

A081699


ktuple abundance recordholders.


4



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

1,1


COMMENTS

A number n is ktuply abundant if it is abundant and either k = 1 or s(n) is (k1)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 ktuply abundant than a(n). 966 is 179tuply abundant.
Lenstra shows that for any k, there is a ktuply abundant number. Hence the sequence is infinite if and only if the CatalanDickson 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 ktuply 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



