OFFSET
1,1
COMMENTS
We consider length(p) = length(q). For example, the primes p = 97, 997, 99999999999999997,...(see A003618) are not in the sequence with q = 2.
Each prime p appears only once in the sequence, but the pair (p, q) is not unique, for example the prime 163 generates two pairs of primes(163, 683) and (163, 863), the prime 283 generates three pairs of primes(283, 167), (283, 617) and (283, 761).
The couples of primes (p, q) are (2, 7), (7, 2), (23, 67), (61, 83), (67, 23), (83, 61), (107, 829), (109, 809), (127, 827),...
In the general case, the digits of p are different from q, but there exists numbers p such that q has the same digits as p, for example (p, q) = (227, 277), (727, 227), (881, 181), ...
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
23 is in the sequence because 9-2 = 7 and 9 - 3 = 6 => 67 is prime, and we obtain the pair (23, 67).
MAPLE
with(numtheory):kk:=0:
for n from 1 to 200 do:
ii:=0:
for k from 1 to 2000 while(ii=0) do:
p1:=ithprime(n):p2:=ithprime(k):
x1:=convert(p1, base, 10):n1:=nops(x1):
x2:=convert(p2, base, 10):n2:=nops(x2):
if n1=n2 then
W:=array(1..n1):U:=array(1..n1):U1:=array(1..n1):
for c from 1 to n1 do:
U1[c]:=x1[c]:od:U:=sort(x1, `<`):V:=sort(x2, `>`):
for j from 1 to n1 do:
W[j]:= 9-V[j]:od:W1:=sort(W, `>`):jj:=0:
for b from 1 to n1 do:
if U[b]=W1[b] then
jj:=jj+1:
else fi:
od:
if jj=n1 then
ii:=1: kk:=kk+1: printf(`%d, `, p1):
else
fi:
fi:
od:
od:
# Alternative:
R:= 2, 7:
for d from 2 to 3 do
P:= select(isprime, [seq(i, i=10^(d-1)+1..10^d-1, 2)]);
nP:= nops(P);
Pd:= map(sort@convert, P, base, 10);
Ps:= convert(map(t -> ListTools:-Reverse([9$d]-t), Pd), set);
S:= select(t -> member(Pd[t], Ps), [$1..nP]);
R:= R, op(P[S]);
od:
R; # Robert Israel, Oct 06 2020
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Aug 28 2013
STATUS
approved