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

1, 1, 2, 3, 5, 7, 10, 13, 16, 21, 26, 31, 36, 41, 48, 55, 62, 69, 76, 83, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 203, 216, 229, 242, 255, 268, 281, 294, 307, 320, 333, 346, 359, 375, 391, 407, 423, 439, 455, 471, 487, 503, 519, 535, 551, 567

Offset: 1

Author

Wagner Martins, Jul 09 2023

Keywords

• easy
• nonn

Comments

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.

Examples

			k denotes the k-th iteration
The sequence is initialized with (1, 1)
For k = 1
Add a(1) = 1 once, you get (1, 1, 2)
For k = 2
Add a(2) = 1 once, you get (1, 1, 2, 3)
For k = 3
Add a(3) = 2 twice, you get (1, 1, 2, 3, 5, 7)
For k = 4
add a(4) = 3 three times, and you get (1, 1, 2, 3, 5, 7, 10, 13, 16)
		

Crossrefs

Cf. A100143.

Programs

Python

def a_list(n):
    if n <= 2:
        return 1
    sequence = [1, 1]
    target_number_index = 0
    times_to_add = sequence[target_number_index]
    for _ in range(n - 2):
        if times_to_add == 0:
            target_number_index += 1
            times_to_add = sequence[target_number_index]
        last_term = sequence[-1]
        sequence.append(last_term + sequence[target_number_index])
        times_to_add -= 1
    return sequence