

A174583


Smallest natural n = n(k) that concatenation (nk)//n (k = 0, 1, 2, ...) is a prime number.


1



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, 27, 27, 29, 31, 33, 31, 33, 39, 47, 37, 39, 37, 39, 39, 41, 51, 47, 47, 61, 47, 49, 49, 53, 49, 51, 51, 59, 57, 57, 61, 57, 57, 61, 63, 63, 71, 63, 63, 67, 67, 77, 67, 69, 77, 71, 73, 71
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OFFSET

1,2


COMMENTS

See comments and references for A174414
10^d*(nk) + n has to be a prime for the smallest ddigit natural n > k (k = 0, 1, 2, ...)
For (nk)//n to be a prime n of course has necessarily end digit 1, 3, 7 or 9
It is conjectured that n(k) = n(k+1) for an infinite number of k's
As n > k the number N of a "possible" n(k) is finite, it is simple to give an upper bound for N
1, 11, ... appear only one time; 3, 9, 13, 19, 21, 23, ...: 2 times; 7, 17: 3 times; ...
Does EACH "possible" candidate n (naturals with end digit 1, 3, 7 or 9) "appear" in the sequence?
Note this interesting "phenomenon", first "occurrence" for case k = 84:
9291013 = prime(620602) = (101384)//1013, n(84) = 1013
2nd such: 9381037 = prime(626219) = (103799)//1037
Let k be a multiple of 7, 11 or 13 => no 3digit n with property (nk)//n a prime exists
Proof: 10^3*(n  k) + n = n * (10^3+1)  k * 10^3 = 7 * 11 * 13 * n  k * 10^3,
not prime as k is a multiple of 7, 11 or 13
Same "phenomenon" for kdigit n with appointed divisors and k > 3: 10^4+1 = 73 * 137, 10^5+1 = 11 * 9091, ...


LINKS

Table of n, a(n) for n=1..72.


EXAMPLE

11 = prime(5) = (10)//1, n(0) = 1
23 = prime(9) = (31)//3, n(1) = 3
13 = prime(6) = (32)//3, n(2) = 3
139 = prime(34) = (3938)//39, n(38) = 39
9109 = prime(1130) = (109100)//109, n(100) = 109


CROSSREFS

A089712, A171154, A173976, A174031, A174414
Sequence in context: A146450 A233810 A263869 * A226781 A147144 A152113
Adjacent sequences: A174580 A174581 A174582 * A174584 A174585 A174586


KEYWORD

base,nonn,uned


AUTHOR

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 23 2010


STATUS

approved



