A057710 -id:A057710 - cp's OEIS frontend

A048138 a(n) = number of m such that sum of proper divisors of m (A001065(m)) is n.

0, 1, 1, 0, 2, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 2, 2, 3, 2, 2, 1, 3, 1, 2, 1, 2, 1, 5, 2, 3, 1, 3, 1, 4, 1, 1, 3, 4, 2, 5, 2, 3, 2, 3, 1, 6, 2, 4, 0, 3, 2, 6, 1, 5, 1, 3, 1, 6, 2, 3, 3, 6, 1, 6, 1, 2, 1, 5, 1, 8, 3, 4, 3, 5, 1, 7, 1, 6, 1, 4, 1, 8, 1, 5, 0, 5, 2, 9, 2, 4, 1, 4, 0, 9, 1, 3, 2, 6, 1, 8, 2, 7, 4

Offset: 2

Naohiro Nomoto

The offset is 2 since there are infinitely many numbers (all the primes) for which A001065 = 1.
The graph of this sequence, shifted by 1, looks similar to that of A061358, which counts Goldbach partitions of n. - T. D. Noe, Dec 05 2008
For n > 2, a(n) <= A000009(n) as all divisor lists must have distinct values. - Roderick MacPhee, Sep 13 2016
The smallest k > 0 such that there are exactly n numbers whose sum of proper divisors is k is A125601(n). - Bernard Schott, Mar 23 2023

			a(6) = 2 since 6 is the sum of the proper divisors of 6 and 25.

T. D. Noe, Table of n, a(n) for n = 2..10000
Carl Pomerance, The first function and its iterates, pp. 125-138 in Connections in Discrete Mathematics, ed. S. Butler et al., Cambridge, 2018.

Cf. A001065, A064440, A125601, A238895, A238896 (records).

Cf. A005114, A057709, A057710, A160133.

with(numtheory): for n from 2 to 150 do count := 0: for m from 1 to n^2 do if sigma(m) - m = n then count := count+1 fi: od: printf(`%d,`,count): od:

list(n)=my(v=vector(n-1),k); for(m=4,n^2, k=sigma(m)-m; if(k>1 & k<=n, v[k-1]++)); v \\ Charles R Greathouse IV, Apr 21 2011

From Bernard Schott, Mar 23 2023: (Start)
a(n) = 0 iff n is in A005114 (untouchable numbers).
a(n) = 1 iff n is in A057709 ("hermit" numbers).
a(n) = 2 iff n is in A057710.
a(n) > 1 iff n is in A160133. (End)

More terms from James Sellers, Feb 19 2001

A057709 Numbers k such that there is a unique m for which the sum of the aliquot parts of m (A001065) is k.

Original entry on oeis.org

3, 4, 7, 9, 10, 11, 12, 18, 24, 26, 28, 30, 34, 36, 38, 39, 48, 56, 58, 60, 66, 68, 70, 72, 78, 80, 82, 84, 86, 94, 98, 102, 112, 116, 118, 122, 126, 128, 132, 138, 142, 144, 158, 160, 164, 168, 172, 174, 178, 180, 190, 192, 204, 208, 212, 220, 222, 224, 228, 250

Offset: 1

Jack Brennen, Oct 24 2000

nonn

Alanen (1972) used the term "hermit" for a number k such that x = k is the only solution to A001065(x) = k. These numbers are the perfect numbers (A000396) in this sequence. Of the first 4 perfect numbers, 6, 28, 496 and 8128, only 28 is a term. - Amiram Eldar, Mar 03 2021

			12 is a member of the sequence because s(121)=12 (and because no other integer m satisfies s(m) = 12).
18 is included because the sum of aliquot parts of 289 = 1+17 = 18, this being the only number with this property. 6 is not included because the sum of aliquot parts of 6 = 1+2+3 = 6 and the sum of aliquot parts of 25 = 1+5 = 6.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
Jack David Alanen, Empirical study of aliquot series, Ph.D Thesis, Yale University, 1972.
Eric Weisstein's World of Mathematics, Restricted Divisor Function.
Eric Weisstein's World of Mathematics, Aliquot Sequence.

Cf. A000396, A001065, A005114, A048138, A057710.

seq[max_] := Module[{s = Table[0, {n, 1, max}], i}, Do[If[(i = DivisorSigma[1, n] - n) <= max, s[[i]]++], {n, 2, (max - 1)^2 }]; Position[s, 1] // Flatten]; seq[250] (* Amiram Eldar, Dec 26 2020 *)

Removed 1 from the sequence. - T. D. Noe, Dec 02 2008

A125601 a(n) is the smallest k > 0 such that there are exactly n numbers whose sum of proper divisors is k.

Original entry on oeis.org

2, 3, 6, 21, 37, 31, 49, 79, 73, 91, 115, 127, 151, 121, 181, 169, 217, 265, 253, 271, 211, 301, 433, 379, 331, 361, 457, 391, 451, 655, 463, 541, 421, 775, 511, 769, 673, 715, 865, 691, 1015, 631, 1069, 1075, 721, 931, 781, 1123, 871, 925, 901, 1177, 991, 1297

Offset: 0

Klaus Brockhaus, Nov 27 2006

nonn

Minimal values for nodes of exact degree in aliquot sequences. Find each node's degree (number of predecessors) in aliquot sequences and choose the smallest value as the sequence member. - Ophir Spector, ospectoro (AT) yahoo.com Nov 25 2007

			a(4) = 37 since there are exactly four numbers (155, 203, 299, 323) whose sum of proper divisors is 37. For k < 37 there are either fewer or more numbers (32, 125, 161, 209, 221 for k = 31) whose sum of proper divisors is k.

Daniel Mondot, Table of n, a(n) for n = 0..5646 (First 157 terms from Ophir Spector. First 1000 terms from Ophir Spector and Donovan Johnson.)
W. Creyaufmueller, Aliquot sequences
Daniel Mondot, A-file, extended version of b-file but with some gaps towards the end.
J. O. M. Pedersen, Tables of Aliquot Cycles [Broken link]
J. O. M. Pedersen, Tables of Aliquot Cycles [Via Internet Archive Wayback-Machine]
J. O. M. Pedersen, Tables of Aliquot Cycles [Cached copy, pdf file only]
Eric Weisstein's World of Mathematics, Aliquot sequence

Cf. A001065, A048138, A070015, A123930, A080907, A115350, A121507, A037020, A126016, A057709, A057710, A063990.

{m=54;z=1500;y=600000;v=vector(z);for(n=2,y,s=sigma(n)-n; if(s

A135244 Largest m such that the sum of the aliquot parts of m (A001065) equals n, or 0 if no such number exists.

Original entry on oeis.org

0, 4, 9, 0, 25, 8, 49, 15, 14, 21, 121, 35, 169, 33, 26, 55, 289, 77, 361, 91, 38, 85, 529, 143, 46, 133, 28, 187, 841, 221, 961, 247, 62, 253, 24, 323, 1369, 217, 81, 391, 1681, 437, 1849, 403, 86, 493, 2209, 551, 94, 589, 0, 667, 2809, 713, 106, 703, 68, 697, 3481

Offset: 2

Ophir Spector (ospectoro(AT)yahoo.com), Nov 25 2007

nonn

Previous name: Aliquot predecessors with the largest values.
Find each node's predecessors in aliquot sequences and choose the largest predecessor.
Climb the aliquot trees on shortest paths (see A135245 = Climb the aliquot trees on thickest branches).
The sequence starts at offset 2, since all primes satisfy sigma(n)-n = 1. - Michel Marcus, Nov 11 2014

			a(25) = 143 since 25 has 3 predecessors (95,119,143), 143 being the largest.
a(5) = 0 since it has no predecessors (see Untouchables - A005114).

Amiram Eldar, Table of n, a(n) for n = 2..10000 (terms 2..150 from Ophir Spector)
Wolfgang Creyaufmueller, Aliquot sequences.
J. O. M. Pedersen, Tables of Aliquot Cycles. [Broken link]
J. O. M. Pedersen, Tables of Aliquot Cycles. [Via Internet Archive Wayback-Machine]
J. O. M. Pedersen, Tables of Aliquot Cycles. [Cached copy, pdf file only]
Eric Weisstein's World of Mathematics, Aliquot sequence.

Cf. A001065, A005114, A125601, A135245, A057709, A057710, A063769, A080907, A121507, A037020, A126016.

seq[max_] := Module[{s = Table[0, {n, 1, max}], i}, Do[If[(i = DivisorSigma[1, n] - n) <= max, s[[i]] = Max[s[[i]], n]], {n, 2, (max - 1)^2}]; Rest @ s]; seq[50]

lista(nn) = {for (n=2, nn, k = (n-1)^2; while(k && (sigma(k)-k != n), k--); print1(k, ", "););} \\ Michel Marcus, Nov 11 2014

a(1)=0 removed and offset set to 2 by Michel Marcus, Nov 11 2014

New name from Michel Marcus, Oct 31 2023

A135245 Aliquot predecessors with the largest degrees.

Original entry on oeis.org

0, 0, 4, 9, 0, 25, 8, 49, 15, 14, 21, 121, 35, 169, 33, 12, 55, 289, 65, 361, 91, 20, 85, 529, 143, 46, 133, 28, 187, 841, 161, 961, 247, 62, 253, 24, 323, 1369, 217, 81, 391, 1681, 341, 1849, 403, 86, 493, 2209, 551, 40, 481, 0, 667, 2809, 533, 106, 703, 68, 697, 3481

Offset: 1

Ophir Spector, ospectoro (AT) yahoo.com, Nov 25 2007

nonn

Find each node's predecessors in aliquot sequences and choose the node with largest number of predecessors.

Climb the aliquot trees on thickest branches (see A135244 = Climb the aliquot trees on shortest paths).

			a(25) = 143 since 25 has 3 predecessors (95,119,143) with degrees (4,5,7), 143 having the largest degree. a(5) = 0 since it has no predecessors (see Untouchables - A005114).

Ophir Spector, Table of n, a(n) for n = 1..150
W. Creyaufmueller, Aliquot sequences
J. O. M. Pedersen, Tables of Aliquot Cycles [Broken link]
J. O. M. Pedersen, Tables of Aliquot Cycles [Via Internet Archive Wayback-Machine]
J. O. M. Pedersen, Tables of Aliquot Cycles [Cached copy, pdf file only]
Eric Weisstein's World of Mathematics, Aliquot sequence

Cf. A001065 A005114 A125601 A135244 A057709 A057710 A063769 A080907 A121507 A037020 A126016.

cp's OEIS Frontend

A048138 a(n) = number of m such that sum of proper divisors of m (A001065(m)) is n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

Extensions

A057709 Numbers k such that there is a unique m for which the sum of the aliquot parts of m (A001065) is k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

A125601 a(n) is the smallest k > 0 such that there are exactly n numbers whose sum of proper divisors is k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A135244 Largest m such that the sum of the aliquot parts of m (A001065) equals n, or 0 if no such number exists.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A135245 Aliquot predecessors with the largest degrees.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs