A081699 - OEIS

%I #12 Jul 01 2021 16:31:44

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

%N k-tuple abundance record-holders.

%C 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.

%C 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

%H H. W. Lenstra, <a href="https://www.jstor.org/stable/2318266">Problem 6064</a>, Amer. Math. Monthly 82 (1975), p. 1016. <a href="https://www.jstor.org/stable/2320042">Solution</a> by the proposer in Amer. Math. Monthly 84 (1977), p. 580.

%e a(1) = 12 because 12 is the first abundant number.

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

%Y Cf. A081700, A081705.

%K nonn,hard,more

%O 1,1

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

%E a(8)-a(12) from _David Wasserman_, Jun 16 2004