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Take n-th prime p, let P = all primes having same digits; a(n) = q-p where q is smallest prime in P >p if q exists; otherwise a(n) = p-r where r is largest prime in P <p if r exists; otherwise a(n) = 0.
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%I #4 Mar 31 2012 10:32:38

%S 0,0,0,0,0,18,54,0,0,0,18,36,0,0,0,0,0,0,0,54,36,18,0,0,18,0,0,594,0,

%T 18,144,180,36,54,270,0,414,450,450,144,18,630,720,54,522,720,0,0,0,0,

%U 0,54,180,270,0,0,0,144,450,540,540,54,0,180,18,144,18,36,396,90,0,234

%N Take n-th prime p, let P = all primes having same digits; a(n) = q-p where q is smallest prime in P >p if q exists; otherwise a(n) = p-r where r is largest prime in P <p if r exists; otherwise a(n) = 0.

%C The primes in P are required to have the same number of digits as p; thus internal 0's must remain internal 0's.

%e a(41)=18 because the 41st prime is 179. The primes having these digits are 179, 197, 719 and 971. The distance from 179 to 197 = 18.

%Y Cf. A052902, A052998, A052999, A053544, A052494, A052507.

%K base,easy,nonn

%O 1,6

%A _Enoch Haga_, Mar 16 2000