|
|
A069157
|
|
Number of positive divisors of n that are divisible by the smallest prime that divides n.
|
|
2
|
|
|
0, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 4, 1, 3, 1, 4, 2, 2, 1, 6, 2, 2, 3, 4, 1, 4, 1, 5, 2, 2, 2, 6, 1, 2, 2, 6, 1, 4, 1, 4, 4, 2, 1, 8, 2, 3, 2, 4, 1, 4, 2, 6, 2, 2, 1, 8, 1, 2, 4, 6, 2, 4, 1, 4, 2, 4, 1, 9, 1, 2, 3, 4, 2, 4, 1, 8, 4, 2, 1, 8, 2, 2, 2, 6, 1, 6, 2, 4, 2, 2, 2, 10, 1, 3, 4, 6
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,4
|
|
LINKS
|
|
|
FORMULA
|
a(n) = A000005(n) * A067029(n)/(1+A067029(n)) = d(n) * e_n/(e_n + 1), where d(n) is the number of positive divisors of n and e_n is the exponent of the smallest prime to divide n in the prime factorization of n.
|
|
EXAMPLE
|
The divisors of 12 which are themselves divisible by 2 (the smallest prime dividing 12) are 2, 4, 6 and 12. So the 12th term is 4.
|
|
MATHEMATICA
|
a[1] = 0; a[n_] := DivisorSigma[0, n] * (e = FactorInteger[n][[1, 2]])/(e + 1); Array[a, 100] (* Amiram Eldar, May 06 2020 *)
|
|
PROG
|
(Python)
from sympy import divisor_count, factorint
def a067029(n): return 0 if n==1 else next(iter(factorint(n).values()))
def a(n): return divisor_count(n)*a067029(n)//(1 + a067029(n))
(PARI) a(n) = if (n==1, 0, my(p=vecmin(factor(n)[, 1])); sumdiv(n, d, ((d % p) == 0))); \\ Michel Marcus, May 06 2020
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|