cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 13 results. Next

A357313 a(n) is the unique number m such that A001065(m) = A057709(n).

Original entry on oeis.org

4, 9, 8, 15, 14, 21, 121, 289, 529, 46, 28, 841, 62, 24, 1369, 217, 2209, 106, 68, 3481, 54, 4489, 134, 5041, 66, 6241, 158, 6889, 166, 116, 9409, 10201, 218, 226, 148, 130, 114, 16129, 17161, 18769, 278, 90, 24649, 314, 26569, 27889, 108, 29929, 225, 32041, 244
Offset: 1

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Author

Amiram Eldar, Sep 23 2022

Keywords

Examples

			a(1) = 4 since 4 is the unique solution m to A001065(m) = A057709(1) = 3.
		

Crossrefs

Programs

  • Mathematica
    seq[max_] := Module[{s = t = Table[0, {n, 1, max}], i, j}, Do[If[(i = DivisorSigma[1, n] - n) <= max, s[[i]]++; t[[i]] = n], {n, 2, (max - 1)^2}]; j = Position[s, 1] // Flatten; t[[j]]]; seq[250]

Formula

A001065(a(n)) = A057709(n).

A005114 Untouchable numbers, also called nonaliquot numbers: impossible values for the sum of aliquot parts function (A001065).

Original entry on oeis.org

2, 5, 52, 88, 96, 120, 124, 146, 162, 188, 206, 210, 216, 238, 246, 248, 262, 268, 276, 288, 290, 292, 304, 306, 322, 324, 326, 336, 342, 372, 406, 408, 426, 430, 448, 472, 474, 498, 516, 518, 520, 530, 540, 552, 556, 562, 576, 584, 612, 624, 626, 628, 658
Offset: 1

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Author

Keywords

Comments

Complement of A078923. - Lekraj Beedassy, Jul 19 2005
Chen & Zhao show that the lower density of this sequence is at least 0.06, improving on te Riele. - Charles R Greathouse IV, Dec 28 2013
Numbers k such that A048138(k) = 0. A048138(k) measures how "touchable" k is. - Jeppe Stig Nielsen, Jan 12 2020
From Amiram Eldar, Feb 13 2021: (Start)
The term "untouchable number" was coined by Alanen (1972). He found the 570 terms below 5000.
Erdős (1973) proved that the lower asymptotic density of untouchable numbers is positive, te Riele (1976) proved that it is > 0.0324, and Banks and Luca (2004, 2005) proved that it is > 1/48.
Pollack and Pomerance (2016) conjectured that the asymptotic density is ~ 0.17. (End)
The upper asymptotic density is less than 1/2 by the 'almost all' binary Goldbach conjecture, independently proved by Nikolai Chudakov, Johannes van der Corput, and Theodor Estermann. (In this context, this shows that the density of the odd numbers of this form is 0 (consider A001065(p*q) for prime p, q); full Goldbach would prove that 5 is the only odd number in this sequence.) - Charles R Greathouse IV, Dec 05 2022

References

  • Richard K. Guy, Unsolved Problems in Number Theory, 3rd ed., Springer, 2004, section B10, pp. 100-101.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 65.
  • József Sándor, Dragoslav S. Mitrinovic, Borislav Crstici, Handbook of Number Theory I, Springer Science & Business Media, 2005, page 93.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 147.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 125.

Crossrefs

Programs

  • Mathematica
    untouchableQ[n_] := Catch[ Do[ If[n == DivisorSigma[1, k]-k, Throw[True]], {k, 0, (n-1)^2}]] === Null; Reap[ Table[ If[ untouchableQ[n], Print[n]; Sow[n]], {n, 2, 700}]][[2, 1]] (* Jean-François Alcover, Jun 29 2012, after Benoit Cloitre *)
  • PARI
    isA078923(n)=if(n==0 || n==1, return(1)); for(m=1,(n-1)^2, if( sigma(m)-m == n, return(1))); 0
    isA005114(n)=!isA078923(n)
    for(n=1,700, if (isA005114(n), print(n))) \\ R. J. Mathar, Aug 10 2006
    
  • PARI
    is(n)=if(n%2 && n<4e18, return(n==5)); forfactored(m=1,(n-1)^2, if(sigma(m)-m[1]==n, return(0))); 1 \\ Charles R Greathouse IV, Dec 05 2022
    
  • Python
    from sympy import divisor_sigma as sigma
    from functools import cache
    @cache
    def f(m): return sigma(m)-m
    def okA005114(n):
        if n < 2: return 0
        return not any(f(m) == n for m in range(1, (n-1)**2+1))
    print([k for k in range(289) if okA005114(k)]) # Michael S. Branicky, Nov 16 2024
    
  • Python
    # faster for intial segment of sequence
    from itertools import count, islice
    from sympy import divisor_sigma as sigma
    def agen(): # generator of terms
        n, touchable, t = 2, {0, 1}, 1
        for m in count(2):
            touchable.add(sigma(m)-m)
            while m > t:
                if n not in touchable:
                    yield n
                else:
                    touchable.discard(n)
                n += 1
                t = (n-1)**2
    print(list(islice(agen(), 20))) # Michael S. Branicky, Nov 16 2024

Extensions

More terms from David W. Wilson

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

Original entry on oeis.org

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

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Author

Keywords

Comments

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

Examples

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

Crossrefs

Programs

  • Maple
    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:
  • PARI
    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

Formula

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)

Extensions

More terms from James Sellers, Feb 19 2001

A057710 Positive integers k with exactly 2 aliquot sequence predecessors. In other words, there are exactly two solutions x for which s(x) = n. The function s(x) here is the sum of all proper divisors of x (A001065).

Original entry on oeis.org

6, 8, 13, 14, 15, 16, 17, 19, 20, 22, 23, 27, 29, 32, 42, 44, 46, 50, 54, 62, 69, 90, 92, 100, 104, 108, 110, 114, 130, 136, 148, 150, 152, 156, 166, 170, 176, 182, 184, 186, 198, 200, 202, 214, 230, 232, 234, 236, 240, 242, 244, 254, 258, 266, 272, 280, 286
Offset: 1

Views

Author

Jack Brennen, Oct 24 2000

Keywords

Examples

			14 is a member of the sequence because s(22) = 14 and s(169) = 14 (and because no other integer x satisfies s(x) = 14).
		

Crossrefs

Programs

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

Views

Author

Klaus Brockhaus, Nov 27 2006

Keywords

Comments

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

Examples

			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.
		

Crossrefs

Programs

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

A357324 Numbers k such that there is a unique m for which the sum of the aliquot unitary divisors of m (A034460) is k.

Original entry on oeis.org

6, 9, 11, 13, 128, 150, 164, 222, 224, 332, 338, 390, 404, 416, 420, 458, 510, 548, 558, 570, 576, 582, 584, 598, 660, 668, 750, 788, 800, 810, 818, 822, 836, 852, 878, 884, 926, 930, 1046, 1118, 1200, 1202, 1230, 1244, 1250, 1260, 1284, 1298, 1304, 1382, 1422, 1472, 1478
Offset: 1

Views

Author

Amiram Eldar, Sep 24 2022

Keywords

Comments

Numbers k such that A324938(k) = 1.

Crossrefs

The unitary version of A057709.

Programs

  • Mathematica
    us[1] = 0; us[n_] := Times @@ (1 + Power @@@ FactorInteger[n]) - n; m = 1500; v = Table[0, {m}]; Do[u = us[k]; If[2 <= u <= m, v[[u]]++], {k, 1, m^2}]; Position[v, 1] // Flatten

Formula

a(n) = A034460(A357325(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

Views

Author

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

Keywords

Comments

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

Examples

			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).
		

Crossrefs

Programs

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

Extensions

a(1)=0 removed and offset set to 2 by Michel Marcus, Nov 11 2014
New name from Michel Marcus, Oct 31 2023

A357325 a(n) is the unique number m such that A034460(m) = A357324(n).

Original entry on oeis.org

6, 15, 21, 35, 250, 138, 4192, 10048, 6112, 748, 20736, 5968, 802, 12256, 41728, 3592, 498, 53632, 8656, 80128, 2284, 2308, 36352, 2372, 10288, 5272, 11728, 84352, 1594, 630, 6472, 48448, 6616, 50368, 1426, 1762, 102016, 172288, 32416, 8872, 2328, 9544, 19408
Offset: 1

Views

Author

Amiram Eldar, Sep 24 2022

Keywords

Crossrefs

The unitary version of A357313.

Programs

  • Mathematica
    us[1] = 0; us[n_] := Times @@ (1 + Power @@@ FactorInteger[n]) - n; m = 1500; v = s = Table[0, {m}]; Do[u = us[k]; If[2 <= u <= m, v[[u]]++; s[[u]] = k], {k, 1, m^2}]; s[[Position[v, 1] // Flatten]]

Formula

A034460(a(n)) = A357324(n).

A372742 Numbers k such that there is a unique number m for which the sum of the aliquot coreful divisors of m (A336563) is k.

Original entry on oeis.org

2, 3, 5, 7, 11, 12, 13, 17, 18, 19, 23, 24, 29, 31, 36, 37, 41, 43, 47, 53, 56, 59, 61, 67, 71, 73, 79, 80, 83, 84, 89, 96, 97, 98, 101, 103, 107, 109, 112, 113, 127, 131, 135, 137, 139, 140, 149, 150, 151, 156, 157, 163, 167, 173, 179, 181, 191, 193, 197, 198
Offset: 1

Views

Author

Amiram Eldar, May 12 2024

Keywords

Comments

A coreful divisor d of n is a divisor that is divisible by every prime that divides n (see also A307958).
Numbers k such that A372739(k) = 1.
The corresponding values of m are in A372743.
Includes all prime numbers.

Crossrefs

A000040 is a subsequence.
Similar sequences: A057709, A357324, A361419.

Programs

  • Mathematica
    f[p_, e_] := (p^(e + 1) - 1)/(p - 1) - 1; s[1] = 0; s[n_] := Times @@ f @@@ FactorInteger[n] - n; seq[max_] := Module[{v = Table[0, {max}], i}, Do[i = s[k]; If[1 <= i <= max, v[[i]]++], {k, 1, max^2}]; Position[v, 1] // Flatten]; seq[200]
  • PARI
    s(n) = {my(f = factor(n)); prod(i = 1, #f~, (f[i, 1]^(f[i, 2] + 1) - 1)/(f[i, 1] -1) - 1) - n;}
    lista(nmax) = {my(v = vector(nmax), i); for(k=1, nmax^2, i = s(k); if(i > 0 && i <= nmax, v[i]++)); for(k = 1, nmax, if(v[k] == 1, print1(k, ", ")));}

Formula

a(n) = A336563(A372743(n)).

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

Views

Author

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

Keywords

Comments

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).

Examples

			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).
		

Crossrefs

Showing 1-10 of 13 results. Next