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 1-2 of 2 results.

A362081 Numbers k achieving record abundance (sigma(k) > 2*k) via a residue-based measure M(k) (see Comments), analogous to superabundant numbers A004394.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 36, 72, 120, 360, 420, 840, 1680, 2520, 4032, 5040, 10080, 25200, 32760, 65520, 98280, 194040, 196560, 388080, 942480, 1801800, 3160080, 3603600, 6320160, 12640320, 24504480, 53721360, 61981920, 73513440, 115315200, 122522400, 189909720, 192099600, 214885440
Offset: 1

Views

Author

Richard Joseph Boland, Apr 08 2023

Keywords

Comments

The residue-based quantifier function, M(k) = (k+1)*(1 - zeta(2)/2) - 1 - ( Sum_{j=1..k} k mod j )/k, measures either abundance (sigma(k) > 2*k), or deficiency (sigma(k) < 2*k), of a positive integer k. It follows from the known facts that Sum_{j=1..k} (sigma(j) + k mod j) = k^2 and that the average order of sigma(k)/k is Pi^2/6 = zeta(2) (see derivation below).
M(k) ~ 0 when sigma(k) ~ 2*k and for sufficiently large k, M(k) is positive when k is an abundant number (A005101) and negative when k is a deficient number (A005100). The terms of this sequence are the abundant k for which M(k) > M(m) for all m < k, analogous to the superabundant numbers A004394, which utilize sigma(k)/k as the measure. However, sigma(k)/k does not give a meaningful measure of deficiency, whereas M(k) does, thus a sensible notion of superdeficient (see A362082).

Examples

			The abundance measure is initially negative, becoming positive for k > 30. Initial measures with factorizations from the Mathematica program:
   1  -0.64493406684822643647   {{1,1}}
   2  -0.46740110027233965471   {{2,1}}
   4  -0.36233516712056609118   {{2,2}}
   6  -0.25726923396879252765   {{2,1},{3,1}}
  12  -0.10873810118013850374   {{2,2},{3,1}}
  24  -0.10334250226949712257   {{2,3},{3,1}}
  30  -0.096478036147509765322  {{2,1},{3,1},{5,1}}
  36   0.068719763307810925260  {{2,2},{3,2}}
  72   0.12657322670640173542   {{2,3},{3,2}}
		

Crossrefs

Programs

  • Mathematica
    Clear[max, Rp, R, seqtable, M];
    max = -1; Rp = 0; seqtable = {};
    Do[R = Rp + 2 k - 1 - DivisorSigma[1, k];
      M = N[(k + 1)*(1 - Zeta[2]/2) - 1 - R/k, 20];
      If[M > max, max = M; Print[k, "   ", max, "   ", FactorInteger[k]];
       AppendTo[seqtable, k]];
      Rp = R, {k, 1, 1000000000}];
    Print[seqtable]
  • PARI
    M(n) = (n+1)*(1 - zeta(2)/2) - 1 - sum(k=2, n, n%k)/n;
    lista(nn) = my(m=-oo, list=List()); for (n=1, nn, my(mm = M(n)); if (mm > m, listput(list, n); m = mm);); Vec(list); \\ Michel Marcus, Apr 21 2023

Formula

Derived starting with lemmas 1-3:
1) Sum_{j=1..k} (sigma(j) + k mod j) = k^2.
2) The average order of sigma(k)/k is Pi^2/6 = zeta(2).
3) R(k) = Sum_{j=1..k} k mod j, so R(k)/k is the average order of (k mod j).
Then:
Sum_{j=1..k} sigma(j) ~ zeta(2)*Sum_{j=1..k} j = zeta(2)*(k^2+k)/2.
R(k)/k ~ k - k*zeta(2)/2 - zeta(2)/2.
0 ~ (k+1)*(1 - zeta(2)/2) - 1 - R(k)/k.
Thus M(k) = (k+1)*(1 - zeta(2)/2) - 1 - R(k)/k is a measure of variance about sigma(k) ~ 2*k corresponding to M(k) ~ 0.

A362082 Numbers k achieving record deficiency via a residue-based measure, M(k) = (k+1)*(1 - zeta(2)/2) - 1 - ( Sum_{j=1..k} k mod j )/k.

Original entry on oeis.org

1, 5, 11, 23, 47, 59, 167, 179, 359, 503, 719, 1439, 5039, 6719, 7559, 15119, 20159, 52919, 75599, 83159, 166319, 415799, 720719, 831599, 1081079, 2162159, 4324319, 5266799, 7900199, 10533599, 18345599, 28274399, 41081039, 136936799, 205405199, 410810399
Offset: 1

Views

Author

Richard Joseph Boland, Apr 17 2023

Keywords

Comments

M(k) = (k+1)*(1 - zeta(2)/2) - 1 - ( Sum_{j=1..k} k mod j )/k is a measure of either abundance (sigma(k) > 2*k), or deficiency (sigma(k) < 2*k), of a positive integer k. The measure follows from the known facts that Sum_{j=1..k} (sigma(j) + k mod j) = k^2 and that the average order of sigma(k)/k is Pi^2/6 = zeta(2) (see derivation below).
M(k) ~ 0 when sigma(k) ~ 2*k and for sufficiently large k, M(k) is positive when k is an abundant number (A005101) and negative when k is a deficient number (A005100).
The terms of this sequence are the deficient k for which M(k) < M(m) for all m < k and may be thought of as "superdeficient", contra-analogous to the superabundant numbers A004394 utilizing sigma(k)/k as the measure of abundance, which is otherwise not particularly meaningful as a deficiency measure.
15119=13*1163 is the first term that is composite and subsequently, up to 1000000000, roughly half of the terms are composite.

Examples

			First few terms with their M(k) measure and factorizations as generated by the Mathematica program:
    1   -0.64493406684822643647   {{1,1}}
    5   -0.73480220054467930942   {{5,1}}
   11   -0.86960440108935861883  {{11,1}}
   23   -1.0000783673961085420   {{23,1}}
   47   -1.0528856894638174541   {{47,1}}
   59   -1.1107338698535727552   {{59,1}}
  167   -1.1984137110594038972  {{167,1}}
  179   -1.2619431113124463216  {{179,1}}
  359   -1.3499704727921791778  {{359,1}}
  503   -1.3722914063892448936  {{503,1}}
  719   -1.4363475145965658088  {{719,1}}
		

Crossrefs

Cf. A362081 (analogous to superabundant A004394).
Cf. A362083 (analogous to A335067, A326393).

Programs

  • Mathematica
    Clear[min, Rp, R, seqtable, M]; min = 1; Rp = 0; seqtable = {};
    Do[R = Rp + 2 k - 1 - DivisorSigma[1, k];
      M = N[(k + 1)*(1 - Zeta[2]/2) - 1 - R/k, 20];
      If[M < min, min = M; Print[k, "   ", min, "   ", FactorInteger[k]];
       AppendTo[seqtable, k]];
      Rp = R, {k, 1, 1000000000}];
    Print[seqtable]
  • PARI
    M(n) = (n+1)*(1 - zeta(2)/2) - 1 - sum(k=2, n, n%k)/n;
    lista(nn) = my(m=+oo, list=List()); for (n=1, nn, my(mm = M(n)); if (mm < m, listput(list, n); m = mm);); Vec(list); \\ Michel Marcus, Apr 21 2023

Formula

Derived starting with lemmas 1-3:
1) Sum_{j=1..k} (sigma(j) + k mod j) = k^2.
2) The average order of sigma(k)/k is Pi^2/6 = zeta(2).
3) R(k) = Sum_{j=1..k} k mod j, so R(k)/k is the average order of (k mod j).
Then:
Sum_{j=1..k} sigma(j) ~ zeta(2)*Sum_{j=1..k} j = zeta(2)*(k^2+k)/2.
R(k)/k ~ k - k*zeta(2)/2 - zeta(2)/2.
0 ~ (k+1)*(1 - zeta(2)/2) - 1 - R(k)/k.
Thus M(k) = (k+1)*(1 - zeta(2)/2) - 1 - R(k)/k is a measure of variance about sigma(k) ~ 2*k corresponding to M(k) ~ 0.
Showing 1-2 of 2 results.