A174583 Smallest natural n = n(k) that concatenation (n-k)//n (k = 0, 1, 2, ...) is a prime number.

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

Offset

1,2

Comments

See comments and references for A174414

10^d*(n-k) + n has to be a prime for the smallest d-digit natural n > k (k = 0, 1, 2, ...)

For (n-k)//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) = (1013-84)//1013, n(84) = 1013

2nd such: 9381037 = prime(626219) = (1037-99)//1037

Let k be a multiple of 7, 11 or 13 => no 3-digit n with property (n-k)//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 k-digit 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) = (1-0)//1, n(0) = 1
23 = prime(9) = (3-1)//3, n(1) = 3
13 = prime(6) = (3-2)//3, n(2) = 3
139 = prime(34) = (39-38)//39, n(38) = 39
9109 = prime(1130) = (109-100)//109, n(100) = 109

Crossrefs

A089712, A171154, A173976, A174031, A174414

Sequence in context: A146450 A233810 A263869 * A370202 A226781 A147144

Adjacent sequences: A174580 A174581 A174582 * A174584 A174585 A174586

Keyword

base,nonn,uned

Author

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

Status

approved