A332999 -id:A332999 - cp's OEIS frontend

A333123 Consider the mapping k -> (k - (k/p)), where p is any of k's prime factors. a(n) is the number of different possible paths from n to 1.

1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 3, 3, 5, 5, 1, 1, 5, 5, 3, 10, 5, 5, 4, 3, 7, 5, 9, 9, 12, 12, 1, 17, 2, 21, 9, 9, 14, 16, 4, 4, 28, 28, 9, 21, 14, 14, 5, 28, 7, 7, 12, 12, 14, 16, 14, 28, 23, 23, 21, 21, 33, 42, 1, 33, 47, 47, 3, 61, 56, 56, 14, 14, 23, 28, 28, 103, 42, 42, 5

Offset: 1

Ali Sada and Robert G. Wilson v, Mar 09 2020

The iteration always terminates at 1, regardless of the prime factor chosen at each step.
Although there may exist multiple paths to 1, their path lengths (A064097) are the same! See A064097 for a proof. Note that this behavior does not hold if we allow any divisor of k.
First occurrence of k or 0 if no such value exists: 1, 6, 12, 24, 14, 96, 26, 85, 28, 21, 578, 30, 194, 38, 164, 39, 33, 104, 1538, 112, 35, 328, 58, 166, ..., .
Records: 1, 2, 3, 5, 10, 12, 17, 21, 28, 33, 42, 47, 61, 103, 168, ..., .
Record indices: 1, 6, 12, 14, 21, 30, 33, 35, 42, 62, 63, 66, 69, ..., .
When viewed as a graded poset, the paths of the said graph are the chains of the corresponding poset. This poset is also a lattice (see Ewan Delanoy's answer to Peter Kagey's question at the Mathematics Stack Exchange link). - Antti Karttunen, May 09 2020

			a(1): {1}, therefore a(1) = 1;
a(6): {6, 4, 2, 1} or {6, 3, 2, 1}, therefore a(6) = 2;
a(12): {12, 8, 4, 2, 1}, {12, 6, 4, 2, 1} or {12, 6, 3, 2, 1}, therefore a(12) = 3;
a(14): {14, 12, 8, 4, 2, 1}, {14, 12, 6, 4, 2, 1}, {14, 12, 6, 3, 2, 1}, {14, 7, 6, 4, 2, 1} or {14, 7, 6, 3, 2, 1}, therefore a(14) = 5.
From _Antti Karttunen_, Apr 05 2020: (Start)
For n=15 we have five alternative paths from 15 to 1: {15, 10, 5, 4, 2, 1}, {15, 10, 8, 4, 2, 1}, {15, 12, 8, 4, 2, 1},  {15, 12, 6, 4, 2, 1},  {15, 12, 6, 3, 2, 1}, therefore a(15) = 5. These form a graph illustrated below:
        15
       / \
      /   \
    10     12
    / \   / \
   /   \ /   \
  5     8     6
   \_   |  __/|
     \__|_/   |
        4     3
         \   /
          \ /
           2
           |
           1
(End)

Antti Karttunen, Table of n, a(n) for n = 1..20000
Peter Kagey, Mathematics Stack Exchange, Does a graded poset on the positive integers generated from subtracting factors define a lattice?

Cf. A064097, A332809 (size of the lattice), A332810.
Cf. A332904 (sum of distinct integers present in such a graph/lattice), A333000 (sum over all paths), A333001, A333785.
Cf. A332992 (max. outdegree), A332999 (max. indegree), A334144 (max. rank level).
Cf. A334230, A334231 (meet and join).
Partial sums of A332903.
Cf. also tables A334111, A334184.

a[n_] := Sum[a[n - n/p], {p, First@# & /@ FactorInteger@n}]; a[1] = 1; (* after PARI coding by Rémy Sigrist *) Array[a, 70]
(* view the various paths *)
f[n_] := Block[{i, j, k, p, q, mtx = {{n}}}, Label[start]; If[mtx[[1, -1]] != 1, j = Length@ mtx;  While[j > 0, k = mtx[[j, -1]]; p = First@# & /@ FactorInteger@k; q = k - k/# & /@ p; pl = Length@p; If[pl > 1, Do[mtx = Insert[mtx, mtx[[j]], j], {pl - 1}]]; i = 1;  While[i < 1 + pl, mtx[[j + i - 1]] = Join[mtx[[j + i - 1]], {q[[i]]}]; i++]; j--]; Goto[start], mtx]]

for (n=1, #a=vector(80), print1 (a[n]=if (n==1, 1, vecsum(apply(p -> a[n-n/p], factor(n)[,1]~)))", ")) \\ Rémy Sigrist, Mar 11 2020

a(n) = 1 iff n is a power of two (A000079) or a Fermat Prime (A019434).
a(p) = a(p-1) if p is prime.
a(n) = Sum_{p prime and dividing n} a(n - n/p) for any n > 1. - Rémy Sigrist, Mar 11 2020

A334144 Consider the mapping k -> (k - (k/p)), where prime p | k. a(n) = maximum distinct terms at any position j among the various paths to 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 3, 1, 1, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 1, 4, 2, 4, 3, 3, 3, 3, 2, 2, 4, 4, 3, 4, 3, 3, 2, 4, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 1, 5, 5, 5, 2, 5, 5, 5, 3, 3, 3, 4, 3, 6, 4, 4, 2, 3, 2, 2, 4, 3, 4, 4, 3, 3, 5, 5, 3, 5, 3, 5, 2, 2, 4, 6, 3, 3, 3, 3, 3, 6, 3

Offset: 1

Michael De Vlieger, Peter Kagey, Antti Karttunen, Apr 15 2020

nonn

Let i = A064097(n) be the common path length and let 1 <= j <= i. Given a path P, we find for any j relatively few distinct values. Regarding a common path length i, see A333123 comment 2, and proof at A064097.

Maximum term in row n of A334184.

			For n=15, the paths are shown vertically at left, and the graph obtained appears at right:
  15   15   15   15   15  =>         15
   |    |    |    |    |            _/ \_
   |    |    |    |    |           /     \
  10   10   12   12   12  =>     10       12
   |    |    |    |    |         | \_   _/ |
   |    |    |    |    |         |   \ /   |
   5    8    6    6    8  =>     5    8    6
   |    |    |    |    |          \_  |  _/|
   |    |    |    |    |            \_|_/  |
   4    4    3    4    4  =>          4    3
   |    |    |    |    |              |  _/
   |    |    |    |    |              |_/
   2    2    2    2    2  =>          2
   |    |    |    |    |              |
   |    |    |    |    |              |
   1    1    1    1    1  =>          1
Because the maximum number of distinct terms in any row is 3, a(15) = 3.

Peter Kagey, Table of n, a(n) for n = 1..10000
Peter Kagey, Math.StackExchange, Does a graded poset on the positive integers generated from subtracting factors define a lattice?
Richard Stanley, MIT OpenCourseWare, Lecture Notes in Algebraic Combinatorics: The Sperner Property
Wikipedia, Sperner property of a partially ordered set

Cf. A064097, A332992, A332999, A333123, A334111, A334184.

Cf. also A096825.

Max[Length@ Union@ # & /@ Transpose@ #] & /@ Nest[Function[{a, n}, Append[a, Join @@ Table[Flatten@ Prepend[#, n] & /@ a[[n - n/p]], {p, FactorInteger[n][[All, 1]]}]]] @@ {#, Length@ # + 1} &, {{{1}}}, 105]
(* Second program: *)
g[n_] := Block[{lst = {{n}}}, While[lst[[-1]] != {1}, lst = Join[lst, {Union@ Flatten[# - #/(First@ # & /@ FactorInteger@ #) & /@ lst[[-1]] ]}]]; Max[Length /@ lst]]; Array[g, 105] (* Robert G. Wilson v, May 08 2020 *)

A332992 Maximum outdegree in the graph formed by a subset of numbers in range 1 .. n with edge relation k -> k - k/p, where p can be any of the prime factors of k.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 1, 3, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 1, 3, 3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 3, 3, 3, 2, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 3, 2, 3, 2, 3, 2, 2, 3, 3, 2, 2, 3, 3, 2, 3

Offset: 1

Antti Karttunen, Apr 04 2020

nonn

Maximum number of distinct prime factors of any one integer encountered on all possible paths from n to 1 when iterating with nondeterministic map k -> k - k/p, where p can be any of the prime factors of k.

			For n=15 we have five alternative paths from 15 to 1: {15, 10, 5, 4, 2, 1}, {15, 10, 8, 4, 2, 1}, {15, 12, 8, 4, 2, 1},  {15, 12, 6, 4, 2, 1},  {15, 12, 6, 3, 2, 1}. These form a lattice illustrated below:
        15
       / \
      /   \
    10     12
    / \   / \
   /   \ /   \
  5     8     6
   \__  |  __/|
      \_|_/   |
        4     3
         \   /
          \ /
           2
           |
           1
With edges going from 15 towards 1, the maximum outdegree is 2, which occurs at nodes 15, 12, 10 and 6, therefore a(15) = 2.

Antti Karttunen, Table of n, a(n) for n = 1..65539

Cf. A001221, A064097, A332809, A332999 (max. indegree), A333123, A334111, A334144, A334184.

Cf. A002110 (positions of records and the first occurrence of each n).

With[{s = Nest[Function[{a, n}, Append[a, Join @@ Table[Flatten@ Prepend[#, n] & /@ a[[n - n/p]], {p, FactorInteger[n][[All, 1]]}]]] @@ {#, Length@ # + 1} &, {{{1}}}, 105]}, Array[If[# == 1, 0, Max@ Tally[#][[All, -1]] &@ Union[Join @@ Map[Partition[#, 2, 1] &, s[[#]] ]][[All, 1]] ] &, Length@ s]] (* Michael De Vlieger, May 02 2020 *)

up_to = 105;
A332992list(up_to) = { my(v=vector(up_to)); v[1] = 0; for(n=2,up_to, v[n] = max(omega(n),vecmax(apply(p -> v[n-n/p], factor(n)[, 1]~)))); (v); };
v332992 = A332992list(up_to);
A332992(n) = v332992[n];

a(n) = max(A001221(n), {Max a(n - n/p), for p prime and dividing n}).
For all odd primes p, a(p) = a(p-1).
For all n >= 0, a(A002110(n)) = n.

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A333123 Consider the mapping k -> (k - (k/p)), where p is any of k's prime factors. a(n) is the number of different possible paths from n to 1.

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A334144 Consider the mapping k -> (k - (k/p)), where prime p | k. a(n) = maximum distinct terms at any position j among the various paths to 1.

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A332992 Maximum outdegree in the graph formed by a subset of numbers in range 1 .. n with edge relation k -> k - k/p, where p can be any of the prime factors of k.

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