A063769 -id:A063769 - cp's OEIS frontend

A127652 Integers whose unitary aliquot sequences are longer than their ordinary aliquot sequences.

25, 28, 36, 40, 50, 68, 70, 74, 94, 95, 98, 116, 119, 134, 142, 143, 154, 162, 170, 175, 182, 189, 190, 200, 220, 226, 242, 245, 262, 273

Offset: 1

Ant King, Jan 24 2007

Here the length of an aliquot sequence is defined to be the length of the transient part of its trajectory + the length of its terminal cycle.

			a(5)=50 because the fifth integer whose unitary aliquot sequence is longer than its ordinary aliquot sequence is 50.

Riele, H. J. J. te; Unitary Aliquot Sequences. MR 139/72, Mathematisch Centrum, 1972, Amsterdam.
Riele, H. J. J. te; Further Results On Unitary Aliquot Sequences. NW 2/73, Mathematisch Centrum, 1973, Amsterdam.

Manuel Benito and Juan L. Varona, Advances In Aliquot Sequences, Mathematics of Computation, Vol. 68, No. 225, (1999), pp. 389-393.
Wolfgang Creyaufmueller, Aliquot Sequences.

Cf. A127161, A127162, A127163, A127164, A063769, A063990, A097032, A098007, A097010, A127653, A098185, A127654, A063991, A127655, A097037, A097036.

UnitaryDivisors[n_Integer?Positive]:=Select[Divisors[n],GCD[ #,n/# ]==1&];sstar[n_]:=Plus@@UnitaryDivisors[n]-n;g[n_] := If[n > 0, sstar[n], 0];UnitaryTrajectory[n_] := Most[NestWhileList[g, n, UnsameQ, All]];s[n_]:=DivisorSigma[1,n]-n;h[n_] := If[n > 0, s[n], 0];OrdinaryTrajectory[n_] := Most[NestWhileList[h, n, UnsameQ, All]];Select[Range[275],Length[UnitaryTrajectory[ # ]]>Length[OrdinaryTrajectory[ # ]] &]

Sequence gives those values of n for which A097032(n)>A098007(n).

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

A371421 Numbers whose aliquot-like sequence based on the largest aliquot divisor of the sum of divisors of n (A371418) terminates in a fixed point.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 43, 45, 46, 47, 49, 50, 51, 52, 53, 55, 57, 58, 59, 61, 63, 64, 67, 68, 71, 73, 74, 79, 80, 81, 82, 89, 93, 96, 97, 98, 100, 101

Offset: 1

Amiram Eldar, Mar 23 2024

nonn

It is unknown whether 222 is a term of this sequence or not (see A371423).

			3 is a term because when we start with 3 and repeatedly apply the mapping x -> A371418(x), we get the sequence 3, 2, 1, 1, 1, ...
40 is a term because when we start with 40 and repeatedly apply the mapping x -> A371418(x), we get the sequence 40, 45, 39, 28, 28, 28, ...

Amiram Eldar, Table of n, a(n) for n = 1..113
Robert D. Carmichael, Empirical Results in the Theory of Numbers, The Mathematics Teacher, Vol. 14, No. 6 (1921), pp. 305-310; alternative link. See p. 309.

Cf. A371418, A371422, A371423.
A023194 is a subsequence.
Cf. A063769, A080907.

r[n_] := n/FactorInteger[n][[1, 1]]; f[n_] := r[DivisorSigma[1, n]]; q[n_] := Module[{m = NestWhileList[f, n, UnsameQ, All][[-1]]}, f[m] == m]; Select[Range[221], q]

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.

A099771 Smallest n-aspiring number. That is, a(n) = smallest k such that s^(n)(k) is perfect but s^(n-1)(k) is not, where s(k) is the sum of the aliquot parts of k and s^(i) means iterate s i times.

Original entry on oeis.org

25, 95, 417, 675, 2541, 3888, 3528, 16256, 13984, 11312, 10648, 10688, 6672, 15364, 20476, 12288, 12636, 32216, 33304, 33896, 34504, 38660, 31824, 15792, 62296, 67304, 49120, 58104, 102740, 82120, 84704, 53680

Offset: 1

Gabriel Cunningham (gcasey(AT)mit.edu), Nov 11 2004

nonn

			a(2) = 95 because s(s(95)) = s(25) = 6, which is perfect.

T. D. Noe, Table of n, a(n) for n=1..100

Cf. A063769.

A347769 a(0) = 0; a(1) = 1; for n > 1, a(n) = A001065(a(n-1)) = sigma(a(n-1)) - a(n-1) (the sum of aliquot parts of a(n-1)) if this is not yet in the sequence; otherwise a(n) is the smallest number missing from the sequence.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 16, 15, 13, 14, 17, 18, 21, 19, 20, 22, 23, 24, 36, 55, 25, 26, 27, 28, 29, 30, 42, 54, 66, 78, 90, 144, 259, 45, 33, 31, 32, 34, 35, 37, 38, 39, 40, 50, 43, 41, 44, 46, 47, 48, 76, 64, 63, 49, 51, 52, 53, 56, 57, 58, 59, 60, 108, 172

Offset: 0

Eric Chen, Sep 13 2021

nonn

This sequence is a permutation of the nonnegative integers iff Catalan's aliquot sequence conjecture (also called Catalan-Dickson conjecture) is true.
a(563) = 276 is the smallest number whose aliquot sequence has not yet been fully determined.
As long as the aliquot sequence of 276 is not known to be finite or eventually periodic, a(563+k) = A008892(k).

			a(0) = 0, a(1) = 1;
since A001065(a(1)) = 0 has already appeared in this sequence, a(2) = 2;
since A001065(a(2)) = 1 has already appeared in this sequence, a(3) = 3;
...
a(11) = 11;
since A001065(a(11)) = 1 has already appeared in this sequence, a(12) = 12;
since A001065(a(12)) = 16 has not yet appeared in this sequence, a(13) = A001065(a(12)) = 16;
since A001065(a(13)) = 15 has not yet appeared in this sequence, a(14) = A001065(a(13)) = 15;
since A001065(a(14)) = 9 has already appeared in this sequence, a(15) = 13;
...

Eric Chen, Table of n, a(n) for n = 0..2703 (all currently known terms, after the b-file of A008892)
Eric Weisstein's World of Mathematics, Aliquot sequence
Eric Weisstein's World of Mathematics, Catalan's Aliquot Sequence Conjecture
Wikipedia, Aliquot sequence

Cf. A032451.
Cf. A001065 (sum of aliquot parts).
Cf. A003023, A044050, A098007, A098008: ("length" of aliquot sequences, four versions).
Cf. A007906.
Cf. A115060 (maximum term of aliquot sequences).
Cf. A115350 (termination of the aliquot sequences).
Cf. A098009, A098010 (records of "length" of aliquot sequences).
Cf. A290141, A290142 (records of maximum term of aliquot sequences).
Cf. A000396, A063990, A122726, A206708, A347770.
Cf. A080907, A063769, A121507, A121508, A131884, A126016.
Aliquot sequences starting at various numbers: A000004 (0), A000007 (1), A033322 (2), A010722 (6), A143090 (12), A143645 (24), A010867 (28), A008885 (30), A143721 (38), A008886 (42), A143722 (48), A143723 (52), A008887 (60), A143733 (62), A143737 (68), A143741 (72), A143754 (75), A143755 (80), A143756 (81), A143757 (82), A143758 (84), A143759 (86), A143767 (87), A143846 (88), A143847 (96), A143919 (100), A008888 (138), A008889 (150), A008890 (168), A008891 (180), A203777 (220), A008892 (276), A014360 (552), A014361 (564), A074907 (570), A014362 (660), A269542 (702), A045477 (840), A014363 (966), A014364 (1074), A014365 (1134), A074906 (1521), A143930 (3630), A072891 (12496), A072890 (14316), A171103 (46758), A072892 (1264460).

A347769_list(N)=print1(0, ", "); if(N>0, print1(1, ", ")); v=[0, 1]; b=1; for(n=2, N, if(setsearch(Set(v), sigma(b)-b), k=1; while(k<=n, if(!setsearch(Set(v), k), b=k; k=n+1, k++)), b=sigma(b)-b); print1(b, ", "); v=concat(v, b))

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A127652 Integers whose unitary aliquot sequences are longer than their ordinary aliquot sequences.

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

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A371421 Numbers whose aliquot-like sequence based on the largest aliquot divisor of the sum of divisors of n (A371418) terminates in a fixed point.

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A135245 Aliquot predecessors with the largest degrees.

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A099771 Smallest n-aspiring number. That is, a(n) = smallest k such that s^(n)(k) is perfect but s^(n-1)(k) is not, where s(k) is the sum of the aliquot parts of k and s^(i) means iterate s i times.

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A347769 a(0) = 0; a(1) = 1; for n > 1, a(n) = A001065(a(n-1)) = sigma(a(n-1)) - a(n-1) (the sum of aliquot parts of a(n-1)) if this is not yet in the sequence; otherwise a(n) is the smallest number missing from the sequence.

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