cp's OEIS Frontend

A116151 a(n) = smallest positive integer x satisfying the system of congruences { x == 1 (mod 2), x == 2 (mod 3), x == 3 (mod 5), x == 5 (mod 7), ..., x == A008578(n) (mod A008578(n+1)) }.

%I A116151 #36 Jul 29 2025 08:38:36
%S A116151 1,5,23,173,2273,2273,452723,6578843,113275433,3682761353,10152454583,
%T A116151 5024164707833,249908523156563,5726413266646343,345878207890067123,
%U A116151 15103232990013860963,1905274424667036455303,111502614383457156882293
%N A116151 a(n) = smallest positive integer x satisfying the system of congruences { x == 1 (mod 2), x == 2 (mod 3), x == 3 (mod 5), x == 5 (mod 7), ..., x == A008578(n) (mod A008578(n+1)) }.
%C A116151 Minimum Chinese Remainder Prime Modulus Ladder: for the n-th term, the number modulus a prime equals the previous prime for the first n primes (the initial term is defined to be 1). - _Fred Schneider_, Oct 21 2007
%H A116151 Harvey P. Dale, <a href="/A116151/b116151.txt">Table of n, a(n) for n = 1..350</a> (First 100 terms from T. D. Noe)
%e A116151 a(3)=23 because that is the smallest number such that n==1 (mod 2), n==2 (mod 3) and n == 3 (mod 5).
%p A116151 Primes:= [1,seq(ithprime(i),i=1..30)]:
%p A116151 seq(chrem(Primes[1..k],Primes[2..k+1]),k=1..30); # _Robert Israel_, Oct 26 2018
%t A116151 Table[ChineseRemainder[Join[{1},Prime[Range[n-1]]],Prime[Range[n]]],{n,20}] (* _Harvey P. Dale_, Mar 30 2018 *)
%o A116151 (PARI) { a(n) = lift(chinese(vector(n,i,Mod(if(i==1,1,prime(i-1)),prime(i))))) }; vector(30,n,a(n)) \\ _Max Alekseyev_, Apr 16 2007
%o A116151 (PARI) my(z=Mod(1,2)); forprime(x=3,100,z=chinese(z,Mod(precprime(x-1),x)); print1(lift(z), ", ")); \\ _Fred Schneider_, Oct 21 2007
%Y A116151 Cf. A070198.
%K A116151 nonn
%O A116151 1,2
%A A116151 Christian Bjartli (cbjartli(AT)gmail.com), Apr 14 2007
%E A116151 More terms from _Max Alekseyev_, Apr 16 2007
%E A116151 Edited by _N. J. A. Sloane_, May 05 2007
%E A116151 Further edited by _N. J. A. Sloane_, Jun 30 2008 at the suggestion of _R. J. Mathar_ and Christian Bjartli.