%I #29 Jul 20 2021 04:02:42
%S 2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,
%T 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,
%U 5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5
%N a(n) is the greatest number k such that k! <= prime(n).
%C Terms 2, 3, 4, 5, ... appear respectively 3, 6, 21, 98, ... times consecutively; indeed, 2 appears A061232(1) + A061232(2) times, then every m >= 3 appears A061232(m) times.
%F a(n)! = A000040(n) - A136437(n).
%e prime(1) = 2 and the greatest k with k! <= 2 is 2, so a(1) = 2.
%e prime(4) = 7 and the greatest k with k! <= 7 is 3, so a(4) = 3.
%e prime(10) = 29 and the greatest k with k! <= 29 is 4 so a(10) = 4.
%e Rows with n, prime(n), greatest k! <=n, a(n) for n = 1..14
%e n 1 2 3 4 5 6 7 8 9 10 11 12 13 14
%e prime(n) 2 3 5 7 11 13 17 19 23 29 31 37 41 43
%e greatest k! 2 2 2 6 6 6 6 6 6 24 24 24 24 24
%e a(n) 2 2 2 3 3 3 3 3 3 4 4 4 4 4
%o (PARI) a(n) = my(k=1); until (k! > prime(n), k++); k-1; \\ _Michel Marcus_, Jul 19 2021
%Y Cf. A000040, A061232, A136437.
%K nonn
%O 1,1
%A _Bernard Schott_, Jul 16 2021