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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.
18

%I #73 Sep 29 2024 09:20:34

%S 33,85,93,141,201,213,217,301,393,445,633,697,921,1041,1137,1261,1345,

%T 1401,1641,1761,1837,1893,1941,1981,2101,2181,2217,2305,2361,2433,

%U 2461,2517,2641,2721,2733,3097,3385,3601,3693,3865,3901,3957,4285,4413,4533,4593,4881,5601

%N Smallest of three consecutive squarefree numbers k, k+1, k+2 of the form p*q where p and q are distinct primes.

%C Equivalently: k, k+1 and k+2 all have 4 divisors.

%C There cannot be four consecutive squarefree numbers as one of them is divisible by 2^2 = 4.

%C 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.

%C Nonsquare terms of A056809. First terms of A056809 absent here are A056809(4)=121=11^2, A056809(14)=841=29^2, A056809(55)=6241=79^2.

%C 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

%C 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

%C a(n) is always 1 or 9 mod 12. - _Charles R Greathouse IV_, Mar 19 2022

%D R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section B18.

%D David Wells, Curious and interesting numbers, Penguin Books, 1986, p. 114.

%H Amiram Eldar, <a href="/A039833/b039833.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Reinhard Zumkeller)

%H Roberto Conti, Pierluigi Contucci, and Vitalii Iudelevich, <a href="https://arxiv.org/abs/2401.03278">Bounds on tree distribution in number theory</a>, arXiv:2401.03278 [math.NT], 2024. See page 13.

%F A008966(a(n)) * A064911(a(n)) * A008966(a(n)+1) * A064911(a(n)+1) * A008966(a(n)+2) * A064911(a(n)+2) = 1. - _Reinhard Zumkeller_, Feb 26 2011

%e 33, 34 and 35 all have 4 divisors.

%e 85 is a term as 85 = 17*5, 86 = 43*2, 87 = 29*3.

%t 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 *)

%t 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 *)

%t SequencePosition[DivisorSigma[0,Range[6000]],{4,4,4}][[All,1]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Aug 17 2017 *)

%o (Haskell)

%o a039833 n = a039833_list !! (n-1)

%o a039833_list = f a006881_list where

%o f (u : vs@(v : w : xs))

%o | v == u+1 && w == v+1 = u : f vs

%o | otherwise = f vs

%o -- _Reinhard Zumkeller_, Aug 07 2011

%o (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

%o (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

%Y Subsequence of A006881 and A056809 and A263990.

%Y Cf. A038456, A039832, A008683, A007675, A063736, A063838, A070552, A045939, A195685, A179502.

%K nonn,nice

%O 1,1

%A _Olivier GĂ©rard_

%E Additional comments from _Amarnath Murthy_, _Vladeta Jovovic_, _Labos Elemer_ and _Benoit Cloitre_, May 08 2002