

A055021


Smallest number k such that n iterations of sigma() are required for the result to be >= 2k.


5




OFFSET

1,1


COMMENTS

There are no more terms: sigma(2*k) is never prime if k is not a power of 2, so an even number needs at most two steps; sigma(k) is odd iff k is a square or twice a square. So A107914 (four recursive steps) contains only odd squares. Assume p prime so sigma(p^2) = p^2 + p + 1 = m^2 never meets the condition with p + 2k = m that (p + 2k)^2 = m^2. This implies the impossibility of a solution for numbers of the form p^(2i) and numbers of the form p^(2i)q^(2i).  Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jun 06 2005
If k is a power of 2 then sigma(sigma(2*k)) = sigma(4*k  1) >= 4*k and so the number of iterations is exactly 2.  David A. Corneth, Mar 18 2024


LINKS



EXAMPLE

sigma(sigma(sigma(9))) = 24 >= 2*9, so a(3)=9.


PROG

(PARI) isok(k, n) = my(kk=k); for (i=1, n, k = sigma(k); if ((i<n) && (k>=2*kk), return(0))); k >= 2*kk;
a(n) = my(k=2); while (!isok(k, n), k++); k; \\ Michel Marcus, Mar 18 2024
(PARI)
seq() = {
my(todo = Set([1, 2, 3, 4]), res = vector(4));
for(i = 2, oo,
t = 1;
s = sigma(i);
while(s < 2*i,
s = sigma(s);
t++
);
if(res[t] == 0,
res[t] = i;
todo = setminus(todo, Set(t));
if(#todo == 0,
return(res)
)
);
)


CROSSREFS



KEYWORD

nonn,fini,full


AUTHOR



STATUS

approved



