A078502 - OEIS

%I #36 Nov 16 2019 01:43:41

%S 3,6,12,12,174600,7224840,10780560,10780560,1086338816640,

%T 50060257410240,7720634052774720,227457297898150320,

%U 7272877497848202240,7272877497848202240

%N a(n) = smallest positive integer N such that (N - k)/k is prime for k = 1, 2, ..., n.

%C The idea for the sequence and first eleven terms are from Ken Wilke.

%C a(n) == 0 (mod 120) for n > 4: because a(n)/2, a(n)/3, a(n)/4 and a(n)/5 must be integer, a(n) == 0 (mod 60); and if a(n) == 60 (mod 120), (a(n)-4)/4 == 14 (mod 120) would not be prime; thus a(n) == 0 (mod 120). A more general result is a(n) == 0 (mod lcm(1,2,...,n)) for all n >= 1, and a(n) == 0 (mod 2*lcm(1,2,...,n)) for n > 4. - _Jean-Christophe Hervé_, Sep 15 2014

%H Carlos Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_206.htm">Puzzle 206. (N-k)/k Primes</a>, Prime Puzzles and Problems Connection.

%H Carlos Rivera, <a href="https://www.primepuzzles.net/puzzles/puzz_977.htm">Puzzle 977. A special set of primes</a>, Prime Puzzles and Problems Connection.

%H G. L. Honaker, Jr. and Chris Caldwell, <a href="https://primes.utm.edu/curios/page.php/7272877497848202240.html">Prime Curios! 7272877497848202240</a>

%F a(n) == 0 (mod A003418(n)) because of the divisibility condition (A003418(n) = lcm(1,2,...,n)). - _Jean-Christophe Hervé_, Sep 15 2014

%e (12-k)/k is prime for k = 1,2,3,4 and 12 is the smallest positive integer satisfying this property. Hence a(4) = 12.

%o (PARI)

%o a(n)=k=1; while(k,c=0; for(i=1,n,if(k%i==0&&isprime(k/i-1),c++)); if(c==n,return(k));k++)

%o n=1;while(n<10,print1(a(n),", ");n++) \\ _Derek Orr_, Sep 15 2014

%Y See A093554 for another version.

%Y Cf. A074200 (equivalent sequence for (N+k)/k prime).

%K nonn,more

%O 1,1

%A _Joseph L. Pe_, Jan 05 2003

%E Corrected and extended by _Jens Kruse Andersen_, Jan 10 2003