%I #15 Jan 15 2023 18:43:02
%S 11,13,12,19,15,33,14,17,16,31,21,37,23,91,24,93,18,39,22,99,25,331,
%T 26,97,29,371,34,71,36,77,41,73,32,79,35,131,27,311,43,111,44,112,132,
%U 133,28,332,134,113,38,135,136,114,115,137,46,116,117,49,171,47
%N a(n) is the least integer > 9 not listed earlier whose first two digits together with a(n1)'s last two digits form two 2digit primes when "read vertically in the same direction"; a(1) = 11.
%C To explain "read vertically in the same direction": let b, c be the last two digits of a(n1), and d, e the first two digits of a(n), then either both of 10b + d and 10c + e, or both of 10d + a and 10e + b must be primes > 9.
%C We are tempted to conjecture that this sequence contains all integers > 10 which don't have '0' as second or one of the last two digits.
%H Éric Angelini, <a href="http://cinquantesignes.blogspot.com/2022/05/primebayonets.html">Prime bayonets</a>, personal web site "Cinquante signes" on blogspot.com, May 03 2022.
%e a(2) together with a(3) form the two 2digit primes 11 and 23, when we write
%e a(2) = 13 and read the two last columns
%e a(3) = 12 both from bottom to top.
%e a(11) = 21 is followed by 37 and not 34 since we cannot write
%e a(11) = 21 and read one column downwards to get the prime 23
%e next = 34 and the other column upwards to get the prime 41.
%o (PARI) u=[a=11]; (ok(x,y)=isprime(x[1]*10+y[1])&&isprime(x[2]*10+y[2])); vector(99, n, if(a>u[1]+1, u=setunion(u, [a]), u[1]=a; while(#u>1 && u[2]==u[1]+1+(u[1]%10==9), u=u[^1])); a=digits(a)[2..1]; my(k=u[1], d); n>1&& until(k++%10 && k%100>9&& !setsearch(u, k) && (d=divrem(k\10^logint(k\10,10),10))[2]&& (ok(a,d)ok(d,a)),); a=k)
%Y Cf. A350831 for a simpler singledigit variant.
%K nonn,base
%O 1,1
%A _M. F. Hasler_ and _Eric Angelini_, May 03 2022
