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n appears {a(1) + a(2) + ... + a(n-1)} times (with a(1) = 1).
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%I #13 Jul 31 2023 10:09:18

%S 1,2,3,3,3,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,7,7,

%T 7,7,7,7,7,7,7,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,8,

%U 9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10

%N n appears {a(1) + a(2) + ... + a(n-1)} times (with a(1) = 1).

%e a(1) = 1 thus 2 appears one time and sequence begins with 1, 2. Then 3 appears a(1) + a(2) = 3 times and sequence continues 1, 2, 3, 3, 3.

%o (PARI) sequence(n) = {

%o my(result = [1]);

%o for (i = 2, n,

%o for (j = 1, sum(k=1, i-1, result[k]), result = concat(result, i))

%o );

%o return(result);

%o } \\ _Benoit Cloitre_, Jul 20 2023

%K nonn

%O 1,2

%A _Benoit Cloitre_, Apr 29 2006