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A216650
Maximum length of each subsequence whose elements are the greatest prime divisors of the integers 2, 3, 4, ... in increasing order.
2
2, 2, 2, 4, 2, 1, 1, 2, 2, 4, 3, 3, 2, 3, 1, 2, 1, 1, 2, 2, 1, 3, 2, 2, 2, 2, 4, 2, 1, 1, 2, 2, 2, 2, 2, 1, 2, 3, 1, 3, 3, 1, 2, 5, 1, 2, 2, 2, 2, 1, 3, 2, 2, 2, 3, 2, 1, 3, 1, 1, 3, 2, 2, 3, 3, 2, 3, 1, 3, 3, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 3, 6, 1, 5, 2, 2, 2
OFFSET
1,1
COMMENTS
Let gpf(m) = A006530(m) be the greatest prime factor of m and the subset E(n) = {m, m+1, ..., m+L-1} such that gpf(m) < gpf(m+1) < ... < gpf(m+L-1) where L is the maximum length of E(n) and n the index such that {E(1) union E(2) union ... } = {2, 3, 4, ...}.
See the examples for the structure of the subsequences of increasing prime divisors.
The growth of a(n) is very slow. See the following smallest values of m such that a(m) = n:
a(6) = 1, a(1) = 2, a(11) = 3, a(4) = 4, a(44) = 5, a(82) = 6, a(4672) = 7, a(23001) = 8, a(360896) = 9.
LINKS
FORMULA
a(n) = A070087(n)-A070087(n-1) for n >= 2. - Pontus von Brömssen, Nov 09 2022
EXAMPLE
Subset 1: {2, 3} obtained with the numbers 2, 3 => a(1) = 2;
Subset 2: {2, 5} obtained with the numbers 4, 5 => a(2) = 2;
Subset 3: {3, 7} obtained with the numbers 6, 7 => a(3) = 2;
Subset 4: {2, 3, 5, 11} obtained with the numbers 8, 9, 10, 11 => a(4) = 4.
MAPLE
with(numtheory):p0:=2:it:=1:for n from 3 to 200 do: x:=factorset(n):n1:=nops(x):p:=x[n1]:if p>p0 then it:=it+1:p0:=p:else printf(`%d, `, it):it:=1:p0:=p:fi:od:
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Sep 12 2012
STATUS
approved