A059955 a(n) = floor( prime(n)!/lcm(1..prime(n)) ) modulo prime(n).

1, 2, 5, 10, 3, 10, 4, 3, 28, 17, 18, 30, 20, 41, 42, 14, 19, 30, 37, 63, 50, 7, 12, 83, 30, 91, 19, 69, 91, 97, 56, 22, 80, 39, 137, 44, 9, 154, 19, 37, 141, 141, 168, 126, 183, 200, 205, 136, 55, 95, 204, 126, 213, 230, 68, 63, 158, 202, 162, 102, 182, 104, 38, 165

Offset: 2

Author

Lekraj Beedassy, Mar 13 2001

Keywords

• nonn

Comments

a(n) gives also the smallest coefficient for which the multiple M of lcm(1 through p(n)-1) satisfies p(n) divides M + 1. This computes the solution of the puzzle requiring the smallest number such that grouping in 2's, 3's, etc. up to the n-th prime,all leave a remainder of one except the last which leaves no remainder.

Examples

			a(2)=1 because prime(2)=3 and floor(3!/lcm(1,2,3)) mod 3 = 1 mod 3 = 1;
a(3)=2 because prime(3)=5 and floor(5!/lcm(1,2,3,4,5)) mod 5 = 2 mod 5 = 2;
a(4)=5 because prime(4)=7 and floor(7!/lcm(1,2,3,4,5,6,7)) mod 7 = 12 mod 7 = 5;
a(7)=10 because prime(7)=17 and floor(17!/lcm(1,2,...,17)) mod 17 = 29030400 mod 17 = 10.
		

Crossrefs

Cf. A003418, A025527, A099796.

Programs

Magma

[Floor( Factorial(p)/Lcm([1..p]) ) mod p: p in PrimesInInterval(3,400)]; // Bruno Berselli, Feb 08 2015

Maple

for n from 2 to 150 do printf(`%d,`, floor(ithprime(n)!/ilcm(i $ i=1..ithprime(n))) mod ithprime(n) ); od:

Extensions

More terms from James Sellers, Mar 15 2001