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%I #18 Sep 05 2018 15:12:35
%S 2,4,6,7,10,16,20,26,28,32,38,40,46,52,56,58,62,68,70,76,80,82,86,88,
%T 92,94,96,97,112,116,118,122,136,140,142,146,160,170,172,176,178,188,
%U 190,202,212,226,230,238,242,248,256,260,266,272,280,290,298,308,316,322,326,338,340,346,352,356,358
%N Numbers which are prime if each digit is replaced by its 9's complement.
%e 32 belongs to this sequence as its 9's complement is 67, which is prime.
%o (Python)
%o nmax=500
%o def is_prime(num):
%o if num == 0 or num == 1: return(0)
%o for k in range(2, num):
%o if (num % k) == 0:
%o return(0)
%o return(1)
%o def c9(num):
%o s=str(num)
%o l=len(str(num))
%o n=""
%o for k in range(l):
%o n = n+str(9-int(s[k]))
%o return(int(n))
%o ris = ""
%o for i in range(2,nmax):
%o if is_prime(c9(i)):
%o ris = ris+str(i)+","
%o print(ris)
%o (PARI) complement(n) = my(d=digits(n)); for(k=1, #d, d[k]=9-d[k]); subst(Pol(d), x, 10)
%o is(n) = ispseudoprime(complement(n)) \\ _Felix Fröhlich_, Sep 03 2018
%Y Cf. A061601 (9's complement of n).
%K nonn,base
%O 1,1
%A _Pierandrea Formusa_, Sep 03 2018