%I #17 Jul 26 2022 13:38:21
%S 1,1,3,9,18,6,30,100,24,12,196,48,20,28,80,60,72,84,90,40,42,8,32,54,
%T 10,729,2,14,81,15,108,21,22,5,26,7,27,33,96,34,56,126,66,320,35,38,
%U 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
%N 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.
%C 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.
%C See A355637 for the fixed points.
%H Scott R. Shannon, <a href="/A355636/a355636.png">Image of the first 250000 terms</a>. The green line is y = n.
%e 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.
%o (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
%o (Python)
%o from sympy import divisor_count
%o from itertools import count, islice
%o def agen():
%o anm1, an, mink, seen = 1, 1, 2, {1}
%o yield 1
%o for n in count(2):
%o yield an
%o k, target, tsum = mink, divisor_count(anm1+an), anm1+an
%o while k in seen or k == tsum 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(), 73))) # _Michael S. Branicky_, Jul 26 2022
%Y Cf. A355637, A000005, A351001, A352768, A352867, A352774.
%K nonn
%O 1,3
%A _Scott R. Shannon_, Jul 11 2022