A081699 -id:A081699 - 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

A081700 k-tuple abundance records.

Original entry on oeis.org

1, 2, 7, 8, 31, 59, 179, 190, 196, 261, 302, 303

Offset: 1

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

This sequence is the dual, of sorts, of the k-tuple abundance record-holders sequence. The numbers in this sequence correspond to the k-tuple abundance of the numbers in the record-holders sequence.

			a(5) = 31 because 31 is the first k-tuple abundance that beats a(4) = 8.

Cf. A081699, A081705.

5 more terms from David Wasserman, Jun 16 2004

A234899 Record holders for lengths of ever-decreasing aliquot sequences.

Original entry on oeis.org

1, 2, 4, 9, 14, 16, 26, 46, 52, 166, 212, 1113, 2343, 4437, 5145, 8535, 10665, 18711, 33682, 64935, 114808, 187232, 228316, 304412, 464132, 556636, 623288, 1230284, 1319956, 1508504, 2897884, 3835556, 7487494, 9446906, 16871648, 22328212, 29668150, 29725184

Offset: 1

Michel Marcus, Jan 01 2014

nonn

If one looks at the lengths of uninterrupted decreasing aliquot sequences, the converse of A081705, one gets a sequence similar to A098008, except for perfect or abundant numbers, but also for numbers that encounter a perfect or abundant numbers in this process.
The current sequence lists the deficient numbers yielding uninterrupted decreasing aliquot sequences that are longer than any previous ones (compare with A081699).
Note that, so far, the lengths of the corresponding sequences are contiguous. Does it remain so for next terms?

			The aliquot sequence starting at 2 decreases as follows 2->1->0 and is longer than the sequence starting at 1. Hence 2 is in the sequence.

Amiram Eldar, Table of n, a(n) for n = 1..59

Cf. A081699, A081705, A098008.

nbdecr(n) = {nb = 0; while (n && ((newn = sigma(n)-n)) < n, n = newn ; nb++); nb;}
lista(nn) = {recab = 0; for (ni = 1, nn, ab = nbdecr(ni); if (ab > recab, recab = ab; print1(ni, ", ")););}

A081751 a(n) is the smallest number that is precisely n-tuply abundant.

Original entry on oeis.org

12, 24, 78, 66, 54, 42, 30, 120, 540, 390, 264, 282, 366, 180, 546, 510, 330, 318, 990, 936, 702, 780, 564, 1290, 870, 528, 312, 168, 222, 150, 138, 5778, 6174, 3432, 3150, 2850, 2190, 8432, 4464, 3472, 2480, 1488, 5430, 6750, 4452, 4396, 4650, 3270, 2712

Offset: 1

Gabriel Cunningham (gcasey(AT)mit.edu), Apr 08 2003

nonn

See A081705 for the definition of n-tuply abundant. - David Wasserman, Jun 24 2004

			a(3)=78 because 78 is the smallest number that is exactly triply abundant, with this aliquot chain: 78->90->144->259->45.

Cf. A081705, A081699.

LIMIT = 50; A = vector(LIMIT); count = 0; i = 1; while (count < LIMIT, i = i + 1; ab = 0; lastn = i; n = sigma(i) - i; while(ab <= LIMIT && n > lastn, ab = ab + 1; lastn = n; n = sigma(lastn) - n); if(ab <= LIMIT && ab > 0 && A[ab] == 0, A[ab] = i; count = count + 1)); A \\ David Wasserman, Jun 24 2004

More terms from David Wasserman, Jun 24 2004

cp's OEIS Frontend

A081705 k-tuple abundance of abundant numbers.

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A081700 k-tuple abundance records.

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A234899 Record holders for lengths of ever-decreasing aliquot sequences.

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PARI

A081751 a(n) is the smallest number that is precisely n-tuply abundant.

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PARI

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