A251417 Lengths of runs of identical terms in A251416.

1, 1, 1, 5, 1, 5, 1, 6, 1, 7, 1, 12, 8, 10, 1, 17, 8, 1, 13, 13, 13, 5, 10, 11, 5, 9, 8, 19, 10, 11, 7, 11, 5, 9, 27, 9, 13, 5, 23, 5, 9, 17, 9, 11, 11, 7, 21, 9, 7, 5, 17, 27, 11, 7, 9, 17, 5, 13, 9, 21, 11, 7, 13, 9, 9

Offset: 1

Author

N. J. A. Sloane, Dec 03 2014

Keywords

• nonn

Comments

It would be nice to have an alternative description of this sequence, one that is not based on A098550.

It appears (conjecture) that a(n)>1 for n>18. - Alexander R. Povolotsky, Dec 07 2014

Conjecture: a(n) = A247253(n-5) for n>12. - Reinhard Zumkeller, Dec 07 2014

The previous conjecture is equivalent to the statement that A251416(n) lists all primes and only primes after a(30)=18. - M. F. Hasler, Dec 08 2014

Examples

			See A251595.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A098550, A251416, A251595, A247253.

Programs

Haskell

import Data.List (group)
a251417 n = a251417_list !! (n-1)
a251417_list = map length $ group a251416_list
-- Reinhard Zumkeller, Dec 05 2014

Mathematica

termsOfA251416 = 700;
f[lst_List] := 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}, termsOfA251416 - 3];
b[1] = 2;
b[n_] := b[n] = For[k = b[n-1], True, k++, If[FreeQ[A098550[[1 ;; n]], k], Return[k]]];
A251416 = Array[b, termsOfA251416];
Length /@ Split[A251416] (* Jean-François Alcover, Aug 01 2018, after Robert G. Wilson v *)

Formula

Let f(n)=A098551(A251595(n)). Then one can prove that A251417(n) = f(n) - f(n-1), n>=2. - Vladimir Shevelev, Dec 09 2014