OFFSET
1,6
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
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?
FORMULA
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
EXAMPLE
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)
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]]
PROG
(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
CROSSREFS
KEYWORD
nonn,look
AUTHOR
Ali Sada and Robert G. Wilson v, Mar 09 2020
STATUS
approved