A387078 Run lengths of A386482(n) mod 2 == n mod 2.

1, 3, 2, 4, 2, 3, 3, 5, 3, 4, 2, 8, 5, 3, 4, 4, 2, 11, 4, 2, 2, 22, 16, 5, 3, 1, 2, 12, 6, 31, 14, 4, 3, 8, 3, 28, 2, 37, 14, 10, 12, 9, 2, 41, 7, 61, 24, 24, 2, 134, 71, 51, 97, 3, 2, 127, 69, 39, 15, 64, 55, 56, 26, 100, 37, 32, 40, 33, 2, 440, 107, 196, 391

Offset: 1

Author

Michael De Vlieger, Aug 15 2025

Keywords

• nonn

Comments

Let S = A386482.

Beginning with S(481) = 948, there are 100 consecutive even terms in S. Starting with S(730076) = 1026330, there are 100869 consecutive even terms in S.

Examples

			S begins as follows, grouping odd terms in brackets [], and even in parentheses ():
   [1], (2, 4, 6), [3, 9], (12, 10, 8, 14), [7, 21], (18, 16, 20), [15, 5, 25], ...
This sequence takes run lengths in the order they appear, therefore a(1) = 1, a(2) = 3, a(3) = 2, a(4) = 4, a(5) = 2, etc. Hence a(n) for odd n pertains to run lengths of odd terms in S, while a(n) for even n pertains to run lengths of even terms in same.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..183 (run lengths available given 2^20 terms of S).

Crossrefs

Cf. A386482.

Programs

Mathematica

Block[{c, j, k, m, p, r, nn},
  nn = 2^12; c[] := False; m[] := 1; j = 2; c[1] = c[2] = True; r = 1;
  {1}~Join~Monitor[Most@ Reap[Do[
    If[PrimePowerQ[j],
      Set[{p, k, m}, {#1, #1^(#2 - 1), #1^(#2 - 1)}] & @@
        FactorInteger[j][[1]]; While[And[c[k*p], k != 0], k--];
        If[k == 0, k = m; While[c[k*p], k++]]; k *= p,
      k = j - 1; While[And[Or[c[k], CoprimeQ[j, k]], k != 1], k--];
        If[k == 1, k += j; While[Or[c[k], CoprimeQ[j, k] ], k++] ] ];
    If[Mod[j, 2] == Mod[k, 2], r++, Sow[r]; r = 1];
    Set[{c[k], j}, {True, k}], {n, 3, nn}] ][[-1, 1]], n] ]