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A205558
(A204898)/2 = (prime(k)-prime(j))/2; A086802 without its zeros.
59
1, 2, 1, 4, 3, 2, 5, 4, 3, 1, 7, 6, 5, 3, 2, 8, 7, 6, 4, 3, 1, 10, 9, 8, 6, 5, 3, 2, 13, 12, 11, 9, 8, 6, 5, 3, 14, 13, 12, 10, 9, 7, 6, 4, 1, 17, 16, 15, 13, 12, 10, 9, 7, 4, 3, 19, 18, 17, 15, 14, 12, 11, 9, 6, 5, 2, 20, 19, 18, 16, 15, 13, 12, 10, 7, 6, 3, 1, 22, 21
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