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%I #22 Mar 29 2023 10:58:25
%S 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,
%T 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,
%U 14,12,11,9,6,5,2,20,19,18,16,15,13,12,10,7,6,3,1,22,21
%N (A204898)/2 = (prime(k)-prime(j))/2; A086802 without its zeros.
%C 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 c....k..........j..........p(k)-p(j).[p(k)-p(j)]/c
%C 2....A133196....A131818....A204898....A205558
%C 3....A205560....A205547....A205557....A205675
%C 4....A205677....A205678....A205681....A205682
%C 5....A205684....A205685....A205688....A205689
%C 6....A205691....A205692....A205695....A205696
%C 7....A205698....A205699....A205702....A205703
%C 8....A205705....A205706....A205709....A205710
%C 9....A205712....A205713....A205716....A205717
%C 10...A205720....A205721....A205724....A205725
%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
%e Writing prime(k) as p(k),
%e p(3)-p(2)=5-3=2
%e p(4)-p(2)=7-3=4
%e p(4)-p(3)=7-5=2
%e p(5)-p(2)=11-3=8
%e p(5)-p(3)=11-5=6
%e p(5)-p(4)=11-7=4,
%e so that the first 6 terms of A205558 are 1,2,1,4,3,2.
%e 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:
%e 1...2...4...5...7...8....10...13...14...17...19...20
%e 1...3...4...6...7...9....12...13...16...18...19...21
%e 2...3...5...6...8...11...12...15...17...18...20...23
%e 1...3...4...6...9...10...13...15...16...18...21...24
%e 2...3...5...8...9...12...14...15...17...20...23...24
%e 1...3...6...7...10..12...13...15...18...21...22...25
%e 2...5...6...9...11..12...14...17...20...21...24...26
%e - _Clark Kimberling_, Sep 29 2013
%t s[n_] := s[n] = Prime[n]; z1 = 200; z2 = 80;
%t f[n_] := f[n] = Floor[(-1 + Sqrt[8 n - 7])/2];
%t Table[s[n], {n, 1, 30}] (* A000040 *)
%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}] (* A204890 *)
%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] = Delete[w[n], Position[w[n], 0]]
%t c = 2; t = d[c] (* A080036 *)
%t k[n_] := k[n] = Floor[(3 + Sqrt[8 t[[n]] - 1])/2]
%t j[n_] := j[n] = t[[n]] - f[t][[n]] (f[t[[n]]] + 1)/2
%t Table[k[n], {n, 1, z2}] (* A133196 *)
%t Table[j[n], {n, 1, z2}] (* A131818 *)
%t Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A204898 *)
%t Table[(s[k[n]] - s[j[n]])/c, {n, 1, z2}] (* A205558 *)
%Y Cf. A205675, A205560, A204892.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jan 30 2012
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