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A364090 Add each term m of the sequence to the last one m times starting with 1, 1.

%I A364090 #37 Aug 06 2023 14:07:21
%S A364090 1,1,2,3,5,7,10,13,16,21,26,31,36,41,48,55,62,69,76,83,90,100,110,120,
%T A364090 130,140,150,160,170,180,190,203,216,229,242,255,268,281,294,307,320,
%U A364090 333,346,359,375,391,407,423,439,455,471,487,503,519,535,551,567
%N A364090 Add each term m of the sequence to the last one m times starting with 1, 1.
%C A364090 a(n) seems to grow as n^c where c is a constant with the value of approximately 1.625, in other words, lim_{n->oo} log_n(a(n)) seems to converge.
%e A364090 k denotes the k-th iteration
%e A364090 The sequence is initialized with (1, 1)
%e A364090 For k = 1
%e A364090 Add a(1) = 1 once, you get (1, 1, 2)
%e A364090 For k = 2
%e A364090 Add a(2) = 1 once, you get (1, 1, 2, 3)
%e A364090 For k = 3
%e A364090 Add a(3) = 2 twice, you get (1, 1, 2, 3, 5, 7)
%e A364090 For k = 4
%e A364090 add a(4) = 3 three times, and you get (1, 1, 2, 3, 5, 7, 10, 13, 16)
%o A364090 (Python)
%o A364090 def a_list(n):
%o A364090     if n <= 2:
%o A364090         return 1
%o A364090     sequence = [1, 1]
%o A364090     target_number_index = 0
%o A364090     times_to_add = sequence[target_number_index]
%o A364090     for _ in range(n - 2):
%o A364090         if times_to_add == 0:
%o A364090             target_number_index += 1
%o A364090             times_to_add = sequence[target_number_index]
%o A364090         last_term = sequence[-1]
%o A364090         sequence.append(last_term + sequence[target_number_index])
%o A364090         times_to_add -= 1
%o A364090     return sequence
%Y A364090 Cf. A100143.
%K A364090 easy,nonn
%O A364090 1,3
%A A364090 _Wagner Martins_, Jul 09 2023