OFFSET
1,2
COMMENTS
Because tau(k)/k is bounded by 2/sqrt(k), this sequence is well-defined.
In the case of ties, terms are sorted from least to greatest.
Let c be the j-th distinct value of tau(a(n))/a(n). Terms of this sequence for which tau(a(n))/a(n) >= c are the proper divisors of A059992(j + 1) for 1 <= j <= 4. True for j = 0 if the 0th value of c is taken to be infinity. Pattern breaks for j > 4.
EXAMPLE
tau(1)/1 = tau(2)/2 = 1
tau(4)/4 = 3/4
tau(3)/3 = tau(6)/6 = 2/3
tau(8)/8 = tau(12)/12 = 1/2
tau(5)/5 = tau(10)/10 = 2/5
tau(9)/9 = tau(18)/18 = tau(24)/24 = 1/3
MATHEMATICA
nmax = 100; s = Sort[Table[{k, DivisorSigma[0, k]/k}, {k, 1, nmax^2}], #1[[2]] >= #2[[2]] &]; Table[s[[j, 1]], {j, 1, nmax}] (* Vaclav Kotesovec, Jan 04 2024 *)
PROG
(Lua)
length = 100
result = {}
for n = 1, length do
local div_count = 0
local root_n = math.sqrt(n)
for d = 1, root_n do
if n % d == 0 then
div_count = div_count + 2
end
end
if (root_n == math.floor(root_n)) then
div_count = div_count - 1
end
result[n] = {n, div_count / n}
end
function compare(a, b)
if a[2] ~= b[2] then
return a[2] > b[2]
else
return a[1] < b[1]
end
end
table.sort(result, compare)
i = 1
bound = 2 / math.sqrt(length)
while result[i][2] >= bound do
io.write(result[i][1] .. ', ')
i = i + 1
end
CROSSREFS
KEYWORD
nonn
AUTHOR
Keith J. Bauer, Dec 28 2023
STATUS
approved