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%I #2 Jul 11 2010 03:00:00
%S 1,3,3,7,7,17,7,9,9,11,13,17,13,19,17,19,21,21,23,27,27,23,43,33,41,
%T 27,27,29,31,33,31,33,39,47,37,39,37,39,39,41,51,47,47,61,47,49,49,53,
%U 49,51,51,59,57,57,61,57,57,61,63,63,71,63,63,67,67,77,67,69,77,71,73,71
%N Smallest natural n = n(k) that concatenation (n-k)//n (k = 0, 1, 2, ...) is a prime number.
%C See comments and references for A174414
%C 10^d*(n-k) + n has to be a prime for the smallest d-digit natural n > k (k = 0, 1, 2, ...)
%C For (n-k)//n to be a prime n of course has necessarily end digit 1, 3, 7 or 9
%C It is conjectured that n(k) = n(k+1) for an infinite number of k's
%C As n > k the number N of a "possible" n(k) is finite, it is simple to give an upper bound for N
%C 1, 11, ... appear only one time; 3, 9, 13, 19, 21, 23, ...: 2 times; 7, 17: 3 times; ...
%C Does EACH "possible" candidate n (naturals with end digit 1, 3, 7 or 9) "appear" in the sequence?
%C Note this interesting "phenomenon", first "occurrence" for case k = 84:
%C 9291013 = prime(620602) = (1013-84)//1013, n(84) = 1013
%C 2nd such: 9381037 = prime(626219) = (1037-99)//1037
%C Let k be a multiple of 7, 11 or 13 => no 3-digit n with property (n-k)//n a prime exists
%C Proof: 10^3*(n - k) + n = n * (10^3+1) - k * 10^3 = 7 * 11 * 13 * n - k * 10^3,
%C not prime as k is a multiple of 7, 11 or 13
%C Same "phenomenon" for k-digit n with appointed divisors and k > 3: 10^4+1 = 73 * 137, 10^5+1 = 11 * 9091, ...
%e 11 = prime(5) = (1-0)//1, n(0) = 1
%e 23 = prime(9) = (3-1)//3, n(1) = 3
%e 13 = prime(6) = (3-2)//3, n(2) = 3
%e 139 = prime(34) = (39-38)//39, n(38) = 39
%e 9109 = prime(1130) = (109-100)//109, n(100) = 109
%Y A089712, A171154, A173976, A174031, A174414
%K base,nonn,uned
%O 1,2
%A Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 23 2010
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