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%I #16 Oct 11 2017 06:56:44
%S 1,1,1,2,7,1,1,6,1,5,1,2,4,1,1,3,1,3,1,2,2,1,7,1,1,1,6,8,5,3,31,2,1,
%T 30,1,1,1,28,5,2,14,4,1,1,3,1,2,1,14,4,1,29,4,1,28,7,4,5,3,1,11,2,1,6,
%U 3,12,1,11,1,1,6,5,27,18,1,1,17,1,2,3,3,1,1,14,4,4,13,1,1,12,2,3,10,1,5,1,4
%N k-tuple abundance of abundant numbers.
%C Note that only increasing steps at the beginning of an aliquot chain count toward k-tuple abundance. I wonder if there are an infinite number of k-tuply abundant numbers for all k? Another interesting question - are there any numbers that are completely abundant (that is, numbers whose aliquot chain increases forever). Though there are several numbers whose aliquot chains aren't yet fully determined, all the ones I've checked have had a finite k-tuple abundance.
%C Lenstra shows that there are in fact infinitely many k-tuply abundant numbers for every k > 0.
%H Paolo P. Lava, <a href="/A081705/b081705.txt">Table of n, a(n) for n = 1..275 (first 97 terms from Gabriel Cunningham)</a>
%H H. W. Lenstra, Jr., <a href="http://www.jstor.org/stable/2320042">Advanced Problems and Solutions, 6064</a>, The American Mathematical Monthly, Vol. 84, No. 7. (Aug. - Sep., 1977), p. 580.
%F a(n) = 0 if n is not abundant, otherwise 1 + (a(sigma(n)-n)) Note, however, that non-abundant numbers are excluded from this sequence.
%F a(n) = number of increasing steps at the start of the aliquot chain of A005101(n).
%e a(4)=2 because the 4th abundant number is 24 which has aliquot sequence 24->36->55->17->1, which has two increasing steps at the beginning.
%p aliqRis := proc(n) local r,a,an ; r := 0 ; a := n; while true do an := numtheory[sigma](a)-a ; if an > a then r := r+1 ; a := an ; else RETURN(r) ; fi ; od ;
%p end proc:
%p A081705 := proc(n)
%p aliqRis(A005101(n)) ;
%p end proc:
%p seq(A081705(n),n=1..100) ; # _R. J. Mathar_, Mar 07 2007
%Y Cf. A081699, A081700.
%K nonn
%O 1,4
%A Gabriel Cunningham (gcasey(AT)mit.edu), Apr 02 2003, Dec 15 2006
%E More terms from _R. J. Mathar_, Mar 07 2007
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