A081705 -id:A081705 - cp's OEIS frontend

A008892 Aliquot sequence starting at 276.

276, 396, 696, 1104, 1872, 3770, 3790, 3050, 2716, 2772, 5964, 10164, 19628, 19684, 22876, 26404, 30044, 33796, 38780, 54628, 54684, 111300, 263676, 465668, 465724, 465780, 1026060, 2325540, 5335260, 11738916, 23117724, 45956820, 121129260, 266485716

Offset: 0

N. J. A. Sloane

nonn

It is an open question whether this sequence ever reaches 0. The trajectory has been calculated to 2145 terms, and is still growing, term 2145 being a 214-digit number (see FactorDB link). - N. J. A. Sloane, Jan 11 2023
The aliquot sequence starting at 306 joins this sequence after one step.
This sequence cannot be extended backwards, since A359132(276) = -1. - N. J. A. Sloane, Jan 10 2023
One can note that the k-tuple abundance of 276 is only 5, since a(6) = 3790 is deficient. On the other hand, the k-tuple abundance of a(8) = 2716 is 164 since a(172) is deficient (see A081705 for definition of k-tuple abundance). - Michel Marcus, Dec 31 2013

K. Chum, R. K. Guy, M. J. Jacobson, Jr., and A. S. Mosunov, Numerical and statistical analysis of aliquot sequences. Exper. Math. 29 (2020), no. 4, 414-425; arXiv:2110.14136, Oct. 2021 [math.NT].
Richard K. Guy, Unsolved Problems in Number Theory, B6.
Richard K. Guy and J. L. Selfridge, Interim report on aliquot series, pp. 557-580 of Proceedings Manitoba Conference on Numerical Mathematics. University of Manitoba, Winnipeg, Oct 1971.

Tyler Busby, Table of n, a(n) for n = 0..2157 (terms 0..2127 from Daniel Suteu, terms 2128..2140 from Jeppe Stig Nielsen)
Christophe Clavier, Aliquot Sequences
Christophe Clavier, Trajectory of 276 - the first 1576 terms and their factorizations
Christophe Clavier, Trajectory of 276 - the first 1576 terms and their factorizations [Cached copy]
Wolfgang Creyaufmüller, Lehmer Five
Paul Erdős, Andrew Granville, Carl Pomerance and Claudia Spiro, On the normal behavior of the iterates of some arithmetic functions, Analytic number theory, Birkhäuser Boston, 1990, pp. 165-204.
Paul Erdos, Andrew Granville, Carl Pomerance and Claudia Spiro, On the normal behavior of the iterates of some arithmetic functions, Analytic number theory, Birkhäuser Boston, 1990, pp. 165-204. [Annotated copy with A-numbers]
FactorDB (factordb.com), Search result for last 20 terms of 276 sequence.
Brady Haran and Ben Sparks, An amazing thing about 276, Numberphile YouTube video, 2024.
N. J. A. Sloane, Three (No, 8) Lovely Problems from the OEIS, Experimental Mathematics Seminar, Rutgers University, Oct 05 2017, Part I, Part 2, Slides. (Mentions this sequence)
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 13.
Paul Zimmermann, Recent information
Index entries for sequences related to aliquot parts.

Cf. A001065, A098007 (length of aliquot sequences).

Cf. A008885 (aliquot sequence starting at 30), ..., A008891 (starting at 180).

f := proc(n) option remember; if n = 0 then 276; else sigma(f(n-1))-f(n-1); fi; end:

NestList[DivisorSigma[1, #] - # &, 276, 50] (* Alonso del Arte, Feb 24 2018 *)

a(n, a=276)={for(i=1,n,a=sigma(a)-a);a} \\ M. F. Hasler, Feb 24 2018

a(n+1) = A001065(a(n)). - R. J. Mathar, Oct 11 2017

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

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, ", ")););}

A234842 Primes that are reached by an ever increasing aliquot sequence.

Original entry on oeis.org

463, 523, 983, 1153, 2851, 2969, 4339, 4507, 6121, 8263, 8893, 10093, 12451, 17911, 18427, 18913, 22807, 22811, 25033, 27961, 33223, 36781, 41849, 42643, 48571, 60091, 64237, 71503, 73303, 74131, 90217, 90481, 103813, 108263, 123601, 124447, 125863, 140443

Offset: 1

Michel Marcus, Dec 31 2013

nonn

Note that the starting point of these aliquot sequences are not in increasing order, since for instance we have: 392->463->1 and 324->523->1, that is, with 392>324 while 463<523.
One can observe that the "ever increasing aliquot" part in the definition is not really necessary. A prime is in the sequence if there is an abundant number whose sum of proper divisors results into this prime. So sequence could also be defined as: Primes resulting from summing up the proper divisors of an abundant number. - Michel Marcus, Jan 05 2014
If we try to build the revert sequence listing the starting points of the aliquot sequences, we would get the following terms in increasing order 324, 392, 784, 800, 2304, 2450, 2704, 3600, 3872. But then for n=5352, we'd hit a sequence that begins 5352->8088->12192->20064 and keeps rising to a point where the factors of the last known term are not known. Then later, there are several other such aliquot sequences like 9336->14064->22392 or 10344->15576->27624 that have the same behavior. Thus the only sure terms of the revert sequence would be the terms listed earlier. - Michel Marcus, Jan 11 2014

			The aliquot sequence that begins with 10712 is always increasing before reaching prime 12451: 10712->11128->11552->12451->1, hence 12451 is in the sequence.
20422951 also belongs here with the aliquot sequence that starts at 14952, so a 13-tuple abundant (see factordb link).
People at the Aliquot Sequences project have found longer sequences that reach higher primes.

Michel Marcus, Table of n, a(n) for n = 1..100
Factordb, Sequence starting with 14952
Michel Marcus, Aliquot sequences leading to these primes
Help needed for a sequence Thread on Aliquot Sequences Forum

Cf. A001065, A005101, A080907, A081705, A098007, A098008, A115350.

prev(n) = {for (i=1, n, if ((sigma(i) - i) == n, return (i));); return (0);}
lista(nn) = {forprime(p=2, nn, if (prev(p), print1(p, ", ");););} \\ simplified by Michel Marcus, Jan 11 2014

A254618 a(n) = k-tuple deficiency of n-th deficient number.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 4, 4, 2, 2, 5, 5, 6, 2, 2, 3, 6, 2, 1, 7, 3, 2, 2, 3, 6, 1, 3, 2, 7, 3, 2, 2, 1, 7, 8, 2, 4, 3, 4, 9, 2, 3, 3, 4, 2, 2, 2, 3, 4, 3, 2, 5, 4, 2, 2, 1, 5, 5, 3, 2, 1, 2, 2, 3, 9, 7, 2, 4, 6, 4, 4, 2, 2, 3, 4, 2, 2, 8, 1, 2, 2, 2, 3, 2, 3, 5

Offset: 1

Paolo P. Lava, Feb 03 2015

nonn

For any deficient number x iterate the process f(x)=sigma(x)-x. Sequence lists how many times f(x) keeps deficient until it reaches zero.
Non-deficient numbers are excluded from this sequence.
k-tuple deficiency records is A000027.
k-tuple deficiency record-holders is A234899.

			a(20) = 1 because the 20th deficient number is 25 and:
1) f(25) = sigma(25) - 25 = 6 < 25.
We must stop here because 6 is abundant.
a(21) = 7 because the 21st deficient number is 26 and:
1) f(26) = sigma(26) - 26 = 16 < 26;
2) f(16) = sigma(16) - 16 = 15 < 16;
3) f(15) = sigma(15) - 15 = 9 < 15;
4) f(9) = sigma(9) - 9 = 4 < 9;
5) f(4) = sigma(4) - 4 = 3 < 4;
6) f(3) = sigma(3) - 3 = 2 < 1;
7) f(1) = sigma(1) - 1 = 0 < 1.
We must stop here because sigma(0) is not defined.

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Cf. A000027, A005100, A081705, A098007, A098008, A234899.

with(numtheory): P:=proc(q) local a,b,n,t;
for n from 1 to q do t:=0; b:=sigma(n)-n; a:=n;
if b

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

A008892 Aliquot sequence starting at 276.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A081699 k-tuple abundance record-holders.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Extensions

A081700 k-tuple abundance records.

Views

Author

Keywords

Comments

Examples

Crossrefs

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A234842 Primes that are reached by an ever increasing aliquot sequence.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A254618 a(n) = k-tuple deficiency of n-th deficient number.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

PARI

Extensions