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A059955 a(n) = floor( prime(n)!/lcm(1..prime(n)) ) modulo prime(n). 1
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

2,2

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.

LINKS

Table of n, a(n) for n=2..65.

EXAMPLE

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.

MAPLE

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

PROG

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

CROSSREFS

Cf. A003418, A025527, A099796.

Sequence in context: A306679 A125974 A218336 * A099796 A022831 A064365

Adjacent sequences:  A059952 A059953 A059954 * A059956 A059957 A059958

KEYWORD

nonn

AUTHOR

Lekraj Beedassy, Mar 13 2001

EXTENSIONS

More terms from James A. Sellers, Mar 15 2001

STATUS

approved

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Last modified October 16 16:08 EDT 2019. Contains 328101 sequences. (Running on oeis4.)