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A174414
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Smallest natural n = n(k) that concatenation (n+k)//n (k = 1, 2, ...) is a prime number.
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1
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3, 1, 1, 3, 1, 1, 3, 3, 1, 9, 19, 1, 3, 1, 7, 3, 1, 1, 3, 1, 13, 17, 9, 1, 3, 1, 1, 3, 7, 1, 9, 1, 23, 3, 3, 19, 17, 7, 1, 3, 1, 1, 3, 21, 1, 11, 3, 1, 3, 7, 1, 9, 1, 7, 21, 1, 7, 3, 1, 7, 3, 1, 1, 3, 1, 17, 9, 1, 1, 3, 3, 7, 9, 1, 1, 9, 13, 7, 3, 1, 1, 3, 3, 11, 3, 7, 1, 27, 7, 1, 9, 3, 1, 9, 3, 1, 9, 1
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OFFSET
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1,1
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COMMENTS
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10^d*(n+k) + n has to be a prime for the smallest d-digit natural n, (k = 1, 2, ...)
n of course has necessarily end digit 1, 3, 7 or 9
Sequence is infinite because of Dirichlet's prime number theorem: (10^d+1) * n + c with constant c = 10^d * k
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REFERENCES
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Theo Kempermann, Zahlentheoretische Kostproben, Harri Deutsch, 2. aktualisierte Auflage 2005
Helmut Kracke, Mathe-musische Knobelisken, Duemmler Bonn, 2. Auflage 1983
Hugo Steinhaus: Studentenfutter, Urania-Verlag Leipzig-Jena-Berlin, 1991
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LINKS
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Table of n, a(n) for n=1..98.
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EXAMPLE
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n=0: 11 = prime(5) = (1+0)//1 is omitted because 0 is no natural
43 = prime(14) = (3+1)//3, n(1) = 3
31 = prime(11) = (1+2)//1, n(2) = 1
41 = prime(13) = (1+3)//1, n(3) = 1
3413 = prime(480) = (13+21)//13, n(21) = 13
11527 = prime(1390) = (27+88)//27, n(88) = 27
Note cases where consecutive values of k give consecutive primes:
k=17: 181 = prime(42) = (1+17)//1, k=18: 191 = prime(43) = (1+18)//1
k=41: 421 = prime(82) = (1+41)//1, k=42: 431 = prime(83) = (1+42)//1
... are there infinitely many of such?
n(11) = 19, 3019 is a resulting "candidate" for k = 301 - 9 = 292, but n(292) = 3 gives 2953 = prime(425)
First twice resulting prime is 5623 = prime(739) = (23+33)//23 = 5623 = (559+3)//3
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CROSSREFS
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A089712, A171154, A173976, A174031
Sequence in context: A049653 A060266 A073310 * A046929 A068695 A110787
Adjacent sequences: A174411 A174412 A174413 * A174415 A174416 A174417
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KEYWORD
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base,nonn,uned
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AUTHOR
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Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 19 2010
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STATUS
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approved
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