%I A059955
%S A059955 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,
%T A059955 91,19,69,91,97,56,22,80,39,137,44,9,154,19,37,141,141,168,126,183,200,
%U A059955 205,136,55,95,204,126,213,230,68,63,158,202,162,102,182,104,38,165
%N A059955 a(n)= [p(n)!/lcm{1,2,...,p(n)}] modulo[p(n)], where p(n) = n-th prime.
%C A059955 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.
%e A059955 a(7)=10 because p(7)=17 and [17!/lcm(1,2,...,17)] mod 17 = 29030400 mod 
               17 = 10
%p A059955 for n from 2 to 150 do printf(`%d,`, floor(ithprime(n)!/ilcm(i $ i=1..ithprime(n))) 
               mod ithprime(n) ); od:
%Y A059955 Cf. A025527, A003418.
%Y A059955 Sequence in context: A109469 A083460 A125974 this_sequence A099796 A022831 
               A064365
%Y A059955 Adjacent sequences: A059952 A059953 A059954 this_sequence A059956 A059957 
               A059958
%K A059955 nonn
%O A059955 2,2
%A A059955 Lekraj Beedassy (blekraj(AT)yahoo.com), Mar 13 2001
%E A059955 More terms from James A. Sellers (sellersj(AT)math.psu.edu), Mar 15 2001