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a(0) = 0, and for n > 0, a(n) = largest k such that A002110(k-1) <= n, where A002110(k) gives the k-th primorial number.
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%I #31 Jul 23 2023 15:43:50

%S 0,1,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,4,4,4,4,

%T 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,

%U 4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4

%N a(0) = 0, and for n > 0, a(n) = largest k such that A002110(k-1) <= n, where A002110(k) gives the k-th primorial number.

%C For n > 0: a(n) = (length of row n in A235168) = A055642(A049345(n)).

%C For n > 0, a(n) gives the length of primorial base expansion of n. Also, after zero, each value n occurs A061720(n-1) times. - _Antti Karttunen_, Oct 19 2019

%H Reinhard Zumkeller, <a href="/A235224/b235224.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Pri#primorialbase">Index entries for sequences related to primorial base</a>

%F From _Antti Karttunen_, Oct 19 2019: (Start)

%F a(n) = A061395(A276086(n)).

%F For all n >= 0, a(n) >= A267263(n).

%F For all n >= 1, A000040(a(n)) > A328114(n). (End)

%p A235224 := proc(n)

%p local k;

%p if n = 0 then

%p 0;

%p else

%p for k from 0 do

%p if A002110(k-1) > n then

%p return k-1 ;

%p end if;

%p end do:

%p end if;

%p end proc: # _R. J. Mathar_, Apr 19 2021

%t primorial[n_] := Times @@ Prime[Range[n]];

%t a[n_] := TakeWhile[primorial /@ Range[0, n], # <= n &] // Length;

%t Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Oct 27 2021 *)

%o (Haskell)

%o a235224 n = length $ takeWhile (<= n) a002110_list

%o (PARI) A235224(n) = { my(s=0, p=2); while(n, s++; n = n\p; p = nextprime(1+p)); (s); }; \\ _Antti Karttunen_, Oct 19 2019

%o (PARI) A235224(n, p=2) = if(!n,n,if(n<p, 1, 1+A235224(n\p, nextprime(p+1)))); \\ (Recursive implementation) - _Antti Karttunen_, Oct 19 2019

%Y Cf. A000040, A002110, A049345, A055642, A061395, A061720, A084558, A267263, A276086, A235168, A328114, A328404, A328405, A328406.

%K nonn

%O 0,3

%A _Reinhard Zumkeller_, Jan 05 2014

%E Name corrected to match the data by _Antti Karttunen_, Oct 19 2019