A346425 - OEIS

%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