cp's OEIS Frontend

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.

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A291528 First term s_n(1) of equivalence classes of prime sequences {s_n(k)} for k > 0 derived by records of first differences of Rowland-like recurrences with increasing even starting values e(n) >= 4.

Original entry on oeis.org

2, 7, 17, 19, 31, 43, 53, 71, 67, 79, 97, 103, 109, 113, 127, 137, 151, 163, 181, 173, 191, 197, 199, 211, 229, 239, 241, 251, 269, 257, 271, 283, 293, 317, 331, 337, 349, 367, 373, 419, 409, 431, 433, 439, 443, 463, 491, 487, 499, 523, 557, 547, 577, 593, 607, 599, 601
Offset: 1

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Author

Ralf Steiner, Aug 25 2017

Keywords

Comments

These kinds of equivalence classes {s_n(k)} were defined by Shevelev, see Crossrefs.
Some equivalence classes of prime sequences {s_n(k)} have the same tail for a constant C_n < k, such as {s_2(k)} = {a(2),...} = {7,13,29,59,131,...} and {s_5(k)} = {a(5),...} = {31,61,131,...} with common tail {131,...}. Thus it seems that all terms are leaves of a kind of an inverse prime-tree with branches in A291620 and the root at infinity.
In each equivalence class {s_n(k)} the terms hold: s_n(k+1)-2*s_n(k) >= -1;
(s_n(k+1)+1)/s_n(k) >= 2; lim_{k -> inf} (s_n(k+1)+1)/s_n(k) = 2.

Examples

			For n=1 the Rowland recurrence with e(1)=4 is A084662 with first differences A134734 and records {2,3,5,11,...} gives the least new prime a(1)=2 as the first term of a first equivalence class {2,3,5,11,...} of prime sequences.
For n=2 with e(2)=8 and records {2,7,13,29,59,...} gives the least new prime a(2)=7 as the first term of a second equivalence class {7,13,29,59,...} of prime sequences.
For n=3 with e(3)=16, a(3)=17 the third equivalence class is {17,41,83,167,...}.
		

Crossrefs

Cf. A291620 (branches), A167168 (equivalence classes), A134734 (first differences of A084662), A134162.

Programs

  • Mathematica
    For[i = 2; pl = {}; fp = {}, i < 350, i++,
    ps = Union@FoldList[Max, 1, Rest@# - Most@#] &@
       FoldList[#1 + GCD[#2, #1] &, 2 i, Range[2, 10^5]];
    p = Select[ps, (i <= #) && ! MemberQ[pl, #] &, 1];
    If[p != {}, fp = Join[fp, {p}];
      pl = Union[pl,
        Drop[ps, -1 + Position[ps, p[[1]]][[1]][[1]]]]]]; Flatten@fp

Formula

a(n) >= 2*n; a(n) > 10*n - 50; a(n) < 12*n.
a(n) >= e(n) - 1, for n > 1; a(n) < e(n) + n.
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