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%I #6 Sep 13 2017 00:03:46
%S 11,13,17,31,37,71,73,79,97,127,163,181,211,257,271,277,293,307,349,
%T 367,431,433,457,491,521,523,541,563,587,631,659,743,839,983,10069,
%U 10151,10337,10429,10559,10889,10973,11059,11251,11329,11411,11437,11471,11617
%N Prime numbers n such that replacing each digit in the decimal expansion of n with the sum of the other digits produces a prime.
%C Let d(1)d(2)...d(q) denote the decimal expansion of a prime number. Replace each digit d(i) in decimal expansion of n with Sum_{j=1..q, j<>i} d(j) such that the result is a prime number.
%C The corresponding primes are 11, 31, 71, 13, 73, 17, 37, 97, 79, 983, 947, 929, 233, 1297, 839, 1499, ...
%e 983 is in the sequence because 983 becomes 111217 which is also prime, where 11=8+3, 12=9+3 and 17=9+8.
%p with(numtheory):T:=array(1..10):
%p for n from 1 to 1000 do:
%p p:=ithprime(n):k:=0:s:=0:j:=0:
%p x:=convert(p,base,10):n1:=nops(x):
%p s1:=sum('x[i]', 'i'=1..n1):
%p for m from n1 by -1 to 1 do:
%p k:=k+1:T[k]:=s1-x[m]:
%p od:
%p n2:=sum('length(T[j])', 'j'=1..n1):s2:=0:
%p for u from 1 to n1 do:
%p s2:=s2+ T[u]*10^(n2-length(T[u])):n2:=n2-length(T[u]):
%p od:
%p if type(s2,prime)=true
%p then
%p printf(`%d, `,p):
%p else
%p fi:
%p od:
%K nonn,base
%O 1,1
%A _Michel Lagneau_, Apr 08 2014
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