A251621 Run lengths in A249943.
1, 1, 1, 1, 1, 1, 3, 1, 2, 4, 1, 1, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8, 4, 2, 4, 2, 4, 14, 4, 6, 2, 10, 2, 6, 6, 4, 6, 6, 2, 10, 2, 4, 2, 12, 12, 4, 2, 4, 6, 2, 10, 6, 6, 6, 2, 6, 4, 2, 10, 14, 4, 2, 4, 14, 6, 10, 2, 4, 6, 8, 6, 6, 4, 6, 8, 4
Offset: 1
Keywords
Examples
From _Vladimir Shevelev_, Dec 11 2014: (Start) For formula for prime(n): 1) n=8, prime(8) = 19; 2) n=9, prime(9) = 19 + a(13) = 19 + 4 = 23; 3) n=10, prime(10)= 19 + a(13) + a(14) = 23 + 6 = 29, etc. (End)
References
- Bradley Klee, Posting to Sequence Fans Mailing List, Dec 07 2014
- Vladimir Shevelev, Postings to Sequence Fans Mailing List, Dec 07, 10 and 11, 2014
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
- David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669 [math.NT], 2015 and J. Int. Seq. 18 (2015) 15.6.7.
Programs
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Haskell
import Data.List (group) a251621 n = a251621_list !! (n-1) a251621_list = map length $ group a249943_list
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Mathematica
f[lst_] := Block[{k = 4}, While[GCD[lst[[-2]], k] == 1 || GCD[lst[[-1]], k] > 1 || MemberQ[lst, k], k++]; Append[lst, k]]; A098550 = Nest[f, {1, 2, 3}, 1000]; runningMax = Rest[FoldList[Max, -Infinity, #]]&; A249943 = runningMax[Take[Ordering[A098550], NestWhile[#+1&, 1, MemberQ[A098550, #] &] - 1]]; Length /@ Split[A249943] (* Jean-François Alcover, Sep 11 2017, using code from Robert G. Wilson v *)
Formula
Connection with prime gaps: conjecturally, for n>=13, we have a(n) = A001223(n-5). - Vladimir Shevelev, Dec 07 2014
Bradley Klee noted that this conjecture and his conjectures in A251416 are equivalent. At least to one side, our conjecture could be deduced from Klee's conjectures by a simple induction. - Vladimir Shevelev, Dec 10 2014
As a corollary, we have an explicit conjectural formula for prime(n), n>=8, essentially based on A098550: prime(n) = 19 + sum{i=9,...,n}a(i+4). - Vladimir Shevelev, Dec 11 2014