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A339793
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a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number not occurring earlier that is a multiple of s(a(n-1)), the sum of the proper divisors of a(n-1).
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2
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1, 2, 3, 4, 6, 12, 16, 15, 9, 8, 7, 5, 10, 24, 36, 55, 17, 11, 13, 14, 20, 22, 28, 56, 64, 63, 41, 18, 21, 33, 30, 42, 54, 66, 78, 90, 144, 259, 45, 99, 57, 23, 19, 25, 48, 76, 128, 127, 26, 32, 31, 27, 39, 34, 40, 50, 43, 29, 35, 52, 46, 104, 106, 112, 136, 134, 70, 74, 80, 212, 166, 86, 92, 152
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OFFSET
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1,2
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COMMENTS
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The sequence is possibly a permutation of the positive integers as when a(n-1) is prime a(n) will be the next smallest number that has not previously occurred. However this will depend on the likelihood of a(n) being a prime as n goes to infinity. For the first 478 terms the last prime is a(144) = 59, while a(478) = 19140499834691254267668, indicating prime values become increasingly rare, and could potentially have a finite number as n->infinity.
The sum of the proper divisors of n is given by A001065(n).
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LINKS
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EXAMPLE
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a(3) = 3 as s(a(2)) = s(2) = 1, and 3 is the smallest multiple of 1 that has not previously occurred.
a(5) = 6 as s(a(4)) = s(4) = 3, and as 3 has already occurred the next lowest multiple is used, being 6.
a(12) = 5 as s(a(11)) = s(7) = 1, and 5 is the smallest multiple of 1 that has not previously occurred.
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PROG
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(Python)
from sympy import divisors
def s(k): return sum(d for d in divisors(k)[:-1])
def aupto(n):
alst, aset = [1, 2], {1, 2}
for k in range(2, n):
ak = sanm1 = s(alst[-1])
while ak in aset: ak += sanm1
alst.append(ak); aset.add(ak)
return alst # use alst[n-1] for a(n)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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