OFFSET
1,1
COMMENTS
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.
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.
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
a(n) is always 1 or 9 mod 12. - Charles R Greathouse IV, Mar 19 2022
REFERENCES
David Wells, Curious and interesting numbers, Penguin Books, 1986, p. 114.
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Reinhard Zumkeller)
Roberto Conti, Pierluigi Contucci, and Vitalii Iudelevich, Bounds on tree distribution in number theory, arXiv:2401.03278 [math.NT], 2024. See page 13.
FORMULA
EXAMPLE
33, 34 and 35 all have 4 divisors.
85 is a term as 85 = 17*5, 86 = 43*2, 87 = 29*3.
MATHEMATICA
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 *)
PROG
(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
-- Reinhard Zumkeller, Aug 07 2011
(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
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
EXTENSIONS
Additional comments from Amarnath Murthy, Vladeta Jovovic, Labos Elemer and Benoit Cloitre, May 08 2002
STATUS
approved