A081700 -id:A081700 - cp's OEIS frontend

A081705 k-tuple abundance of abundant numbers.

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, 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, 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

Offset: 1

Gabriel Cunningham (gcasey(AT)mit.edu), Apr 02 2003, Dec 15 2006

nonn

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.

Lenstra shows that there are in fact infinitely many k-tuply abundant numbers for every k > 0.

			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.

Paolo P. Lava, Table of n, a(n) for n = 1..275 (first 97 terms from Gabriel Cunningham)
H. W. Lenstra, Jr., Advanced Problems and Solutions, 6064, The American Mathematical Monthly, Vol. 84, No. 7. (Aug. - Sep., 1977), p. 580.

Cf. A081699, A081700.

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 ;
end proc:
A081705 := proc(n)
    aliqRis(A005101(n)) ;
end proc:
seq(A081705(n),n=1..100) ; # R. J. Mathar, Mar 07 2007

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.

a(n) = number of increasing steps at the start of the aliquot chain of A005101(n).

More terms from R. J. Mathar, Mar 07 2007

A081699 k-tuple abundance record-holders.

Original entry on oeis.org

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

Offset: 1

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

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

			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).

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

Cf. A081700, A081705.

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

cp's OEIS Frontend

A081705 k-tuple abundance of abundant numbers.

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