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A364090
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Add each term m of the sequence to the last one m times starting with 1, 1.
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0
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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)
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PROG
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(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
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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