A110969 Length of the runs of ones in A014963.
1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 4, 3, 1, 3, 1, 3, 5, 1, 2, 2, 3, 1, 5, 1, 1, 5, 7, 3, 1, 3, 1, 3, 7, 3, 1, 2, 5, 1, 9, 1, 5, 5, 3, 1, 3, 5, 1, 9, 1, 3, 1, 11, 11, 3, 1, 3, 5, 1, 1, 7, 4, 5, 5, 1, 5, 3, 1, 5, 3, 13, 3, 1, 3, 13, 5, 5, 3, 1, 3, 5, 1, 5, 5, 5, 3, 5, 7, 3, 7
Offset: 1
Keywords
Examples
a(5)=2 because the fifth run of ones in A014963 is of length 2.
Links
- Andrew Howroyd, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Mathematica
Length /@ SplitBy[Table[Exp[MangoldtLambda[n]], {n, 400}], # != 1 &][[ ;; -1 ;; 2]] (* Michael De Vlieger, Mar 21 2024 *) DeleteCases[Differences[Select[Range[100],PrimePowerQ]]-1,0] (* Gus Wiseman, Jun 18 2024 *)
-
PARI
\\ b(n) returns boolean of A014963(n) == 1. b(n)={my(t); !isprime(if(ispower(n, ,&t), t, n))} seq(n)={my(k=1, i=0, L=List()); while(#L
k, listput(L, i-k)); k = i+1)); Vec(L)} \\ Andrew Howroyd, Jan 02 2020
Extensions
Terms a(41) and beyond from Andrew Howroyd, Jan 02 2020
Comments