A355647 - OEIS

%I #17 Jul 26 2022 13:38:29

%S 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,

%T 15,41,30,43,47,60,53,59,48,61,67,40,71,21,44,22,42,64,26,72,45,50,27,

%U 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

%N 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).

%C 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.

%C See A355648 for the fixed points.

%H Scott R. Shannon, <a href="/A355647/a355647.png">Image of the first 500000 terms</a>. The green line is y = n.

%e 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.

%o (Python)

%o from sympy import divisor_count

%o from itertools import count, islice

%o def agen():

%o     anm1, an, mink, seen = 1, 2, 3, {1, 2}

%o     yield 1

%o     for n in count(2):

%o         yield an

%o         k, target = mink, divisor_count(anm1+an)

%o         while k in seen or divisor_count(k) != target: k += 1

%o         while mink in seen: mink += 1

%o         anm1, an = an, k

%o         seen.add(an)

%o print(list(islice(agen(), 76))) # _Michael S. Branicky_, Jul 26 2022

%Y Cf. A355648, A355636, A000005, A351001, A352768, A352867, A352774.

%K nonn

%O 1,2

%A _Scott R. Shannon_, Jul 12 2022