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%I #16 Nov 09 2022 19:09:12
%S 1,2,2,2,1,1,2,4,2,2,1,1,2,1,2,1,2,2,1,3,4,2,3,1,2,2,2,2,2,1,1,2,4,2,
%T 2,2,2,3,2,1,3,1,2,1,3,2,1,1,1,3,2,2,2,3,1,2,2,2,2,1,2,3,1,4,1,2,2,2,
%U 1,2,1,2,2,1,3,1,2,1,2,4,2,4,2,3,1,2,1
%N Lengths of decreasing blocks of A006530, the greatest prime factor of n, starting from the second term.
%C Let gpf(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, ...}.
%C The growth of a(n) is very slow. See the following smallest values of m such that a(m) = n:
%C a(1) = 1, a(2) = 2, a(20) = 3, a(8) = 4, a(251) = 5, a(936) = 6, a(15553) = 7, a(6380) = 8, a(54838)=9, a(293548) = 10.
%H Robert Israel, <a href="/A216651/b216651.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A070089(n+1)-A070089(n). - _Pontus von Brömssen_, Nov 09 2022
%e A006530 with decreasing blocks marked: (2), (3, 2), (5, 3), (7, 2), (3), (5), (11, 3), (13, 7, 5, 2), .... Thus the terms of this sequence are 1, 2, 2, 2, 1, 1, 2, 4, ....
%p N:= 1000: # to use A006530(1..N)
%p L:= map(max @ numtheory:-factorset, [$1..N]):
%p DL:= L[2..-1]-L[1..-2]:
%p R:= select(t -> DL[t]>= 0, [$1..N-1]):
%p R[2..-1]-R[1..-2]; # _Robert Israel_, Mar 02 2018
%Y Cf. A006530, A216650.
%Y First differences of A070089.
%K nonn
%O 1,2
%A _Michel Lagneau_, Sep 12 2012
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