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)) }.

1, 5, 23, 173, 2273, 2273, 452723, 6578843, 113275433, 3682761353, 10152454583, 5024164707833, 249908523156563, 5726413266646343, 345878207890067123, 15103232990013860963, 1905274424667036455303, 111502614383457156882293

Offset: 1

Author

Christian Bjartli (cbjartli(AT)gmail.com), Apr 14 2007

Keywords

• nonn

Comments

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

Examples

			a(3)=23 because that is the smallest number such that n==1 (mod 2), n==2 (mod 3) and n == 3 (mod 5).
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..350 (First 100 terms from T. D. Noe)

Crossrefs

Cf. A070198.

Programs

Maple

Primes:= [1,seq(ithprime(i),i=1..30)]:
seq(chrem(Primes[1..k],Primes[2..k+1]),k=1..30); # Robert Israel, Oct 26 2018

Mathematica

Table[ChineseRemainder[Join[{1},Prime[Range[n-1]]],Prime[Range[n]]],{n,20}] (* Harvey P. Dale, Mar 30 2018 *)

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

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

Extensions

More terms from Max Alekseyev, Apr 16 2007

Edited by N. J. A. Sloane, May 05 2007

Further edited by N. J. A. Sloane, Jun 30 2008 at the suggestion of R. J. Mathar and Christian Bjartli.