A081705 -id:A081705

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A008892

Aliquot sequence starting at 276.

+10
25

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,1

COMMENTS

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

REFERENCES

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.

LINKS

Tyler Busby, Table of n, a(n) for n = 0..2146 (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.

FORMULA

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

MAPLE

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

MATHEMATICA

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

PROG

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

CROSSREFS

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

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

KEYWORD

nonn

AUTHOR

N. J. A. Sloane

STATUS

approved

A081699

k-tuple abundance record-holders.

+10
4

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

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

LINKS

Table of n, a(n) for n=1..12.

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

EXAMPLE

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).

CROSSREFS

Cf. A081700, A081705.

KEYWORD

nonn,hard,more

AUTHOR

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

EXTENSIONS

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

STATUS

approved

A081700

k-tuple abundance records.

+10
2

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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.

LINKS

Table of n, a(n) for n=1..12.

EXAMPLE

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

CROSSREFS

Cf. A081699, A081705.

KEYWORD

hard,nonn

AUTHOR

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

EXTENSIONS

5 more terms from David Wasserman, Jun 16 2004

STATUS

approved

A234899

Record holders for lengths of ever-decreasing aliquot sequences.

+10
2

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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?

LINKS

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

EXAMPLE

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.

PROG

(PARI) 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, ", ")); ); }

CROSSREFS

Cf. A081699, A081705, A098008.

KEYWORD

nonn

AUTHOR

Michel Marcus, Jan 01 2014

STATUS

approved

A234842

Primes that are reached by an ever increasing aliquot sequence.

+10
1

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

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

LINKS

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

EXAMPLE

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.

PROG

(PARI) 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

CROSSREFS

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

KEYWORD

nonn

AUTHOR

Michel Marcus, Dec 31 2013

STATUS

approved

A254618

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

+10
1

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

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.

LINKS

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

EXAMPLE

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.

MAPLE

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<a then while b<a do t:=t+1; a:=b; b:=sigma(b)-b; od;

print(t); fi; od; end: P(10^3);

CROSSREFS

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

KEYWORD

nonn

AUTHOR

Paolo P. Lava, Feb 03 2015

STATUS

approved

A081751

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

+10
0

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

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

LINKS

Table of n, a(n) for n=1..49.

EXAMPLE

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

PROG

(PARI) 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

CROSSREFS

Cf. A081705, A081699.

KEYWORD

nonn

AUTHOR

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

EXTENSIONS

More terms from David Wasserman, Jun 24 2004

STATUS

approved

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