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a(1)=1; otherwise, find the first digit from the left in a(n-1) which is 1, 3, 7 or 9. Omitting the digits before this one, we reverse the remaining digits, obtaining s, say. Then a(n) is the smallest prime which ends with s and has not already appeared.
3

%I #47 Nov 07 2023 03:16:42

%S 1,11,211,311,2113,2311,5113,6311,6113,8311,11113,131111,3111131,

%T 21311113,63111131,31311113,31111313,531311113,1131111313,

%U 273131111311,311311113137,5731311113113,6311311113137,12731311113113,2331131111313721

%N a(1)=1; otherwise, find the first digit from the left in a(n-1) which is 1, 3, 7 or 9. Omitting the digits before this one, we reverse the remaining digits, obtaining s, say. Then a(n) is the smallest prime which ends with s and has not already appeared.

%C The sequence is infinite. Indeed, by [Sierpiński] (see also Theorem 21 in [Trost]) for given decimal digits c_1..c_m such that c_m equals 1,3,7 or 9, there are infinitely many primes ending with c_1..c_m.

%D W. Sierpiński, Sur l'existence de nombres premiers avec une suite arbitraire de chiffres initiaux, Le Matematiche Catania, 1951.

%D E. Trost, Primzahlen, Verlag Birkhäuser, 1953, Theorems 20 - 21.

%e Let n=6. Since a(5)=2113, then, omitting 2, we obtain the number 113 whose reverse is s=311. The smallest prime ending with 311 is 2311. So a(6)=2311.

%Y Cf. A000040, A262373.

%K nonn,base

%O 1,2

%A _Vladimir Shevelev_, Sep 20 2015