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 1-8 of 8 results.

A008848 Squares whose sum of divisors is a square.

Original entry on oeis.org

1, 81, 400, 32400, 1705636, 3648100, 138156516, 295496100, 1055340196, 1476326929, 2263475776, 2323432804, 2592846400, 2661528100, 7036525456, 10994571025, 17604513124, 39415749156, 61436066769, 85482555876, 90526367376, 97577515876, 98551417041
Offset: 1

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Author

Keywords

Comments

Solutions to sigma(x^2) = (2k+1)^2. - Labos Elemer, Aug 22 2002
Intersection of A006532 and A000290. The product of any two coprime terms is also in this sequence. - Charles R Greathouse IV, May 10 2011
Also intersection of A069070 and A000290. - Michel Marcus, Oct 06 2013
Conjectures: (1) a(2) = 81 is the only prime power (A246655) in this sequence. (2) 81 and 400 are only terms x for which sigma(x) is in A246655. (3) x = 1 is the only such term that sigma(x) is also a term. See also comments in A074386, A336547 and A350072. - Antti Karttunen, Jul 03 2023, (2) corrected in May 11 2024

Examples

			n=81: sigma(81) = 1+3+9+27+81 = 121 = 11^2.
n=400: sigma(400) = sigma(16)*sigma(25) = 31*31 = 961.
n=32400 (= 81*400): sigma(32400) = 116281 = 341^2 = 121*961.
		

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 10.
  • I. Kaplansky, The challenges of Fermat, Wallis and Ozanam (and several related challenges): II. Fermat's second challenge, Preprint, 2002.

Crossrefs

Terms of A008847 squared.
Subsequence of A000290, of A006532, and of A069070.

Programs

  • Mathematica
    Do[s=DivisorSigma[1, n^2]; If[IntegerQ[Sqrt[s]]&&Mod[s, 2]==1, Print[n^2]], {n, 1, 10000000}] (* Labos Elemer *)
    Select[Range[320000]^2,IntegerQ[Sqrt[DivisorSigma[1,#]]]&] (* Harvey P. Dale, Feb 22 2015 *)
  • PARI
    for(n=1,1e6,if(issquare(sigma(n^2)), print1(n^2", "))) \\ Charles R Greathouse IV, May 10 2011

Formula

a(n) = A008847(n)^2.

A350072 a(n) = sigma(n^2) / gcd(sigma(n^2), A003961(n^2)), where A003961 shifts the prime factorization of n one step towards larger primes, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 7, 13, 31, 31, 91, 57, 127, 121, 31, 133, 403, 183, 133, 403, 511, 307, 847, 381, 961, 741, 931, 553, 1651, 781, 427, 1093, 589, 871, 403, 993, 2047, 133, 2149, 1767, 3751, 1407, 889, 2379, 3937, 1723, 1729, 1893, 4123, 3751, 3871, 2257, 6643, 2801, 781, 3991, 1891, 2863, 7651, 589, 2413, 4953, 6097, 3541, 12493
Offset: 1

Views

Author

Antti Karttunen, Dec 12 2021

Keywords

Comments

Conjecture: There are no 1's after the initial term. Remark: If there were some k = x^2 > 1, for which a(x) = 1, then sigma(k) would be a divisor of A003961(k). In other words, d = A350073(k) = A064989(sigma(k)) would be a divisor of k. Then, if that divisor were also a unitary divisor [with gcd(d,k/d) = 1], it would need to satisfy the equation sigma(k) = sigma(d) * sigma(k/d) = sigma(A064989(sigma(k))) * sigma(k/A064989(sigma(k))), because sigma is a multiplicative function. (Minor correction by Antti Karttunen, Jul 11 2023)
Note that if d = A064989(sigma(k)) were a unitary divisor of a square k, then sigma(k) would also be a square, the cases which are quite rare (see A008848 and A336547). Also compare to A349756. - Antti Karttunen, Jul 24 2022

Crossrefs

Programs

  • Mathematica
    f1[p_, e_] := (p^(2*e + 1) - 1)/(p - 1); f2[p_, e_] := NextPrime[p]^(2*e); a[1] = 1; a[n_] := (s = Times @@ f1 @@@ (f = FactorInteger[n])) / GCD[s, Times @@ f2 @@@ f]; Array[a, 60] (* Amiram Eldar, Dec 12 2021 *)
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A349162(n) = { my(s=sigma(n)); (s/gcd(s,A003961(n))); };
    A350072(n) = A349162(n^2);

Formula

a(n) = A349162(n^2).
a(n) = A065764(n) / A350071(n).

A230043 Numbers whose abundancy sigma(n)/n is a rational cube.

Original entry on oeis.org

1, 8232, 32640, 265825, 3846879, 6517665, 14705145, 16926000, 31441920, 56471688, 146475000, 211421364, 277368000, 369022500, 662518050, 679568670, 968353620, 2166699360, 3091750900, 3755367252, 4122716598, 6536970000, 9740587500, 10066738500, 12423246290
Offset: 1

Views

Author

Michel Marcus, Oct 06 2013

Keywords

Comments

All terms listed in the data section are deficient, but all 8-multiperfect numbers (which are abundant...) also belong to this sequence.
As with A230538, it is possible to find larger numbers with same ratio sigma(n)/n, in some cases using perfect numbers A000396 (see a230043.txt link). - Michel Marcus, Oct 30 2013
One motivation for this sequence lies in the fact that n*sigma(n) is a square (A069070) if and only if sigma(n)/n is a rational square. But this does not hold for higher powers: If sigma(n)/n = (p/q)^k then n*sigma(n) = (pq)^k (n/q^k)^2. - M. F. Hasler, Nov 02 2013
In his post to NMBRTHRY, Michiel Kosters gives a 233-digit number x such that sigma(x^3) is a cube. Actually this x^3 also belongs to the sequence, although there are no cubes in the current data. He has found many others such cubes that belong here, the smallest of which is 3590918978816938469301573291605^3, x having 31 digits, and x^3 92 digits. Is it possible to find the smallest such cube, or even a smaller one? - Michel Marcus, Jan 02 2014

Examples

			For n=8232, sigma(n)/n = 1000/343 = (10/7)^3.
		

Crossrefs

Cf. A069070 (abundancy is a square).

Programs

  • Maple
    isQcube := proc(r)
        isA000578(numer(r)) and isA000578(denom(r)) ;
    end proc: # see A000578 for isA000578()
    isA230043 := proc(n)
        abu := numtheory[sigma](n)/n ;
        isQcube(abu) ;
    end proc:
    for n from 1 do
        if isA230043(n) then
            printf("%d,\n",n);
        end if;
    end do: # R. J. Mathar, Oct 08 2013
  • PARI
    is_A230043(n) = ispower(sigma(n)/n, 3);

Extensions

a(11)-a(25) from Donovan Johnson, Oct 10 2013
a(26)-a(33) from Donovan Johnson, Dec 22 2013

A355929 Difference between the squarefree part of the sum of divisors of n and the squarefree part of n.

Original entry on oeis.org

0, 1, -2, 6, 1, -3, -5, 13, 12, -8, -8, 4, 1, -8, -9, 30, -15, 37, -14, 37, -19, -21, -17, 9, 30, 16, 7, 7, 1, -28, -29, 5, -30, -28, -32, 90, 1, -23, -25, 0, 1, -36, -32, 10, 73, -44, -44, 28, 56, 91, -49, -11, -47, 24, -53, 16, -52, -48, -44, 27, 1, -56, 19, 126, -44, -65, -50, -3, -63, -69, -69, 193, 1, 40, 28, 16
Offset: 1

Views

Author

Antti Karttunen, Jul 24 2022

Keywords

Crossrefs

Cf. A000203, A007913, A069070 (positions of 0's), A355928.

Programs

  • PARI
    A355929(n) = (core(sigma(n))-core(n));

Formula

a(n) = A355928(n) - A007913(n) = A007913(A000203(n)) - A007913(n).

A230538 Numbers whose abundancy sigma(n)/n is a rational fourth power.

Original entry on oeis.org

1, 976250, 6276690, 6542085247225
Offset: 1

Views

Author

Michel Marcus, Oct 23 2013

Keywords

Comments

Subsequence of A069070.
Note that there exist several other large numbers with the same abundancy as a(3), that is sigma(6276690)/6276690 = 19837440/6276690 = 256/81. For this, consider the two numbers 559625737239 (3^10*23*107*3851) and 1373356918809 (3^6*23*137*547*1093), both of which have sigma(n)/n = 128/81. As they are coprime to the perfect numbers, except 6, it suffices to multiply them by those terms of A000396 to get an abundancy of 2*128/81 = 256/81. The smallest of these is the 14-digit number 15669520642692. - Michel Marcus, Oct 29 2013
It is also possible to get higher powers for sigma(n)/n, for instance, 1024/243 = (4/3)^5 with n=1556619120, 4096/729 = (4/3)^6 with n=1526227435825092000, 279936/78125 = (6/5)^7 with n=553131046875000, 1679616/390625 = (6/5)^8 with n=15487669312500000. - Michel Marcus, Oct 30 2013
6542085247225 is a term. - Hiroaki Yamanouchi, Sep 22 2014
a(5) > 10^13. - Giovanni Resta, Jun 16 2015

Examples

			For n = 976250, sigma(n)/n = 2024352/976250 = 1296/625 = (6/5)^4.
		

Crossrefs

Cf. A069070 (square), A230043 (cube).

Programs

  • PARI
    isok(n) = ispower(sigma(n)/n, 4); \\ Michel Marcus, Oct 23 2013

Extensions

a(4) from Giovanni Resta, Jun 14 2015

A318084 Numbers m such that sigma(sigma(m))/m is a square.

Original entry on oeis.org

1, 15, 50, 100, 168, 1023, 1444, 1470, 1600, 1944, 3179, 3822, 4000, 5120, 5776, 6174, 9025, 10752, 12348, 14440, 15125, 21970, 26250, 28416, 28665, 29127, 37544, 39200, 45630, 47151, 49392, 52500, 60984, 66125, 67200, 69819, 71485, 77175, 80000, 90250, 100254, 102300, 102400
Offset: 1

Views

Author

Michel Marcus, Aug 15 2018

Keywords

Comments

This is a necessary condition to have sigma(sigma(m))/sigma(m) = sigma(m)/m.
Are there other integers than 1, for which this is satisfied?
If m is an odd number such that sigma(sigma(m^2))/2 is a square, and p is in A000043 such that 2^p-1 does not divide sigma(m^2), then 2^(p-1)*m^2 is in the sequence. Such m include 5, 19, 161, 543, 1031, 1899, 3035, 6673. Thus if A000043 is infinite, so is this sequence. - Robert Israel, Aug 17 2018

Crossrefs

Programs

  • Maple
    filter:= proc(n) local t; t:= (numtheory:-sigma @@2)(n)/n; issqr(numer(t)) and issqr(denom(t)) end proc:select(filter, [$1..200000]); # Robert Israel, Aug 17 2018
  • Mathematica
    Select[Range[10^5], IntegerQ@ Sqrt[ DivisorSigma[1, DivisorSigma[1, #]] #] &] (* Giovanni Resta, Aug 19 2018 *)
  • PARI
    isok(n) = issquare(sigma(sigma(n))/n);

A383482 Integers k such that sigma(k)/k - 1 is a rational square.

Original entry on oeis.org

1, 6, 9, 28, 216, 360, 496, 2016, 2401, 8128, 16758, 182520, 884736, 1022112, 1352328, 1571328, 1935360, 2678400, 33550336, 54758400, 101382400, 119533176, 136808280, 163298502, 198288000, 618591192, 691022088, 782481673, 796663296, 1137067008, 1275418369, 1303102080
Offset: 1

Views

Author

Michel Marcus, Apr 28 2025

Keywords

Examples

			6 is a term because sigma(6)/6 - 1 = 2 - 1 = 1, a square; like for all perfect numbers.
9 is a term because sigma(9)/9 - 1 = 13/9 - 1 = 4/9, a square.
		

Crossrefs

Cf. A000203 (sigma), A069070.
Subsequences: A000396 (perfect numbers), A046060 (5-multiperfect numbers), A381321.
Cf. A218404 (for those terms with sigma(x)/x = 13/4).

Programs

  • Mathematica
    q[k_] := And @@ IntegerQ /@ Sqrt[NumeratorDenominator[DivisorSigma[-1, k] - 1]]; Select[Range[2*10^6], q] (* Amiram Eldar, Apr 28 2025 *)
  • PARI
    isok(k) = issquare(sigma(k)/k - 1);

Extensions

a(30)-a(32) from Jinyuan Wang, Apr 28 2025

A349687 Numbers whose numerator and denominator of their abundancy index are both Fibonacci numbers.

Original entry on oeis.org

1, 2, 6, 15, 24, 26, 28, 84, 90, 96, 120, 270, 330, 496, 672, 1335, 1488, 1540, 1638, 8128, 24384, 27280, 44109, 68200, 131040, 447040, 523776, 18506880, 22256640, 33550336, 36197280, 38257095, 65688320, 91963648, 95472000, 100651008, 102136320, 176432256, 197308800
Offset: 1

Views

Author

Amiram Eldar, Nov 25 2021

Keywords

Comments

This sequence includes all the perfect numbers (A000396), 3-perfect numbers (A005820) and 5-perfect numbers (A046060).
The deficient terms, 1, 2, 15, 26, 1335, 44109, 38257095, ..., have an abundancy index which is a ratio of two consecutive Fibonacci numbers, 1/1, 3/2, 8/5, 21/13, 144/89, 610/377, 46368/28657, ..., which approaches the golden ratio phi = 1.618... (A001622) as the numerators and denominators get larger.

Examples

			2 is a term since sigma(2)/2 = 3/2 = Fibonacci(4)/Fibonacci(3).
15 is a term since sigma(15)/15 = 8/5 = Fibonacci(6)/Fibonacci(5).
		

Crossrefs

Subsequences: A000396, A005820, A046060.
Similar sequences: A069070, A216780, A247086, A348658.

Programs

  • Mathematica
    fibQ[n_] := Or @@ IntegerQ /@ Sqrt[{5 n^2 - 4, 5 n^2 + 4}]; ai[n_] := DivisorSigma[1, n]/n; q[n_] := fibQ[Numerator[(ain = ai[n])]] && fibQ[Denominator[ain]]; Select[Range[10^6], q]
  • PARI
    isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
    isok(n) = my(q=sigma(n)/n); isfib(numerator(q)) && isfib(denominator(q)); \\ Michel Marcus, Nov 25 2021
Showing 1-8 of 8 results.