OFFSET
1,2
COMMENTS
Let p(n) denote the n-th prime. If c is a positive integer, there are infinitely many pairs (k,j) such that c divides p(k)-p(j). The set of differences p(k)-p(j) is ordered as a sequence at A204890. Guide to related sequences:
c....k..........j..........p(k)-p(j).[p(k)-p(j)]/c
It appears that, as rectangular array, this sequence can be described by A(n,k) is the least m such that there are k primes in the set prime(n) + 2*i for {i=1..n}. - Michel Marcus, Mar 29 2023
EXAMPLE
Writing prime(k) as p(k),
p(3)-p(2)=5-3=2
p(4)-p(2)=7-3=4
p(4)-p(3)=7-5=2
p(5)-p(2)=11-3=8
p(5)-p(3)=11-5=6
p(5)-p(4)=11-7=4,
so that the first 6 terms of A205558 are 1,2,1,4,3,2.
The sequence can be regarded as a rectangular array in which row n is given by [prime(n+2+k)-prime(n+1)]/2; a northwest corner follows:
1...2...4...5...7...8....10...13...14...17...19...20
1...3...4...6...7...9....12...13...16...18...19...21
2...3...5...6...8...11...12...15...17...18...20...23
1...3...4...6...9...10...13...15...16...18...21...24
2...3...5...8...9...12...14...15...17...20...23...24
1...3...6...7...10..12...13...15...18...21...22...25
2...5...6...9...11..12...14...17...20...21...24...26
- Clark Kimberling, Sep 29 2013
MATHEMATICA
s[n_] := s[n] = Prime[n]; z1 = 200; z2 = 80;
f[n_] := f[n] = Floor[(-1 + Sqrt[8 n - 7])/2];
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] = Delete[w[n], Position[w[n], 0]]
c = 2; t = d[c] (* A080036 *)
k[n_] := k[n] = Floor[(3 + Sqrt[8 t[[n]] - 1])/2]
j[n_] := j[n] = t[[n]] - f[t][[n]] (f[t[[n]]] + 1)/2
Table[k[n], {n, 1, z2}] (* A133196 *)
Table[j[n], {n, 1, z2}] (* A131818 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A204898 *)
Table[(s[k[n]] - s[j[n]])/c, {n, 1, z2}] (* A205558 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 30 2012
STATUS
approved