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

Author

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

Keywords

• nonn
• look

Comments

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

Examples

			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)
		

Links

• 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?

Crossrefs

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.

Programs

Mathematica

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]]

PARI

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

Formula

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