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.

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A048698 Nonprime numbers k such that sum of aliquot divisors of k is a cube.

Original entry on oeis.org

1, 10, 49, 56, 69, 76, 122, 133, 568, 578, 1001, 1018, 1243, 1324, 1431, 1611, 1685, 1819, 1994, 2296, 2323, 3344, 3403, 3627, 3641, 3763, 3981, 4336, 5482, 8186, 9077, 9641, 10113, 10688, 13471, 14188, 14509, 14727, 15940, 16697, 17141, 17619, 19241, 19637
Offset: 1

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Author

Keywords

Comments

The sum of the aliquot divisors of a prime is exactly 1. - Martin Renner, Oct 13 2011

Examples

			a(4) = 56: the aliquot divisors 1,2,4,7,8,14,28 sum to 64, a cube.
		

Crossrefs

Programs

  • Maple
    a := []; for n from 1 to 1000 do if sigma(n) <> n+1 and type( simplify((sigma(n)-n)^(1/3)), `integer`) then a := [op(a), n]; fi; od: a;
  • Mathematica
    Select[Range[20000], !PrimeQ[#] && IntegerQ @ Surd[DivisorSigma[1, #] - #, 3] &] (* Amiram Eldar, Feb 23 2020 *)
  • PARI
    c=0; for(n=1, 13127239, if(isprime(n)==0, if(ispower(sigma(n)-n, 3), c++; write("b048698.txt", c " " n)))) /* Donovan Johnson, Mar 10 2013 */

A187771 Numbers whose sum of divisors is the cube of the sum of its prime divisors.

Original entry on oeis.org

245180, 612408, 639198, 1698862, 1721182, 5154168, 7824284, 15817596, 20441848, 25969788, 27688078, 28404862, 35860609, 67149432, 77378782, 91397838, 96462862, 179302264, 191550135, 289772221, 306901244, 311657084, 392802179, 441839706, 572673855, 652117774, 988918364
Offset: 1

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Author

Manuel Valdivia, Jan 04 2013

Keywords

Comments

This sequence and A187824 and A187761 are winners in the contest held at the 2013 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 14 2013
The identity sigma(k) = (sopf(k))^m only occurs for m = 3 (this sequence) in the given range, however it is likely that it also occurs for other powers m in higher numbers.
The smallest k such that sigma(k) = sopf(k)^m, for m=4,5,6 are 1056331752 (A221262), 213556659624 (A221263) and 45770980141656, respectively. - Giovanni Resta, Jan 07 2013
Prime divisors are taken without multiplicity. - Harvey P. Dale, Dec 17 2016

Examples

			a(13) = 35860609 = 41 * 71 * 97 * 127, then sigma(35860609) = 37933056 = (41 + 71 + 97 + 127)^3.
		

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 38.

Crossrefs

Cf. A221262 (sigma(k)=sopf(k)^4), A221263 (sigma(k)=sopf(k)^5).

Programs

  • Mathematica
    d[n_]:= If[Plus@@Divisors[n]-Power[Plus@@Select[Divisors[n], PrimeQ], 3]==0, n]; Select[Range[2,10^9], #==d[#]&]
    Select[Range[2, 10^9],DivisorSigma[1,#]==Total[FactorInteger[#][[All, 1]]]^3&] (* Harvey P. Dale, Dec 17 2016 *)
  • PARI
    is(n)=my(f=factor(n));sum(i=1,#f~,f[i,1])^3==sigma(n) \\ Charles R Greathouse IV, Jun 29 2013

Formula

a(n) = k if sigma(k) = (sopf(k))^3, where sigma(k) = A000203(k) and sopf(k) = A008472(k).

A048253 a(n) is the number of integers whose sum of divisors is 6^n.

Original entry on oeis.org

1, 1, 1, 5, 11, 18, 30, 48, 85, 148, 250, 415, 669, 1066, 1697, 2635, 4036, 6111, 9137, 13540, 19930, 29098, 42184, 60655, 86598, 122821, 173314, 243469, 340329, 473221, 654779, 901741, 1236668, 1689322, 2298592, 3115200, 4206016, 5658677, 7588039
Offset: 0

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Author

Keywords

Examples

			For n=3, sigma(1,k) = 6^3 = 216 for each of 5 integers: 102, 110, 142, 159, and 187, so a(3) = 5.
		

Crossrefs

Programs

  • Mathematica
    With[{s = Array[DivisorSigma[1, #] &, 6^8]}, Array[Count[s, 6^#] &, Log[6, Length@ s] + 1, 0]] (* Michael De Vlieger, May 14 2018 *)
  • PARI
    a(n) = sum(k=1, 6^n, sigma(k)==6^n); \\ Michel Marcus, May 14 2018

Extensions

a(9)-a(14) from Donovan Johnson, Sep 02 2008
Edited and extended by Ray Chandler, Sep 01 2010

A063869 Least k such that sigma(k)=m^n for some m>1.

Original entry on oeis.org

2, 3, 7, 217, 21, 2667, 93, 217, 381, 651, 2752491, 2667, 8191, 11811, 24573, 57337, 82677, 172011, 393213, 761763, 1572861, 2752491, 5332341, 11010027, 21845397, 48758691, 85327221, 199753347, 341310837, 677207307, 1398273429, 3220807683
Offset: 1

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Author

Labos Elemer, Aug 27 2001

Keywords

Comments

For n=2 to 20 sigma(a(n)) = m^n with m=2 or m=4. Computed terms are products of Mersenne primes (A000668). Is this true for larger n? Validity of a(11) was tested individually.
The Nagell-Ljunggren conjecture implies that sigma(x) is never 3^n for n>1. If this is true, then m=2 and m=4 are the smallest possible solutions. When A063883(n)>0, we can take m=2 and, as explained by Brown, find k to be a product of Mersenne primes (i.e. one of the numbers in A046528). When A063883(n)=0, which is true for the n in A078426, then m=4 and we have a(n)=a(2n) because 4=2^2. - T. D. Noe, Oct 18 2006
Sierpiński says that he proved sigma(x) is never 3^r for r>1. Hence m=2 and m=4 are the smallest possible solutions. When A063883(n)>0, we can take m=2 and, as explained by Brown, find k to be a product of Mersenne primes (i.e. one of the numbers in A046528). When A063883(n)=0, which is true for the n in A078426, then m=4 and we have a(n)=a(2n) because 4=2^2. - T. D. Noe, Oct 18 2006

Examples

			For n = 11, sigma(a(n)) = sigma(2752491) = sigma(3 * 7 * 131071) = 4^11.
		

Crossrefs

Programs

Formula

a(n) = Min{x : A000203(x)=m^n} for some m.

Extensions

a(24) corrected by T. D. Noe, Oct 15 2006

A194948 Numbers k such that sum of aliquot divisors of k, sigma(k) - k, is a cube.

Original entry on oeis.org

1, 2, 3, 5, 7, 10, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 56, 59, 61, 67, 69, 71, 73, 76, 79, 83, 89, 97, 101, 103, 107, 109, 113, 122, 127, 131, 133, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233
Offset: 1

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Author

Martin Renner, Oct 13 2011

Keywords

Comments

For prime numbers, the sum of their aliquot divisors is exactly 1 = 1^3.

Examples

			a(6) = 10, since the sum of aliquot divisors 1 + 2 + 5 = 8 = 2^3.
		

Crossrefs

Union of A000040 and A048698.

Programs

  • Maple
    for n do s:=numtheory[sigma](n)-n; if root(s,3)=trunc(root(s,3)) then print(n); fi; od:
  • Mathematica
    Select[Range[250],IntegerQ[Power[DivisorSigma[1,#]-#, (3)^-1]]&] (* Harvey P. Dale, Nov 25 2011 *)

A334339 Least positive integer m such that sigma(m * n) is a cube, where sigma(k) is the sum of the divisors of k.

Original entry on oeis.org

1, 51, 34, 291, 22, 17, 1, 1347, 597, 11, 10, 97, 892, 51, 46, 1758, 6, 3540, 343, 1649, 34, 5, 30, 449, 2928, 446, 199, 291, 472, 23, 34, 879, 235, 3, 22, 1770, 8661, 356, 3007, 1593, 884, 17, 241, 298, 1416, 15, 22, 586, 133, 1464, 2, 223, 3, 1180, 2, 1347, 711, 236, 232, 1062, 1200, 17, 597, 96771, 586, 265, 577, 485, 10, 11
Offset: 1

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Author

Zhi-Wei Sun, Apr 23 2020

Keywords

Comments

Conjecture: a(n) exists for any n > 0. In other words, for any positive integer n, there is a positive integer m with sigma(m * n) equal to a cube.
The author's conjecture in A259915 implies that for each n = 1, 2, 3, ... there is a positive integer m with sigma(m * n) equal to a square.
See also A334337 for a similar conjecture.

Examples

			a(2) = 51 with sigma(2*51) = 216 = 6^3.
a(4) = 291 with sigma(4*291) = 2744 = 14^3.
a(578) = 34312749 with sigma(578*34312749) = 42144192000 = 3480^3.
a(673) = 49061802 with sigma(673*49061802) = 66135317184 = 4044^3.
		

Crossrefs

Programs

  • Mathematica
    cubeQ[n_] := cubeQ[n] = IntegerQ[n^(1/3)];
    sigma[n_] := sigma[n] = DivisorSigma[1, n];
    tab = {}; Do[m = 0; Label[aa]; m = m + 1; If[cubeQ[sigma[m * n]], tab = Append[tab, m], Goto[aa]], {n, 70}]; tab
    lpi[n_]:=Module[{k=1},While[!IntegerQ[Surd[DivisorSigma[1,n*k],3]],k++]; k]; Array[lpi,70] (* Harvey P. Dale, Nov 05 2020 *)
  • PARI
    a(n) = my(m=1); while (!ispower(sigma(n*m), 3), m++); m; \\ Michel Marcus, Apr 23 2020

Extensions

Corrected and extended by Harvey P. Dale, Nov 05 2020

A048255 Integers whose sum of divisors is 6^5 = 7776.

Original entry on oeis.org

3210, 3498, 3710, 3882, 3910, 4310, 4922, 4982, 5182, 5457, 5885, 6035, 6095, 6307, 6797, 7117, 7327, 7597
Offset: 1

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Author

Keywords

Comments

Sequence has A048253(5)=18 terms from A048251(5)=3210 to A048252(5)=7597. - Ray Chandler

Examples

			Divisors of 7597 are {1,71,107,7597}, whose sum is 7776, so 7597 is in the sequence.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[7600],DivisorSigma[1,#]==7776&] (* Harvey P. Dale, Jun 04 2016 *)
  • PARI
    for(i=1,t=6^5, sigma(i)==t & print1(i",")) \\ M. F. Hasler, Dec 09 2009

Formula

A048255 = { n | A000203(n)=6^5 }. - M. F. Hasler, Dec 09 2009

Extensions

Minor edits, keywords added, and values checked with given PARI code by M. F. Hasler, Dec 09 2009

A244428 Sum of divisors of n and product of divisors of n are both perfect cubes.

Original entry on oeis.org

1, 1164, 8148, 11596, 12028, 28128, 32980, 34144, 34528, 36244, 38764, 39916, 41164, 41516, 73200, 75252, 81172, 84196, 94023, 100348, 181948, 182430, 192175, 193380, 193612, 194044, 195780, 196896, 200574, 204180, 208416, 211620, 214176, 217668, 220116, 225696, 230860, 235716
Offset: 1

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Author

Derek Orr, Jun 27 2014

Keywords

Comments

This is also the intersection of A020477 and A048944.
Numbers m such that sigma(m) is a cube and (m is a cube or number of divisors of m is a multiple of 3). - Chai Wah Wu, Mar 10 2016

Examples

			The divisors of 1164 are {1, 2, 3, 4, 6, 12, 97, 194, 291, 388, 582, 1164}. 1*2*3*4*6*12*97*194*291*388*582*1164 = 2487241979165915136 = 1354896^3 = (1164^2)^3. 1+2+3+4+6+12+97+194+291+388+582+1164 = 2744 = 14^3. Thus, since both the sum of divisors and the product of divisors are perfect cubes, 1164 is a member of this sequence.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[236000],AllTrue[{CubeRoot[DivisorSigma[1,#]],CubeRoot[Times@@Divisors[#]]},IntegerQ]&] (* Harvey P. Dale, Nov 26 2024 *)
  • PARI
    for(n=1,10^6,d=divisors(n);s=sum(i=1,#d,d[i]);p=prod(j=1,#d,d[j]);if(ispower(s,3)&&ispower(p,3),print1(n,", ")))
    
  • Python
    from gmpy2 import iroot
    from sympy import divisor_sigma
    A244428_list = [i for i in range(1,10**4) if (iroot(i,3)[1] or not divisor_sigma(i,0) % 3) and iroot(int(divisor_sigma(i,1)),3)[1]] # Chai Wah Wu, Mar 10 2016

A284284 Let x be the sum of the divisors d_i of k such that d_i | sigma(k). Sequence lists the numbers k for which x^3 = sigma(k).

Original entry on oeis.org

1, 690, 714, 75432, 81172, 81192, 81624, 82248, 84196, 305320, 312040, 315880, 619542, 639198, 646758, 665874, 684342, 737694, 743958, 750114, 751626, 761454, 762966, 763614, 4349280, 4651680, 4789920, 4939680, 4981920, 5259936, 5325216, 5428896, 5474976
Offset: 1

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Author

Paolo P. Lava, Mar 24 2017

Keywords

Comments

Subset of A020477.

Examples

			Divisors of 690 are 1, 2, 3, 5, 6, 10, 15, 23, 30, 46, 69, 115, 138, 230, 345, 690 and sigma(690) = 1728. Then:
1728 / 1 = 1728, 1728 / 2 = 864, 1728 /  3 = 576, 1728 / 6 = 288 and (1 + 2 + 3 + 6)^2 = 12^3 = 1728.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local a,k,n,x;
    for n from 1 to q do a:=sort([op(divisors(n))]); x:=0;
    for k from 1 to nops(a)-1 do if type(sigma(n)/a[k],integer) then x:=x+a[k]; fi; od;
    if x^3=sigma(n) then print(n); fi; od; end: P(10^6);
  • Mathematica
    Select[Range[10^5], (d = DivisorSigma[1, #]; IntegerQ[ d^(1/3)] && d == DivisorSigma[1, GCD[d, #]]^3) &] (* Giovanni Resta, Mar 28 2017 *)

Extensions

a(1), a(25)-a(33) from Giovanni Resta, Mar 28 2017

A329239 Numbers k such that the sum of unitary divisors of k is a cube.

Original entry on oeis.org

1, 7, 102, 110, 120, 142, 159, 184, 187, 381, 588, 684, 690, 714, 770, 796, 840, 931, 994, 1034, 1054, 1065, 1113, 1128, 1173, 1240, 1265, 1288, 1293, 1309, 1528, 1633, 1643, 2619, 2667, 3638, 3937, 4280, 4505, 4664, 4788, 4830, 4855, 5176, 5572, 5671, 5730
Offset: 1

Views

Author

Amiram Eldar, Mar 13 2020

Keywords

Examples

			7 is a term since usigma(7) = 8 = 2^3, where usigma is A034448.
102 is a term since usigma(102) = 216 = 6^3.
		

Crossrefs

The unitary version of A020477.

Programs

  • Mathematica
    usigma[1] = 1; usigma[n_] := Times @@ (1 + Power @@@ FactorInteger[n]); Select[Range[10000], IntegerQ @ Surd[usigma [#], 3] &]
  • PARI
    isok(m) = ispower(sumdiv(m, d, if(gcd(d, m/d)==1, d)), 3); \\ Michel Marcus, Mar 15 2020
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