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A355636
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a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of divisors as the sum a(n-2) + a(n-1) but does not equal the sum.
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4
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1, 1, 3, 9, 18, 6, 30, 100, 24, 12, 196, 48, 20, 28, 80, 60, 72, 84, 90, 40, 42, 8, 32, 54, 10, 729, 2, 14, 81, 15, 108, 21, 22, 5, 26, 7, 27, 33, 96, 34, 56, 126, 66, 320, 35, 38, 11, 4, 39, 13, 44, 46, 132, 51, 55, 57, 162, 58, 140, 150, 70, 156, 62, 65, 17, 69, 74, 77, 19, 160, 23, 82, 78
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OFFSET
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1,3
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COMMENTS
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In the first 250000 terms the smallest numbers that have not appeared are 64, 1024, 11664, 15625. It is unknown if these and all other numbers eventually appear.
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LINKS
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EXAMPLE
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a(6) = 6 as a(4) + a(5) = 9 + 18 = 27 which has four divisors, and 6 is the smallest unused number that does not equal 27 and has four divisors.
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PROG
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(PARI) listm(nn) = my(va = vector(nn)); va[1] = 1; va[2] = 1; my(m = Map()); mapput(m, 1, 1); for (n=3, nn, my(s=va[n-2]+va[n-1], d=numdiv(s), k=1, vs=Vec(va, n-1)); while (mapisdefined(m, k) || (k==s) || (numdiv(k)!=d), k++); va[n] = k; mapput(m, k, n); ); va; \\ Michel Marcus, Jul 11 2022
(Python)
from sympy import divisor_count
from itertools import count, islice
def agen():
anm1, an, mink, seen = 1, 1, 2, {1}
yield 1
for n in count(2):
yield an
k, target, tsum = mink, divisor_count(anm1+an), anm1+an
while k in seen or k == tsum or divisor_count(k) != target: k += 1
while mink in seen: mink += 1
anm1, an = an, k
seen.add(an)
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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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