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%I #35 Sep 27 2018 12:36:22
%S 1,1,2,3,12,20,60,105,280,504,2520,27720,27720,51480,72072,45045,
%T 720720,1361360,12252240,46558512,33256080,21162960,232792560,
%U 5354228880,1070845776,2059318800,2974571600,11473347600,80313433200,2329089562800,2329089562800,4512611027925
%N a(n) is the period of the periodic k-sequence q_k=lcm(k+1,k+2,...,k+n)/(n*binomial(k+n,n)).
%C For n>0, k>=0, the k-sequence q_k=lcm(k+1,k+2,...,k+n)/(n*binomial(k+n,n)) is a periodic integer sequence with period a(n).
%C a(n) is a divisor of A003418(n-1) and a multiple of A003418(n)/n.
%C a(n) = A003418(n-1) if n is a member of A027854 (a mutinous number), otherwise a(n) = A003418(n)/q^v where q^v is the highest prime power which divides n.
%C a(n) = A003418(n-1) iff n is a mutinous number or n is a prime number.
%C a(n) = A003418(n) iff n is a mutinous number.
%C lcm(k+1,k+2,...,k+n)/(n*binomial(k+n,n)) is a divisor of lcm(1,2,...,n)/n, therefore a(n) is also the period of the periodic k-sequence r_k= binomial(k+n,n)*lcm(1,2,...,n)/lcm(k+1,k+2,...,k+n).
%C Let g be the smallest multiple of A003418(n)/n such that r_g=r_0=1 and r_{g+1}=r_1=gcd(m+1,A003418(n)), then a(n)=g.
%C a(n+j) is a multiple of binomial(n+j-1,j).
%C All these statements require proofs.
%e For n = 5, a(5) = 12 since from k>=0, we have lcm(k+1,k+2,k+3,k+4,k+5)/5/binomial(k+5,5) = 12,2,4,3,4,2,12,1,4,6,4,1,12,2,4,3,4,2,12,1,4,6,4,1,12,2,4,3,4,2,12,1,4,6,4,1,12,..., etc. a periodic sequence of period 12.
%t ll2[n0_, m0_] :=
%t Module[{f, g, i, n = n0, m = m0}, g = 1;
%t If[1 <= m <= n, Do[f = LCM[g, n - i]; g = f, {i, 0, m - 1}], f = 1];f]
%t list3 = {1};
%t Do[i = 0; ll = ll2[m, m]/m; b = {1, ll };a = {0, 0 };
%t While[ a != b, i = i + ll;
%t a = { ll2[m + i - 1, m]/(m*Binomial[m + i - 1, m]), ll2[m + i, m]/(
%t m*Binomial[m + i, m])}]; AppendTo[list3, i], {m, 2, 50}]; Print[list3]
%Y Cf. A003418, A027854.
%K nonn
%O 1,3
%A _René Gy_, Sep 18 2018
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