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

1,4

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

Links

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

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

Example

See A251595.

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 *)

Prog

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

Crossrefs

Cf. A098550, A251416, A251595, A247253.

Sequence in context: A056957 A224834 A095118 * A100947 A096940 A141345

Adjacent sequences: A251414 A251415 A251416 * A251418 A251419 A251420

Keyword

nonn

Author

N. J. A. Sloane, Dec 03 2014

Status

approved