

A358657


Numbers such that the three numbers before and the three numbers after are squarefree semiprimes.


2



216, 143100, 194760, 206136, 273420, 684900, 807660, 1373940, 1391760, 1516536, 1591596, 1611000, 1774800, 1882980, 1891764, 2046456, 2051496, 2163420, 2163960, 2338056, 2359980, 2522520, 2913840, 3108204, 4221756, 4297320, 4334940, 4866120, 4988880, 5108796, 5247144, 5606244, 5996844
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OFFSET

1,1


COMMENTS

All numbers in this sequence are divisible by 36. Proof: Suppose k is odd and in this sequence; then either k1 or k3 is divisible by 4, creating a contradiction. Suppose k is even, but not divisible by k; then k2 is divisible by 4, creating a contradiction. Suppose k is not divisible by 3. Then there exists a number j such that 3*j and 3*(j+1) are among squarefree semiprimes surrounding k; one of them is divisible by 6, creating a contradiction. Suppose k is divisible by 3, but not by 9; then one of the squarefree semiprimes k3 or k+3 is divisible by 9, creating a contradiction.
Since each term k is divisible by 36, it follows that (k3)/3, (k2)/2, (k+2)/2, and (k+3)/3 are primes. Additionally, none of the six integers nearest to k can be the cube of a prime: for any prime p > 3, p^3 == {+1, +17} (mod 36), so only k1 or k+1 could be the cube of a prime, yet in either of those cases, that cube's two nearest neighbors, p^3  1 and p^3 + 1, would both be factorable (i.e., p^3  1 = (p^2 + p + 1)*(p  1) and p^3 + 1 = (p^2  p + 1)*(p + 1)), and neither would be a semiprime. Thus, since neither k1 nor k+1 can be the cube of a prime, testing whether each has four divisors (see the Magma code below) is equivalent to testing whether each is a squarefree semiprime.  Jon E. Schoenfield, Nov 26 2023


LINKS



FORMULA



EXAMPLE

The following numbers are squarefree semiprimes: 213 = 3*71, 214 = 2*107, 215 = 5*43, 217 = 7*31, 218 = 2*109, and 219 = 3*73. Thus, 216 is in this sequence.


MAPLE

N:= 10^6: # for terms <= N
P:= select(isprime, [2, seq(i, i=3..N/2, 2)]):
S:= NULL:
for i from 1 to nops(P) do
p:= P[i];
r:= ListTools:BinaryPlace(P, N/p);
if r <= i then break fi;
S:= S, op(p * P[i+1 .. r]);
od:
S:= sort([S]):
J:= select(t > S[t+5] = S[t]+6, [$1..nops(S)5]):


MATHEMATICA

Select[Range[10000000], Transpose[FactorInteger[#  3]][[2]] == {1, 1} && Transpose[FactorInteger[#  2]][[2]] == {1, 1} && Transpose[FactorInteger[#  1]][[2]] == {1, 1} && Transpose[FactorInteger[# + 3]][[2]] == {1, 1} && Transpose[FactorInteger[# + 2]][[2]] == {1, 1} && Transpose[FactorInteger[# + 1]][[2]] == {1, 1} &]
36*Flatten@Position[({1, 1}==Last@Transpose@FactorInteger@# &/@ {#3, #2, #1, #+1, #+2, #+3}) & /@ (36*Range@(10^6)), {True ..}] (* Hans Rudolf Widmer, Aug 01 2024 *)


PROG

(Python)
from itertools import count, islice
from sympy import isprime, factorint
def issfsemiprime(n): return list(factorint(n).values()) == [1, 1] if n&1 else isprime(n//2)
def ok(n): return all(issfsemiprime(n+i) for i in (2, 2, 3, 1, 1, 3))
def agen(): yield from (k for k in count(36, 36) if ok(k))
(Magma)
a:=[];
IsP:=IsPrime;
Tau:=NumberOfDivisors;
for m in [1..170000] do
t:=36*m;
if IsP((t3) div 3)
and IsP((t+3) div 3)
and IsP((t2) div 2)
and IsP((t+2) div 2)
and Tau(t1) eq 4
and Tau(t+1) eq 4 then
a:=a cat [t];
end if;
end for;


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



