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A287170
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a(n) = number of runs of consecutive prime numbers among the prime divisors of n.
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29
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0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2
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OFFSET
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1,10
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COMMENTS
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a(n) = 0 iff n = 1.
a(p) = 1 for any prime p.
a(n!) = 1 for any n > 1.
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.
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LINKS
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FORMULA
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EXAMPLE
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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
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MATHEMATICA
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Table[Length[Select[First/@If[n==1, {}, FactorInteger[n]], !Divisible[n, NextPrime[#]]&]], {n, 30}] (* Gus Wiseman, Sep 03 2022 *)
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PROG
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(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
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CROSSREFS
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Positions of first appearances are A066205.
These are the row-lengths of A356226 and A356234. Other statistics are:
A003963 multiplies together the prime indices.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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