A287170 a(n) = number of runs of consecutive prime numbers among the prime divisors of n.
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
Offset: 1
Keywords
Examples
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
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
- Rémy Sigrist, Illustration of the first terms
- Rémy Sigrist, Scatterplot of the ordinal transform of the first 1000000 terms
Crossrefs
Programs
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Mathematica
Table[Length[Select[First/@If[n==1,{},FactorInteger[n]],!Divisible[n,NextPrime[#]]&]],{n,30}] (* Gus Wiseman, Sep 03 2022 *)
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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))))
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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
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