|
|
A086708
|
|
Primes p such that p-1 and p+1 are both divisible by cubes.
|
|
9
|
|
|
271, 487, 593, 751, 809, 919, 1249, 1567, 1783, 1889, 1999, 2647, 2663, 2753, 2969, 3079, 3511, 3617, 3727, 3833, 3943, 4049, 4159, 4481, 4591, 4751, 4801, 5023, 6857, 6967, 7937, 8263, 8369, 9127, 9343, 10289, 10313, 10529, 10639, 11071, 11177
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
LINKS
|
|
|
FORMULA
|
|
|
MAPLE
|
isA086708 := proc(n)
if isprime(n) then
isA046099(n-1) and isA046099(n+1) ;
else
false;
end if;
end proc:
n := 1:
for c from 1 to 50000 do
if isA086708(c) then
printf("%d %d\n", n, c) ;
n := n+1 ;
end if;
Res:= NULL: count:= 0:
p:= 1:
while count < 100 do
p:= nextprime(p);
if max(seq(t[2], t=ifactors(p-1)[2]))>=3 and max(seq(t[2], t=ifactors(p+1)[2]))>=3 then
count:= count+1; Res:= Res, p;
fi
od:
|
|
MATHEMATICA
|
f[n_]:=Max[Last/@FactorInteger[n]]; lst={}; Do[p=Prime[n]; If[f[p-1]>=3&&f[p+1]>=3, AppendTo[lst, p]], {n, 6!}]; lst (* Vladimir Joseph Stephan Orlovsky, Oct 03 2009 *)
|
|
PROG
|
(PARI)
\\ Input no. of iterations n, power p and number to subtract and add k.
powerfreep4(n, p, k) = { c=0; pc=0; forprime(x=2, n, pc++; if(!ispowerfree(x-k, p) && !ispowerfree(x+k, p), c++; print1(x", "); ) ); print(); print(c", "pc", "c/pc+.0) }
ispowerfree(m, p1) = { flag=1; y=component(factor(m), 2); for(i=1, length(y), if(y[i] >= p1, flag=0; break); ); return(flag) } \\ Cino Hilliard, Dec 08 2003
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|