A358657 Numbers such that the three numbers before and the three numbers after are squarefree semiprimes.

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

Offset

1,1

Comments

All numbers in this sequence are divisible by 36. Proof: Suppose k is odd and in this sequence; then either k-1 or k-3 is divisible by 4, creating a contradiction. Suppose k is even, but not divisible by k; then k-2 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 k-3 or k+3 is divisible by 9, creating a contradiction.

Since each term k is divisible by 36, it follows that (k-3)/3, (k-2)/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 k-1 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 k-1 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

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (first 169 terms from Robert Israel)

Formula

a(n) = 2*(A158476(n) + 1). - Hugo Pfoertner, Dec 12 2022

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]):
map(t -> S[t+2]+1, J); # Robert Israel, Nov 26 2023

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))
print(list(islice(agen(), 20))) # Michael S. Branicky, Nov 26 2022
(Magma)
a:=[];
IsP:=IsPrime;
Tau:=NumberOfDivisors;
for m in [1..170000] do
   t:=36*m;
   if IsP((t-3) div 3)
         and IsP((t+3) div 3)
         and IsP((t-2) div 2)
         and IsP((t+2) div 2)
         and Tau(t-1) eq 4
         and Tau(t+1) eq 4 then
      a:=a cat [t];
   end if;
end for;
a; // Jon E. Schoenfield, Nov 26 2023

Crossrefs

Cf. A001358, A158476, A350101, A358666.

Sequence in context: A167127 A269821 A268364 * A048100 A222337 A008697

Adjacent sequences: A358654 A358655 A358656 * A358658 A358659 A358660

Keyword

nonn,changed

Author

Tanya Khovanova and Massimo Kofler, Nov 25 2022

Status

approved