A071322 Alternating sum of all prime factors of n; primes nonincreasing, starting with the largest prime factor: A006530(n).

0, 2, 3, 0, 5, 1, 7, 2, 0, 3, 11, 3, 13, 5, 2, 0, 17, 2, 19, 5, 4, 9, 23, 1, 0, 11, 3, 7, 29, 4, 31, 2, 8, 15, 2, 0, 37, 17, 10, 3, 41, 6, 43, 11, 5, 21, 47, 3, 0, 2, 14, 13, 53, 1, 6, 5, 16, 27, 59, 2, 61, 29, 7, 0, 8, 10, 67, 17, 20, 4, 71, 2, 73, 35

Offset: 1

Author

Reinhard Zumkeller, May 18 2002

Keywords

• nonn

Comments

a(n) = 0 iff n square, a(A000290(n)) = 0;

a(n) = n iff n prime, a(A000040(n)) = A000040(n).

Examples

			72 = 2*2*2*3*3, therefore a(72) = 3 - 3 + 2 - 2 + 2 = 2;
90 = 2*3*3*5, therefore a(90) = 5 - 3 + 3 - 2 = 3.
		

Links

• Shreyansh Jaiswal, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale).

Crossrefs

Cf. A001414, A071324, a(n) = abs(A071321(n)).

Programs

Mathematica

aspf[n_]:=Total[Times@@@Partition[Riffle[Reverse[Flatten[Table[#[[1]],{#[[2]]}]&/@FactorInteger[n]]],{1,-1},{2,-1,2}],2]]; Join[{0}, Array[ aspf,80,2]] (* Harvey P. Dale, Apr 19 2015 *)

Python

from sympy import factorint
def c(n):
    fs = factorint(n, multiple=True)
    return sum(fs[::2])-sum(fs[1::2])
for n in range(1,75):
    print(abs(c(n)), end=", ") #Shreyansh Jaiswal, Apr 16 2025