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.
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
Keywords
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}}
Links
- Jeffrey C. Lagarias, An Elementary Problem Equivalent to the Riemmann Hypothesis, arXiv:math/0008177 [math.NT], 2000-2001; Amer. Math. Monthly, 109 (2002), 534-543.
Crossrefs
Programs
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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]
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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.
Comments