A319404 a(n) is the period of the periodic k-sequence q_k=lcm(k+1,k+2,...,k+n)/(n*binomial(k+n,n)).

1, 1, 2, 3, 12, 20, 60, 105, 280, 504, 2520, 27720, 27720, 51480, 72072, 45045, 720720, 1361360, 12252240, 46558512, 33256080, 21162960, 232792560, 5354228880, 1070845776, 2059318800, 2974571600, 11473347600, 80313433200, 2329089562800, 2329089562800, 4512611027925

Offset: 1

Author

René Gy, Sep 18 2018

Keywords

• nonn

Comments

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).

a(n) is a divisor of A003418(n-1) and a multiple of A003418(n)/n.

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.

a(n) = A003418(n-1) iff n is a mutinous number or n is a prime number.

a(n) = A003418(n) iff n is a mutinous number.

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).

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.

a(n+j) is a multiple of binomial(n+j-1,j).

All these statements require proofs.

Examples

			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.
		

Crossrefs

Cf. A003418, A027854.

Programs

Mathematica

ll2[n0_, m0_] :=
Module[{f, g, i, n = n0, m = m0}, g = 1;
  If[1 <= m <= n, Do[f = LCM[g, n - i]; g = f, {i, 0, m - 1}], f = 1];f]
list3 = {1};
Do[i = 0; ll = ll2[m, m]/m; b = {1, ll };a = {0, 0 };
  While[ a != b, i = i + ll;
   a = { ll2[m + i - 1, m]/(m*Binomial[m + i - 1, m]), ll2[m + i, m]/(
     m*Binomial[m + i, m])}]; AppendTo[list3, i], {m, 2, 50}]; Print[list3]