A081699 k-tuple abundance record-holders.

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

Offset: 1

Author

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

Keywords

• nonn
• hard
• more

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

Examples

			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).
		

Links

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

Crossrefs

Cf. A081700, A081705.

Extensions

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