A081699 k-tuple abundance record-holders.

12, 24, 30, 120, 138, 858, 966, 1134, 1218, 1476, 2514, 4494

Offset

1,1

Comments

A number n is k-tuply abundant if it is abundant and either k = 1 or s(n) is (k-1)-tuply abundant. Thus 24 is doubly abundant: its aliquot chain is 24->36->55->17->1. a(n+1) is defined as the smallest number that is more k-tuply abundant than a(n). 966 is 179-tuply abundant.

Lenstra shows that for any k, there is a k-tuply abundant number. Hence the sequence is infinite if and only if the Catalan-Dickson conjecture holds: for all n, the aliquot sequence n, s(n), s(s(n)), ... either terminates at 0 or is periodic. - Charles R Greathouse IV, Jun 28 2021

Links

Table of n, a(n) for n=1..12.

H. W. Lenstra, Problem 6064, Amer. Math. Monthly 82 (1975), p. 1016. Solution by the proposer in Amer. Math. Monthly 84 (1977), p. 580.

Example

a(1) = 12 because 12 is the first abundant number.
a(3) = 30 because 30 is the first number more k-tuply abundant than a(2).

Crossrefs

Cf. A081700, A081705.

Sequence in context: A328632 A261435 A103590 * A120570 A164014 A336772

Adjacent sequences: A081696 A081697 A081698 * A081700 A081701 A081702

Keyword

nonn,hard,more

Author

Gabriel Cunningham (gcasey(AT)mit.edu), Apr 02 2003

Extensions

a(8)-a(12) from David Wasserman, Jun 16 2004

Status

approved