OFFSET
1,4
COMMENTS
This sequence is similar to A087207; here we encode the exponents, there the prime numbers appearing in the prime factorization of a number.
The binary representation of a(n) shows which exponents appear in the prime factorization of n, but without multiplicities:
- for any prime number p and k > 0, if p^k divides n but p^(k+1) does not divide n, then a(n) AND 2^(k-1) = 2^(k-1) (where AND denotes the bitwise AND operator),
- conversely, if a(n) AND 2^(k-1) = 2^(k-1) for some k > 0, then there is prime number p such that p^k divides n but p^(k+1) does not divide n.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
FORMULA
a(p^k) = 2^(k-1) for any prime number p and k > 0.
a(n^2) = A000695(2 * a(n)) / 2 for any n > 0.
a(n) <= 1 iff n is squarefree (A005117).
a(n) <= 3 iff n is cubefree (A004709).
a(n) is odd iff n belongs to A052485 (weak numbers).
a(n) is even iff n belongs to A001694 (powerful numbers).
a(n) AND 2 = 2 iff n belongs to A038109 (where AND denotes the bitwise AND operator).
If gcd(m, n) = 1, then a(m * n) = a(m) OR a(n) (where OR denotes the bitwise OR operator).
a(n) = a(A328400(n)). - Peter Munn, Oct 02 2023
EXAMPLE
For n = 90:
- 90 = 5^1 * 3^2 * 2^1,
- the exponents appearing in the prime factorization of 90 are 1 and 2,
- hence a(90) = 2^(1-1) + 2^(2-1) = 3.
MATHEMATICA
Array[Total@ Map[2^(# - 1) &, Union[FactorInteger[#][[All, -1]] ]] - Boole[# == 1] &, 86] (* Michael De Vlieger, Dec 29 2017 *)
PROG
(PARI) a(n) = my (x=Set(factor(n)[, 2]~)); sum(i=1, #x, 2^(x[i]))/2
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
Rémy Sigrist, Dec 29 2017
STATUS
approved