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.

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, 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, 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

Offset: 1

Author

Antti Karttunen, Dec 27 2018

Keywords

• sign

Examples

			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.
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.
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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Cf. A001222, A005117, A285328, A322817.

Programs

PARI

A007947(n) = factorback(factorint(n)[, 1]); \\ From A007947
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)); };
A322818(n) = (bigomega(n)-bigomega(A285328(n)));

Formula

a(n) = A001222(n) - A001222(A285328(n)).

a(A005117(n)) = A001222(A005117(n)).