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%I A007675 M3824 #99 Jul 16 2024 12:51:39
%S A007675 1,5,13,21,29,33,37,41,57,65,69,77,85,93,101,105,109,113,129,137,141,
%T A007675 157,165,177,181,185,193,201,209,213,217,221,229,237,253,257,265,281,
%U A007675 285,301,309,317,321,329,345,353,357,365,381,389,393,397,401,409,417,429,433,437,445,453
%N A007675 Numbers m such that m, m+1 and m+2 are squarefree.
%C A007675 Four categories: all terms are composites like {33, 34, 35}; first term only is prime like {37, 38, 39}; third term only is prime like {57, 58, 59}; first and third are primes like {29, 30, 31}. - _Labos Elemer_
%C A007675 Four consecutive integers cannot be squarefree as one of them is divisible by 2^2 = 4. - _Amarnath Murthy_, Feb 18 2002
%C A007675 Numbers m such that m^3 + 3m^2 + 2m is squarefree. See proof below. - _Charles R Greathouse IV_, Mar 05 2013
%C A007675 There are kx + O(x/log x) terms of this sequence below x, where k = A206256. - _Charles R Greathouse IV_, Mar 05 2013
%C A007675 Proof: m^3 + 3m^2 + 2m = m*(m+1)*(m+2) and the factors are pairwise relatively prime, because (m+1) is even. - _Thomas Ordowski_, Apr 20 2013
%C A007675 Conjecture: for every prime p, the numbers p# - 1, p#, p# + 1 are squarefree, where primorial p# = product of all primes <= p. - _Thomas Ordowski_, Apr 21 2013
%C A007675 Let f(m) = abs(mu(m*(m+1)*(m+2))), where mu(m) is the Moebius function, then the sum S(m) = f(1) + f(2) + ... + f(m) ~ k*m with the constant k = A206256 = 0.12548698.... - _Thomas Ordowski_, Apr 22 2013
%C A007675 All terms are congruent to 1 (mod 4). - _Zak Seidov_, Dec 22 2014
%D A007675 P. R. Halmos, Problems for Mathematicians Young and Old. Math. Assoc. America, 1991, p. 28.
%D A007675 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A007675 Zak Seidov, <a href="/A007675/b007675.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from T. D. Noe)
%F A007675 Numbers m such that g(m)*g(m+1)*g(m+2) = 1, where g(w) = abs(mu(w)). - _Labos Elemer_
%F A007675 a(n) ~ c*n with c = 7.96895... = 1/A206256. - _Charles R Greathouse IV_, Mar 05 2013
%e A007675 85 is a term as 85 = 17*5, 86 = 43*2, 87 = 29*3.
%p A007675 select(t -> andmap(NumberTheory:-IsSquareFree,[t,t+1,t+2]), [seq(i,i=1..1000,4)]); # _Robert Israel_, Jul 16 2024
%t A007675 Select[Range[1000], SquareFreeQ[#(# + 1)(# + 2)] &] (* _Vladimir Joseph Stephan Orlovsky_, Mar 30 2011 *)
%t A007675 Transpose[Select[Partition[Select[Range[400], SquareFreeQ], 3, 1], Differences[#] == {1, 1} &]][[1]] (* _Harvey P. Dale_, Apr 11 2012 *)
%t A007675 Select[Range[1, 499, 2], MoebiusMu[#^3 + 3#^2 + 2#] != 0 &] (* _Alonso del Arte_, Jan 16 2014 *)
%t A007675 SequencePosition[Table[If[SquareFreeQ[n],1,0],{n,500}],{1,1,1}][[All,1]] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Dec 14 2017 *)
%o A007675 (Haskell)
%o A007675 a007675 n = a007675_list !! (n-1)
%o A007675 a007675_list = f 1 a008966_list where
%o A007675    f n (u:xs'@(v:w:x:xs)) | u == 1 && w == 1 && v == 1 = n : f (n+4) xs
%o A007675                           | otherwise = f (n+1) xs'
%o A007675 -- _Reinhard Zumkeller_, Nov 05 2011
%o A007675 (PARI) is(n)=issquarefree(n)&&issquarefree(n+1)&&issquarefree(n+2) \\ _Charles R Greathouse IV_, Mar 05 2013
%Y A007675 Subsequence of A007674, A016813, and A056911.
%Y A007675 Cf. A005117, A013929, A008966, A007531.
%K A007675 nonn,easy,nice
%O A007675 1,2
%A A007675 _N. J. A. Sloane_, _Robert G. Wilson v_

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