A204982 - OEIS

%I #10 Apr 15 2020 11:41:06

%S 2,3,4,5,4,4,4,6,6,6,6,5,5,5,6,8,7,7,7,6,8,12,6,8,10,7,10,8,14,7,8,10,

%T 6,7,9,10,12,10,9,6,8,8,15,12,6,6,6,8,11,11,7,7,17,10,12,8,7,16,9,9

%N Least k such that n divides k!! - j!! for some j satisfying 1 <= j < k.

%C See A204892 for a discussion and guide to related sequences.

%e Example 1. Using 1!! = 1, 2!! = 2, 3!! = 3, 4!! = 8, we verify that a(5) = 5 as follows: The values of 4!!-j!! for j = 1,2,3 are 7,6,5, respectively, so 5 divides 4!! - 3!!, and so for k = 4 there is a number j as required.  On the other hand, it is easy to check that for k = 1,2,3, there is no such j.

%e Example 2. To see that a(6) = 4, we already noted that 6 divides 4!!-2!! in Example 1, and it is easy to check that for k = 1,2,3, the number 6 does not divide k!! - j!! for any j satisfying 1 <=j < k.

%t s[n_] := s[n] = n!!; z1 = 400; z2 = 60;

%t Table[s[n], {n, 1, 30}]  (* A006882 *)

%t u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]

%t Table[u[m], {m, 1, z1}]  (* A204912 *)

%t v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]

%t w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]

%t d[n_] := d[n] = First[Delete[w[n], Position[w[n], 0]]]

%t Table[d[n], {n, 1, z2}]     (* A204913 *)

%t k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]

%t m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]

%t j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2

%t Table[k[n], {n, 1, z2}]     (* A204982 *)

%t Table[j[n], {n, 1, z2}]     (* A205100 *)

%t Table[s[k[n]], {n, 1, z2}]  (* A205101 *)

%t Table[s[j[n]], {n, 1, z2}]  (* A205102 *)

%t Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A205103 *)

%t Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A205104 *)

%Y Cf. A006882, A204892, A205104.

%K nonn

%O 1,1

%A _Clark Kimberling_, Jan 22 2012

%E Edited by _Clark Kimberling_, Apr 15 2020