A229222 - OEIS

%I #19 Sep 20 2013 10:46:03

%S 2,23,113,23,191,223,191,41,223,113,157,191,137,113,113,43,137,191,

%T 179,337,577,223,227,113,263,113,199,229,263,199,467,89,223,179,223,

%U 113,443,683,1279,337,661,463,827,2281,577,223,223,661,137,229,11399,461,577

%N Smallest prime p such that p contains a digit larger than 1 and the sum of the n-th powers of the decimal digits of p is a prime number.

%C We impose the condition that p is not in A020449 in order to avoid trivial sequences with infinite repetitions with the numbers 11 if p>1, or 101 if p>11, or 101111 if p > 101, ... for example if p > 1 the sequence is {2, 11, 11, 11, ...}, if p > 11 the sequence is {23, 23, 101, 23, 101, 101, 41, 101, 101, 101, 101, 101, ...}.

%C a(n) is an unification of a family of sequences mentioned hereafter:

%C A082101: primes of the form 2^n+3^n => 23 is in the sequence;

%C A057735: primes of the form 3^n+2 => 113 is in the sequence;

%C A153133: primes of the form 2^n+3^(n-1) => 223 is in the sequence;

%C A228034: primes of the form 9^n+2 => 191 is in the sequence;

%C A057733: primes of the form 2^n+3 => 2111 is in the sequence;

%C A228026: primes of the form 4^n+3 => 4111 is in the sequence;

%C A228034: primes of the form 9^n+2 => 191 is in the sequence;

%C A182330: primes of the form 5^n+2 => 151 is in the sequence;

%C A111974: primes of the form 2*3^n+1 => 313 is in the sequence;

%C A102903: primes of the form 3^n+4 => 11113 is in the sequence.

%C In this sequence, we observe repetitions of numbers such that 23, 113, 223, 191, 199, 223,... and this problem is very difficult, because it is probable that there exists both finite and infinite repetitions according to the numbers: for example, if we consider the number 23 of this sequence, it is probable that the number of element "23" is finite (see the comment in A082101 for the primes of form 2^k + 3^k). But, if we consider the number 113 of this sequence, is the number of the elements "113" infinite ? (see A057735 with the primes of the form 2+3^n). We observe that a(n) = 113 for n = 3, 14, 15, 24, 26,..., 123, 126, 139,..., 386, 391, 494, ....

%H Michel Lagneau, <a href="/A229222/b229222.txt">Table of n, a(n) for n = 1..500</a>

%e a(3) = 113 because 1^3+1^3+3^3 = 29 is prime.

%p with(numtheory) :lst:={11, 101, 101111, 10011101, 10101101, 10110011, 10111001, 11000111, 11100101, 11110111, 11111101 }:for n from 1 to 300 do :ii:=0:for k from 1 to 10^8 while(ii=0) do:x:=convert(k,base,10):n1:=nops(x):it:=0:jj:=0:s:= sum('x[i]^n', 'i'=1..n1):lst1:={k} intersect lst:if type(k,prime)=true and type(s,prime)=true and (lst1<>{k}) then ii:=1: printf(`%d, `,k):else fi:od:od:

%t Table[p = 2; While[d = IntegerDigits[p]; Union[d][[-1]] < 2 || ! PrimeQ[Total[d^n]],  p = NextPrime[p]]; p, {n, 60}]

%o (PARI) a(n)=forprime(p=2,,my(d=digits(p)); if(vecmax(d)>1 && isprime(sum(i=1,#d,d[i]^n)), return(p))) \\ _Charles R Greathouse IV_, Sep 19 2013

%Y Cf. A020449.

%K nonn,base

%O 1,1

%A _Michel Lagneau_, Sep 16 2013