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 A055021 Smallest number k such that n iterations of sigma() are required for the result to be >= 2k. 5
 6, 2, 9, 81 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 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 LINKS Table of n, a(n) for n=1..4. 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=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 CROSSREFS Cf. A000203, A055020, A107912, A107913, A107914, A060800. Sequence in context: A339410 A235362 A018801 * A260329 A141379 A357396 Adjacent sequences: A055018 A055019 A055020 * A055022 A055023 A055024 KEYWORD nonn,fini,full AUTHOR Jud McCranie, May 31 2000 STATUS approved

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Last modified August 14 07:07 EDT 2024. Contains 375146 sequences. (Running on oeis4.)