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A204982
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Least k such that n divides k!! - j!! for some j satisfying 1 <= j < k.
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9
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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, 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
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OFFSET
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1,1
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COMMENTS
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See A204892 for a discussion and guide to related sequences.
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LINKS
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EXAMPLE
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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.
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.
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MATHEMATICA
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s[n_] := s[n] = n!!; z1 = 400; z2 = 60;
Table[s[n], {n, 1, 30}] (* A006882 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}] (* A204912 *)
v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]
w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]
d[n_] := d[n] = First[Delete[w[n], Position[w[n], 0]]]
Table[d[n], {n, 1, z2}] (* A204913 *)
k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]
m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]
j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2
Table[k[n], {n, 1, z2}] (* A204982 *)
Table[j[n], {n, 1, z2}] (* A205100 *)
Table[s[k[n]], {n, 1, z2}] (* A205101 *)
Table[s[j[n]], {n, 1, z2}] (* A205102 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A205103 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A205104 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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