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A328489 Odd numbers k such that the four consecutive odd numbers starting with k have a total of 5 prime factors counting multiplicity.

%I A328489 #13 Oct 20 2019 02:03:43
%S A328489 3,5,7,11,13,17,37,67,107,307,877,1297,2267,2657,3457,3847,3917,4787,
%T A328489 4967,5737,11827,12037,14627,16447,18127,18517,19417,20477,20747,
%U A328489 20897,21377,21557,22567,22637,23057,23557,23627,25577,29567,31387,32057,33347,33767,34757,35797,36467,36787,37307
%N A328489 Odd numbers k such that the four consecutive odd numbers starting with k have a total of 5 prime factors counting multiplicity.
%C A328489 Numbers k such that A001222(A190577(k))=5.
%C A328489 There are three cases:
%C A328489   k=3.
%C A328489   k, k+4 and k+6 are primes while k+2 is 3 times a prime.
%C A328489   k, k+2 and k+6 are primes while k+4 is 3 times a prime.
%C A328489 All terms > 13 have final digit 7.
%C A328489 The first n for which a(n+1)-a(n)=10 is 7538. - _Robert Israel_, Oct 19 2019
%H A328489 Robert Israel, <a href="/A328489/b328489.txt">Table of n, a(n) for n = 1..10000</a>
%e A328489 a(3)=7 is in the sequence because 7*9*11*13 is the product of exactly 5 primes: 3*3*7*11*13.
%p A328489 A1:= select(t -> isprime((t+2)/3) and isprime(t) and isprime(t+4) and isprime(t+6), {seq(i,i=7..100000,30)}):
%p A328489 A2:= select(t -> isprime((t+4)/3) and isprime(t) and isprime(t+2) and isprime(t+6), {seq(i,i=17..100000,30)});
%p A328489 sort(convert({3,5,11,13} union A1 union A2,list));
%Y A328489 Cf. A001222, A190577.
%K A328489 nonn
%O A328489 1,1
%A A328489 _J. M. Bergot_ and _Robert Israel_, Oct 16 2019