A323327 Numbers that start an unbounded aliquot-like sequence based on Dedekind psi function (A001615).

318, 330, 498, 510, 534, 546, 636, 660, 786, 798, 942, 954, 978, 990, 996, 1020, 1068, 1092, 1110, 1122, 1254, 1272, 1320, 1398, 1410, 1470, 1494, 1506, 1518, 1530, 1572, 1596, 1602, 1614, 1626, 1638, 1734, 1884, 1908, 1938, 1950, 1956, 1980, 1992, 2040, 2046

Offset: 1

Author

Amiram Eldar, Jan 11 2019

Keywords

• nonn

Comments

Let t(k) = psi(k) - k = A001615(k) - k be the sum of aliquot divisors d of k, such that k/d is squarefree. Penney & Pomerance proposed a problem to show that the aliquot-like sequence related to this function, i.e., the trajectory of an integer k under the repeated application of the map k -> t(k), can be unbounded. Since t(m^j * k) = m^j * t(k) if m|k, then if in the sequence a_0 = k, a_1 = t(k), a_2 = t(t(k)), ... there is a term a_{i1} = m^j * a_0 such that m|k and j > 0 then a_{i+i1} = m^j * a_i for all i and thus the sequence is unbounded.

Examples

			318 is in the sequence since t(318) = psi(318) - 318 = 330, t(330) = 534, etc., and this repeated mapping yields an unbounded sequence.
		

References

• Jean-Marie De Koninck, Those Fascinating Numbers, Amer. Math. Soc., 2009, page 71, entry 318.

Links

• Kevin Brown and Charles Vanden Eynden, Pseudo-aliquot Sequences, Solution to Problem 10323, The American Mathematical Monthly, Volume 103, No. 8 (1996), pp. 697-698.
• David E. Penney and Carl Pomerance, Problem 10323, The American Mathematical Monthly, Volume 100, No. 7 (1993), p. 688.
• Eric Weisstein's World of Mathematics, Aliquot Sequence.
• Wikipedia, Aliquot sequence.

Crossrefs

Cf. A001615, A098007, A187778.

Programs

Mathematica

t[1]=0; t[n_] := (Times@@(1+1/Transpose[FactorInteger[n]][[1]])-1)*n; rt[n_] := Module[{f=FactorInteger[n]}, e=GCD@@f[[;;,2]]; Surd[n,e]]; divrootQ[n_, m_] := Divisible[n, rt[m]]; divQ[s_, n_] := If[n==0, 0, If[MemberQ[s, n], 1, If[ Length[Select[s, Divisible[n,#] && divrootQ[#, n/#] &]] > 0, 2, 3]]]; seqQ[n_] := Module[{n1=n}, s={}; While[divQ[s, n1] ==3, AppendTo[s, n1]; n1=t[n1]]; divQ[s, n1]] == 2; Select[Range[10000], seqQ]