This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
(* See the program at A204982 *)
Let s(k)=prime(k). As in A204890, the ordering of differences s(k)-s(j), follows from the arrangement shown here: k...........1..2..3..4..5...6...7...8...9 s(k)........2..3..5..7..11..13..17..19..23 ... s(k)-s(1)......1..3..5..9..11..15..17..21..27 s(k)-s(2).........2..4..8..10..14..16..20..26 s(k)-s(3)............2..6..8...12..14..18..24 s(k)-s(4)...............4..6...10..12..16..22 ... least (k,j) such that 1 divides s(k)-s(j) for some j is (2,1), so a(1)=2. least (k,j) such that 2 divides s(k)-s(j): (3,2), so a(2)=3. least (k,j) such that 3 divides s(k)-s(j): (3,1), so a(3)=3.
s[n_] := s[n] = Prime[n]; z1 = 400; z2 = 50;
Table[s[n], {n, 1, 30}] (* A000040 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j],
{k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}] (* A204890 *)
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}] (* A204891 *)
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}] (* A204892 *)
Table[j[n], {n, 1, z2}] (* A204893 *)
Table[s[k[n]], {n, 1, z2}] (* A204894 *)
Table[s[j[n]], {n, 1, z2}] (* A204895 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A204896 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204897 *)
(* Program 2: generates A204892 and A204893 rapidly *)
s = Array[Prime[#] &, 120];
lk = Table[NestWhile[# + 1 &, 1, Min[Table[Mod[s[[#]] - s[[j]], z], {j, 1, # - 1}]] =!= 0 &], {z, 1, Length[s]}]
Table[NestWhile[# + 1 &, 1, Mod[s[[lk[[j]]]] - s[[#]], j] =!= 0 &], {j, 1, Length[lk]}]
(* Peter J. C. Moses, Jan 27 2012 *)
a(n)=forprime(p=n+2,,forstep(k=p%n,p-1,n,if(isprime(k), return(primepi(p))))) \\ Charles R Greathouse IV, Mar 20 2013
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.
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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