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Integers k such that each of k, k+1, k+2, k+4, k+5, k+6 is the product of two distinct primes.
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%I #72 Jul 14 2021 01:56:17

%S 213,143097,194757,206133,273417,684897,807657,1373937,1391757,

%T 1516533,1591593,1610997,1774797,1882977,1891761,2046453,2051493,

%U 2163417,2163957,2338053,2359977,2522517,2913837,3108201,4221753

%N Integers k such that each of k, k+1, k+2, k+4, k+5, k+6 is the product of two distinct primes.

%C A remarkable gap occurs between the initial two members, and the sequence seems to be rather sparse compared to the related A242805.

%C Here, the first member k of the sextet is the reference, whereas in A068088 the center k+3 is selected as reference. Observe that k+3 must be divisible by the square 4.

%C All terms are congruent to 9 (mod 12). - _Zak Seidov_, Apr 14 2015

%C From _Robert Israel_, Apr 15 2015: (Start)

%C All terms are congruent to 33 (mod 36).

%C Numbers k in A039833 such that k+4 is in A039833. (End)

%C From _Robert G. Wilson v_, Apr 15 2015: (Start)

%C k is congruent to 33 (mod 36) so one of its factors is 3 and the other is == 11 (mod 12);

%C k+1 is congruent to 34 (mod 36) so one of its factors is 2 and the other is == 17 (mod 18);

%C k+2 is congruent to 35 (mod 36) so its factors are == +-1 (mod 6);

%C k+4 is congruent to 1 (mod 36) so its factors are == +-1 (mod 6);

%C k+5 is congruent to 2 (mod 36) so one of its factors is 2 and the other is == 1 (mod 18);

%C k+6 is congruent to 3 (mod 36) so one of its factors is 3 and the other is == 1 (mod 12). (End).

%C Number of terms < 10^m: 0, 0, 1, 1, 1, 7, 39, 169, 882, 4852, 27479, ...,. - _Robert G. Wilson v_, Apr 15 2015

%C Or, numbers k such that k, k+1 and k+2 are terms in A175648. - _Zak Seidov_, Dec 08 2015

%H Zak Seidov and Robert G. Wilson v, <a href="/A242804/b242804.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A202319(n) - 1. - _Jon Maiga_, Jul 10 2021

%e 213=3*71, 214=2*107, 215=5*43, 217=7*31, 218=2*109, 219=3*73.

%p f:= t -> numtheory:-issqrfree(t) and (numtheory:-bigomega(t) = 2):

%p select(t -> andmap(f, [t,t+1,t+2,t+4,t+5,t+6]), [seq(36*k+33,k=0..10^6)]); # _Robert Israel_, Apr 15 2015

%t fQ[n_] := PrimeQ[n/3] && PrimeQ[(n + 1)/2] && PrimeQ[(n + 5)/2] && PrimeQ[(n + 6)/3] && PrimeNu[{n + 2, n + 4}] == {2, 2} == PrimeOmega[{n + 2, n + 4}]; k = 33; lst = {}; While[k < 10^8, If[fQ@ k, AppendTo[lst, k]]; k += 36]; lst (* _Robert G. Wilson v_, Apr 14 2015 and revised Apr 15 2015 after _Zak Seidov_ and _Robert Israel_ *)

%o (PARI) default(primelimit, 1000M); i=0; j=0; k=0; l=0; m=0; loc=0; lb=2; ub=9*10^9; o=2; for(n=lb, ub, if(issquarefree(n)&&(o==omega(n)), loc=loc+1; if(1==loc, i=n; ); if(2==loc, if(i+1==n, j=n; ); if(i+1<n, loc=1; i=n; ); ); if(3==loc, if(j+1==n, k=n; ); if(j+1<n, loc=1; i=n; ); ); if(4==loc, if(k+2==n, l=n; ); if(k+2<n, loc=1; i=n; ); ); if(5==loc, if(l+1==n, m=n; ); if(l+1<n, loc=1; i=n; ); ); if(6==loc, if(m+1==n, print1(i,","); loc=0; ); if(m+1<n, loc=1; i=n))))

%o (PARI) forstep(x=213,4221753,12, if( isprime(x/3) && isprime((x+1)/2) && 2==omega(x+2) && 2==bigomega(x+2) && 2==omega(x+4) && 2==bigomega(x+4) && isprime((x+5)/2) && isprime((x+6)/3), print1(x", "))) \\ _Zak Seidov_, Apr 14 2015

%Y Cf. A242793 (minima for two, three and more prime divisors) and A068088 (arbitrary squarefree integers).

%Y Cf. A242805, A039833, A175648, A202319.

%K nonn

%O 1,1

%A _Daniel Constantin Mayer_, May 23 2014