1,6,18,40,75,126,196,288,405,550,726},{1,7,25,65,140,266,462,750,1155,1705,2431},{1,8,30,80,175,336,588,960,1485,2200,3146},{1,9,36,100,225,441,784,1296,2025,3025,4356},{1,10,42,120,275,546,980,1632,2565,3850,5566},{1,11,53,173,448,994,1974,3606,6171,10021,15587},{1,12,59,194,504,1120,2226,4068,6963,11308,17589},{1,13,72,266,770,1890,4116,8184,15147,26455,44044}}
In[]:=
ff=Table[LinearRecurrence[Table[(-1)^(k+1)Binomial[PrimePi[m]+1,k],{k,1,PrimePi[m]+1}],Prepend[Length/@NestList[Union[Times@@#&/@Tuples[{Range[m],#}]]&,Range[m],PrimePi[m]-1],1],11],{m,2,13}]
Out[]=
{{1,2,3,4,5,6,7,8,9,10,11},{1,3,6,10,15,21,28,36,45,55,66},{1,4,9,16,25,36,49,64,81,100,121},{1,5,14,30,55,91,140,204,285,385,506},{1,6,18,40,75,126,196,288,405,550,726},{1,7,25,65,140,266,462,750,1155,1705,2431},{1,8,30,80,175,336,588,960,1485,2200,3146},{1,9,36,100,225,441,784,1296,2025,3025,4356},{1,10,42,120,275,546,980,1632,2565,3850,5566},{1,11,53,173,448,994,1974,3606,6171,10021,15587},{1,12,59,194,504,1120,2226,4068,6963,11308,17589},{1,13,72,266,770,1890,4116,8184,15147,26455,44044}}
In[]:=
Grid[ff,Frame->All,AlignmentRight]
Out[]=
1
2
3
4
5
6
7
8
9
10
11
1
3
6
10
15
21
28
36
45
55
66
1
4
9
16
25
36
49
64
81
100
121
1
5
14
30
55
91
140
204
285
385
506
1
6
18
40
75
126
196
288
405
550
726
1
7
25
65
140
266
462
750
1155
1705
2431
1
8
30
80
175
336
588
960
1485
2200
3146
1
9
36
100
225
441
784
1296
2025
3025
4356
1
10
42
120
275
546
980
1632
2565
3850
5566
1
11
53
173
448
994
1974
3606
6171
10021
15587
1
12
59
194
504
1120
2226
4068
6963
11308
17589
1
13
72
266
770
1890
4116
8184
15147
26455
44044
s across the top ... t down
The sth column is the dimension s multiplication table
In[]:=
Differences/@ff
Out[]=
{{1,1,1,1,1,1,1,1,1,1},{2,3,4,5,6,7,8,9,10,11},{3,5,7,9,11,13,15,17,19,21},{4,9,16,25,36,49,64,81,100,121},{5,12,22,35,51,70,92,117,145,176},{6,18,40,75,126,196,288,405,550,726},{7,22,50,95,161,252,372,525,715,946},{8,27,64,125,216,343,512,729,1000,1331},{9,32,78,155,271,434,652,933,1285,1716},{10,42,120,275,546,980,1632,2565,3850,5566},{11,47,135,310,616,1106,1842,2895,4345,6281},{12,59,194,504,1120,2226,4068,6963,11308,17589}}
Number of multiplicands across the top; number of steps down
In[]:=
Grid[%,Frame->All,AlignmentRight]
Out[]=
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
8
9
10
11
3
5
7
9
11
13
15
17
19
21
4
9
16
25
36
49
64
81
100
121
5
12
22
35
51
70
92
117
145
176
6
18
40
75
126
196
288
405
550
726
7
22
50
95
161
252
372
525
715
946
8
27
64
125
216
343
512
729
1000
1331
9
32
78
155
271
434
652
933
1285
1716
10
42
120
275
546
980
1632
2565
3850
5566
11
47
135
310
616
1106
1842
2895
4345
6281
12
59
194
504
1120
2226
4068
6963
11308
17589
In[]:=
Table[Binomial[n,3],{n,10}]
Out[]=
{0,0,1,4,10,20,35,56,84,120}
Number of numbers that appear exactly at a given row...
In[]:=
Graph[ResourceFunction["NestGraphTagged"][n|->{2n,3n},1,10,"StateLabeling"->True,"RuleStyling"->None],GraphLayout"LayeredDigraphEmbedding",AspectRatio1/2]
Out[]=
In[]:=
Graph[ResourceFunction["NestGraphTagged"][n|->{n,2n,3n},1,10,"StateLabeling"->True,"RuleStyling"->None],GraphLayout"LayeredDigraphEmbedding",AspectRatio1/2]
Out[]=
In[]:=
Graph3D[ResourceFunction["NestGraphTagged"][n|->{2n,3n,4n},1,6,"StateLabeling"->True,"RuleStyling"->None]]
Out[]=

After 2 steps, one gets the distinct numbers in the s×s multiplication table:

[ “numbers of low multiplicative complexity” ]

When a number can be generated in more ways, it adds an edge in the graph, but what matters to the geodesic volume is the first appearance of each number....

How many distinct numbers appear? [Given s and t, how long will be list be?] In the “free” case with no merging, there will be s^t such numbers. But many can be generated in multiple ways, and some can’t be generated

With 1 multiplier, just get powers of 2.
With 2 multipliers [[ volume of ball is triangular numbers ]]
3 multipliers
[[ volume of ball is the squares ]]
4 multipliers
[ Squares ]

When Does a Number First Occur?

Any given number

As a function of s, how fast does the ball grow in t?

As a function of s, the exponent of t is PrimePi[s]
The fastest way to generate numbers is by combining primes......
At each step, you pick one of the q available numbers.... t^q

Branchial Graphs

Each edge could be labeled by the common ancestor that led to it..... [[[[ Ed: need to actually label them ]]]]

Stacked Plot

Rulial Space

Making a square

Linear Recurrences