|
|
A355647
|
|
a(1) = 1, a(2) = 2; 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).
|
|
5
|
|
|
1, 2, 3, 5, 6, 7, 11, 12, 13, 4, 17, 8, 9, 19, 18, 23, 29, 20, 25, 28, 31, 37, 32, 10, 24, 14, 15, 41, 30, 43, 47, 60, 53, 59, 48, 61, 67, 40, 71, 21, 44, 22, 42, 64, 26, 72, 45, 50, 27, 33, 84, 52, 54, 34, 56, 90, 35, 38, 73, 39, 80, 46, 96, 51, 63, 66, 55, 49, 70, 57, 79, 78, 83, 58, 62, 120
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
In the first 500000 terms the smallest numbers that have not appeared are 15625, 25600, 28561, 36864. It is unknown if these and all other numbers eventually appear. In the same range on eighty-two occasions a(n) equals the sum of the previous two terms, these values begin 3, 5, 17, 64, 90, 73, 120, 144, 192.
|
|
LINKS
|
|
|
EXAMPLE
|
a(5) = 6 as a(3) + a(4) = 3 + 5 = 8 which has four divisors, and 6 is the smallest unused number that has four divisors.
|
|
PROG
|
(Python)
from sympy import divisor_count
from itertools import count, islice
def agen():
anm1, an, mink, seen = 1, 2, 3, {1, 2}
yield 1
for n in count(2):
yield an
k, target = mink, divisor_count(anm1+an)
while k in seen or divisor_count(k) != target: k += 1
while mink in seen: mink += 1
anm1, an = an, k
seen.add(an)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|