A055021 Smallest number k such that n iterations of sigma() are required for the result to be >= 2k.

6, 2, 9, 81

Offset: 1

Author

Jud McCranie, May 31 2000

Keywords

• nonn
• fini
• full

Comments

These are the first terms of A023196, A107912, A107913, A107914. - Jud McCranie, May 28 2005

a(5) > 4*10^9, if it exists. - Jud McCranie, May 28 2005

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

Examples

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

Crossrefs

Cf. A000203, A055020, A107912, A107913, A107914, A060800.

Programs

PARI

isok(k, n) = my(kk=k); for (i=1, n, k = sigma(k); if ((i=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)
			)
		);	
	)
} \\ David A. Corneth, Mar 18 2024