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A365901 Irregular triangle read by rows giving trajectory beginning with A002182(n) under recursion of x -> f(x) until reaching 1, where f(x) = x/rad(x), rad(x) = A007947(x). 1
1, 2, 1, 4, 2, 1, 6, 1, 12, 2, 1, 24, 4, 2, 1, 36, 6, 1, 48, 8, 4, 2, 1, 60, 2, 1, 120, 4, 2, 1, 180, 6, 1, 240, 8, 4, 2, 1, 360, 12, 2, 1, 720, 24, 4, 2, 1, 840, 4, 2, 1, 1260, 6, 1, 1680, 8, 4, 2, 1, 2520, 12, 2, 1, 5040, 24, 4, 2, 1, 7560, 36, 6, 1 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Let h(n) = A002182(n).
Since highly composite numbers h(n) are products of primorials (i.e., in A025487), the squarefree kernel is always a primorial (i.e., in A002110), and the trajectory always reaches 1.
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..11397 (rows n = 1..1200, flattened)
Michael De Vlieger, Log log scatterplot of log_10(a(n)), n = 1..1053678, showing 2^16 rows.
FORMULA
Length of row n is A112779(n)+1.
T(n,2) = A301413(n), n > 1.
EXAMPLE
Row 1 = {1} since h(1) = 1, already 1.
Row 2 = {2, 1} since h(2) = 2, 2/rad(2) = 2/2 = 1, reaching 1.
Row 3 = {4, 2, 1} since h(3) = 4, 4/rad(4) = 4/2 = 2, and we follow the trajectory of 2 thereafter.
Row 6 = {24, 4, 2, 1} since h(6) = 24, 24/rad(24) = 24/6 = 4, and we follow the trajectory of 4 thereafter.
Row 14 = {720, 24, 4, 2, 1} since h(14) = 720, 720/rad(720) = 720/30 = 24, which appends row 6 thereafter.
.
First rows of this sequence:
row 1: 1
2: 2, 1
3: 4, 2, 1
4: 6, 1
5: 12, 2, 1
6: 24, 4, 2, 1
7: 36, 6, 1
8: 48, 8, 4, 2, 1
9: 60, 2, 1
10: 120, 4, 2, 1
11: 180, 6, 1
12: 240, 8, 4, 2, 1
...
MATHEMATICA
a2182 = Import["https://oeis.org/A002182/b002182.txt", "Data"][[All, -1]]; Array[NestWhileList[#/(Times @@ FactorInteger[#][[All, 1]]) &, a2182[[#]], # > 1 &] &, 20] // TableForm
CROSSREFS
Sequence in context: A329504 A147542 A347069 * A325309 A211956 A128177
KEYWORD
nonn,tabf
AUTHOR
Michael De Vlieger, Oct 11 2023
STATUS
approved

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Last modified July 21 18:08 EDT 2024. Contains 374475 sequences. (Running on oeis4.)