OFFSET
1,10
COMMENTS
a(n) = 0 iff n = 1.
a(n) = 1 iff n belongs to A073491.
a(p) = 1 for any prime p.
a(A002110(n)) = 1 for any n > 0.
a(n!) = 1 for any n > 1.
a(A066205(n)) = n for any n > 0.
a(n) = a(A007947(n)) for any n > 0.
a(n) = a(A003961(n)) for any n > 0.
a(n*m) <= a(n) + a(m) for any n > 0 and m > 0.
Each number n can be uniquely represented as a product of a(n) distinct terms from A073491; this representation is minimal relative to the number of terms.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Illustration of the first terms
EXAMPLE
See illustration of the first terms in the Links section.
The prime indices of 18564 are {1,1,2,4,6,7}, which separate into maximal gapless submultisets {1,1,2}, {4}, {6,7}, so a(18564) = 3; this corresponds to the ordered factorization 18564 = 12 * 7 * 221. - Gus Wiseman, Sep 03 2022
MATHEMATICA
Table[Length[Select[First/@If[n==1, {}, FactorInteger[n]], !Divisible[n, NextPrime[#]]&]], {n, 30}] (* Gus Wiseman, Sep 03 2022 *)
PROG
(PARI) a(n) = my (f=factor(n)); if (#f~==0, return (0), return (#f~ - sum(i=1, #f~-1, if (primepi(f[i, 1])+1 == primepi(f[i+1, 1]), 1, 0))))
(Python)
from sympy import factorint, primepi
def a087207(n):
f=factorint(n)
return sum([2**primepi(i - 1) for i in f])
def a069010(n): return sum(1 for d in bin(n)[2:].split('0') if len(d)) # this function from Chai Wah Wu
def a(n): return a069010(a087207(n)) # Indranil Ghosh, Jun 06 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Rémy Sigrist, Jun 04 2017
STATUS
approved