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A039833
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Smallest of three consecutive squarefree numbers k, k+1, k+2 of the form p*q where p and q are distinct primes.
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18
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33, 85, 93, 141, 201, 213, 217, 301, 393, 445, 633, 697, 921, 1041, 1137, 1261, 1345, 1401, 1641, 1761, 1837, 1893, 1941, 1981, 2101, 2181, 2217, 2305, 2361, 2433, 2461, 2517, 2641, 2721, 2733, 3097, 3385, 3601, 3693, 3865, 3901, 3957, 4285, 4413, 4533, 4593, 4881, 5601
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OFFSET
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1,1
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COMMENTS
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Equivalently: k, k+1 and k+2 all have 4 divisors.
There cannot be four consecutive squarefree numbers as one of them is divisible by 2^2 = 4.
These 3 consecutive squarefree numbers of the form p*q have altogether 6 prime factors always including 2 and 3. E.g., if k = 99985, the six prime factors are {2,3,5,19997,33329,49993}. The middle term is even and not divisible by 3.
Cf. A179502 (Numbers k with the property that k^2, k^2+1 and k^2+2 are all semiprimes). - Zak Seidov, Oct 27 2015
The numbers k, k+1, k+2 have the form 2p-1, 2p, 2p+1 where p is an odd prime. A195685 gives the sequence of odd primes that generates these maximal runs of three consecutive integers with four positive divisors. - Timothy L. Tiffin, Jul 05 2016
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REFERENCES
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David Wells, Curious and interesting numbers, Penguin Books, 1986, p. 114.
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LINKS
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FORMULA
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EXAMPLE
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33, 34 and 35 all have 4 divisors.
85 is a term as 85 = 17*5, 86 = 43*2, 87 = 29*3.
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MATHEMATICA
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lst = {}; Do[z = n^3 + 3*n^2 + 2*n; If[PrimeOmega[z/n] == PrimeOmega[z/(n + 2)] == 4 && PrimeNu[z] == 6, AppendTo[lst, n]], {n, 1, 5601, 2}]; lst (* Arkadiusz Wesolowski, Dec 11 2011 *)
okQ[n_]:=Module[{cl={n, n+1, n+2}}, And@@SquareFreeQ/@cl && Union[ DivisorSigma[ 0, cl]]=={4}]; Select[Range[1, 6001, 2], okQ] (* Harvey P. Dale, Dec 17 2011 *)
SequencePosition[DivisorSigma[0, Range[6000]], {4, 4, 4}][[All, 1]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 17 2017 *)
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PROG
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(Haskell)
a039833 n = a039833_list !! (n-1)
a039833_list = f a006881_list where
f (u : vs@(v : w : xs))
| v == u+1 && w == v+1 = u : f vs
| otherwise = f vs
(PARI) is(n)=n%4==1 && factor(n)[, 2]==[1, 1]~ && factor(n+1)[, 2]==[1, 1]~ && factor(n+2)[, 2]==[1, 1]~ \\ Charles R Greathouse IV, Aug 29 2016
(PARI) is(n)=my(t=n%12); if(t==1, isprime((n+2)/3) && isprime((n+1)/2) && factor(n)[, 2]==[1, 1]~, t==9 && isprime(n/3) && isprime((n+1)/2) && factor(n+2)[, 2]==[1, 1]~) \\ Charles R Greathouse IV, Mar 19 2022
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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