cp's OEIS Frontend

A322818 a(n) = A001222(n) - A001222(A285328(n)), where A285328(n) gives the next smaller m that has same prime factors as n (ignoring multiplicity), or 1 if n is squarefree, and A001222 gives the number of prime factors, when counted with multiplicity.

%I A322818 #7 Dec 27 2018 17:36:37
%S A322818 0,1,1,1,1,2,1,1,1,2,1,1,1,2,2,1,1,0,1,1,2,2,1,1,1,2,1,1,1,3,1,1,2,2,
%T A322818 2,0,1,2,2,1,1,3,1,1,1,2,1,1,1,-1,2,1,1,-1,2,1,2,2,1,1,1,2,1,1,2,3,1,
%U A322818 1,2,3,1,1,1,2,0,1,2,3,1,2,1,2,1,1,2,2,2,1,1,0,2,1,2,2,2,1,1,-1,1,-1,1,3,1,1,3
%N A322818 a(n) = A001222(n) - A001222(A285328(n)), where A285328(n) gives the next smaller m that has same prime factors as n (ignoring multiplicity), or 1 if n is squarefree, and A001222 gives the number of prime factors, when counted with multiplicity.
%H A322818 Antti Karttunen, <a href="/A322818/b322818.txt">Table of n, a(n) for n = 1..20000</a>
%F A322818 a(n) = A001222(n) - A001222(A285328(n)).
%F A322818 a(A005117(n)) = A001222(A005117(n)).
%e A322818 For n = 6 = 2*3, there is no smaller number with only the prime factors 2 and 3 as 6 is squarefree, thus A285328(6) = 1, and a(6) = A001222(6) = 2.
%e A322818 For n = 40 = 2^3 * 5^1, the next smaller number with the same prime factors is 20 = 2^2 * 5^1. While 40 has 3+1 = 4 prime factors in total, 20 has 2+1 = 3, thus a(40) = 4-3 = 1.
%e A322818 For n = 50 = 2^1 * 5^2, the next smaller number with the same prime factors is 40 = 2^3 * 5^1, thus a(50) = (1+2)-(3+1) = -1.
%o A322818 (PARI)
%o A322818 A007947(n) = factorback(factorint(n)[, 1]); \\ From A007947
%o A322818 A285328(n) = { my(r); if((n > 1 && !bitand(n,(n-1))), (n/2), r=A007947(n); if(r==n,1,n = n-r; while(A007947(n) <> r, n = n-r); n)); };
%o A322818 A322818(n) = (bigomega(n)-bigomega(A285328(n)));
%Y A322818 Cf. A001222, A005117, A285328, A322817.
%K A322818 sign
%O A322818 1,6
%A A322818 _Antti Karttunen_, Dec 27 2018