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A355636
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.
4
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
OFFSET
1,3
COMMENTS
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.
See A355637 for the fixed points.
LINKS
Scott R. Shannon, Image of the first 250000 terms. The green line is y = n.
EXAMPLE
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.
PROG
(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)
print(list(islice(agen(), 73))) # Michael S. Branicky, Jul 26 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Jul 11 2022
STATUS
approved