%I #6 Mar 02 2019 23:35:32
%S 4,3,2,2,3,2,2,2,2,2,1,3,2,0,2,2,2,1,2,2,1,3,1,1,1,3,2,1,1,2,2,2,1,0,
%T 3,1,1,1,2,2,1,1,2,1,2,1,2,1,2,2,1,3,0,1,2,2,1,1,1,0,2,1,1,2,1,2,2,1,
%U 3,0,1,3,1,2,1,2,1,1,1,1,1,1,2,1,2,2,0,1,1,1,1,3,1,1,1,3,0,2,2,0,1,1,1,1,2
%N Frequency of n in A155213.
%C Since prime gaps increase on the average, this sequence fades out with more and more zeros as the indices increase.
%e There are 4 zeros, 3 ones, 2 twos, etc. in A155213.
%p := proc(n) floor( ithprime(n)/9) ; end: a155213 := [seq(A155213(n),n=1..200)] ;
%p A155462 := proc(n) global a155213; ListTools[Occurrences](n,a155213) ; end:
%p seq(A155462(n),n=0..max(op(a155213))) ; # _R. J. Mathar_, Jul 23 2009
%K nonn
%O 0,1
%A _Paul Curtz_, Jan 22 2009
%E Edited and extended by _R. J. Mathar_, Jul 23 2009