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%I #24 Mar 18 2024 13:49:15
%S 6,2,9,81
%N Smallest number k such that n iterations of sigma() are required for the result to be >= 2k.
%C These are the first terms of A023196, A107912, A107913, A107914. - _Jud McCranie_, May 28 2005
%C a(5) > 4*10^9, if it exists. - _Jud McCranie_, May 28 2005
%C 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
%C 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
%e sigma(sigma(sigma(9))) = 24 >= 2*9, so a(3)=9.
%o (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;
%o a(n) = my(k=2); while (!isok(k, n), k++); k; \\ _Michel Marcus_, Mar 18 2024
%o (PARI)
%o seq() = {
%o my(todo = Set([1,2,3,4]), res = vector(4));
%o for(i = 2, oo,
%o t = 1;
%o s = sigma(i);
%o while(s < 2*i,
%o s = sigma(s);
%o t++
%o );
%o if(res[t] == 0,
%o res[t] = i;
%o todo = setminus(todo, Set(t));
%o if(#todo == 0,
%o return(res)
%o )
%o );
%o )
%o } \\ _David A. Corneth_, Mar 18 2024
%Y Cf. A000203, A055020, A107912, A107913, A107914, A060800.
%K nonn,fini,full
%O 1,1
%A _Jud McCranie_, May 31 2000
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