%I #23 Nov 26 2022 12:34:37
%S 8,21,23,30,35,50,53,54,75,98,111,158,174,210,230,260,284,315,336,350,
%T 410,440,459,473,485,495,525,545,554,576,590,608,615,629,660,680,683,
%U 774,846,900,923,966,975,989,1071,1103,1133,1148,1220,1400,1430,1463,1499
%N Numbers k such that 4k+1, 4k+2, and 4k+3 are all semiprimes.
%C 4k+4 = 4*(k+1) = 2*2*(k+1) cannot be semiprime as well, as it has at least 3 prime factors with multiplicity. Thus there are no four consecutive semiprimes.
%H Harvey P. Dale, <a href="/A123255/b123255.txt">Table of n, a(n) for n = 1..1000</a>
%F {k: 4k+1 is in A001358 AND 4k+2 is in A001358 AND 4k+3 is in A001358}.
%F {k: 4k+1 is in A070552 AND 4k+2 is in A070552}.
%F {(A056809(i)-1)/4}.
%e a(1) = 8 because 4*8+1 = 33 = 3*11 is semiprime and 4*8+2 = 34 = 2*17 is semiprime and 4*8+3 = 35 = 3*5 is semiprime.
%t Select[Range[1100],Union[PrimeOmega[4#+{1,2,3}]]=={2}&] (* _Harvey P. Dale_, Feb 02 2015 *)
%o (Magma) IsSemiprime:=func< n | &+[k[2]: k in Factorization(n)] eq 2 >; [ n: n in [2..1500] | IsSemiprime(4*n+1) and IsSemiprime(4*n+2) and IsSemiprime(4*n+3) ]; // _Vincenzo Librandi_, Dec 22 2010
%o (Python)
%o from sympy import factorint, isprime
%o def issemiprime(n):
%o return sum(factorint(n).values()) == 2 if n&1 else isprime(n//2)
%o def ok(n): return all(issemiprime(4*n+i) for i in (2, 1, 3))
%o print([k for k in range(1500) if ok(k)]) # _Michael S. Branicky_, Nov 26 2022
%Y Cf. A001358, A056809, A070552.
%K easy,nonn
%O 1,1
%A _Jonathan Vos Post_, Oct 09 2006
%E 336 and 680 added by _Vincenzo Librandi_, Dec 22 2010