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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.

User: Jianqiang Zhao

Jianqiang Zhao's wiki page.

Jianqiang Zhao has authored 3 sequences.

A387045 Positive numbers k with property that the largest circle on the xy-plane enclosing exactly k lattice points in its interior does not exist.

Original entry on oeis.org

5, 6, 17, 18, 33, 34, 35, 36, 38, 50, 53, 54, 63, 70, 71, 72, 73, 89, 90, 97, 98, 102, 109, 110, 125, 126, 127, 128, 129, 150, 151, 165, 167, 168, 178, 188, 198, 209, 210, 217, 218, 219, 220, 221, 222, 242, 243, 257, 258, 259, 260, 277, 278, 285, 286, 294
Offset: 1

Views

Author

Jianqiang Zhao, Aug 14 2025

Keywords

Comments

Conjecture: This sequence is infinite.

Examples

			It can be proved that the largest circle enclosing exactly 5 or 6 lattice points in the interior on the xy-plane does not exist. Number 5 is the smallest nonnegative integer having this property and 6 is the next. Therefore, a(1)=5 and a(2)=6.
Here is a brief argument. For details, please see my arxiv paper 2505.06234.
First, let C be the circle going through (-1,0) centered at (1/2,1/2). It passes exactly 8 lattice points and encloses exactly 4. Now with (-1,0) fixed on the circle we can shrink C by an infinitesimal amount to circle C' so that C' only goes through one lattice point (-1,0). Then another infinitesimal perturbation will move C' to include exactly 5 lattice points in its interior. Another infinitesimal perturbation will move C' to include exactly 6 lattice points in its interior. Therefore, if the largest circle enclosing exactly 5 or 6 interior lattice points exists, then its radius is at least sqrt(10)/2.
Second, a geometric argument shows that if the radius of a circle is at least sqrt(10)/2 then it encloses either exactly 4 interior lattice points or at least 7 interior lattice points.

Links

Crossrefs

Cf. A387044, complement of A387045; A192493, A192494, A128006, A128007.

A079404 Let G(n) be the set of numbers between 2^(n-1) and 2^n-1, inclusive. There is a unique number m(n) in G(n) so that the denominator of the m(n)-th partial sum of the double harmonic series is divisible by smaller 2-power than that of others in G(n). This power is defined to be a(n).

Original entry on oeis.org

0, 1, 1, 3, 4, 3, 3, 5, 7, 9, 10, 9, 10, 12, 14, 13, 13, 15, 17, 19, 19
Offset: 2

Views

Author

Jianqiang Zhao (jqz(AT)math.upenn.edu), Jan 06 2003

Keywords

Comments

The sequence is conjectured to go to positive infinity.

Examples

			a(3)=1 because G(3)={4,5,6,7} and among Sum_{1 <= k < l <= 4} 1/(kl) = 35/24, Sum_{1 <= k < l <= 5} 1/(kl) = 15/8, Sum_{1 <= k < l <= 6} 1/(kl) = 203/90, Sum_{1 <= k < l <= 7} 1/(kl) = 469/180, 90 has the smallest 2-power factor among the denominators.

References

  • Partial sums of multiple zeta value series II: finiteness of p-divisible sets.

Links

Crossrefs

Cf. A079403.

Programs

  • Maple
    sequ := proc(T) local b,counter,A,n,t,psum,innersum; psum := 0; innersum := 0; A := array(1..T-1); for t to T-1 do for n from 2^(t) to 2^(t+1)-1 do innersum := innersum+1/(n-1); psum := psum+innersum/n; if 2^(2*t)*psum mod 2^(2*t+1)=0 then print(`The conjecture that 2 never divides the numerators of partial sums of double harmonic series is wrong.`); else b := 0; counter := 2*t; while b=0 do b := 2^counter*psum mod 2; counter := counter-1; od; if counter
  • Mathematica
    nmax = 15; dhs = Array[HarmonicNumber[# - 1 ]/# &, 2^nmax] // Accumulate; Print["dhs finished"];
f[s_] := IntegerExponent[s // Denominator, 2];
a[n_] := Table[{f[dhs[[k]] ], k}, {k, 2^(n - 1), 2^n - 1}] // Sort // First // First;
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 2, nmax}] (* Jean-François Alcover, Jan 22 2018 *)

Extensions

Typo in data corrected by Jean-François Alcover, Jan 22 2018

A079403 Let G(t) be the set of numbers between 2^(t-1) and 2^t-1, inclusive. There is a unique number a(t) in G(t) so that the denominator of the a(t)-th partial sum of the double harmonic series is divisible by smaller 2-powers than its neighbors.

Original entry on oeis.org

3, 6, 13, 27, 54, 109, 219, 439, 879, 1759, 3518, 7037, 14075, 28151, 56303, 112606, 225212, 450424, 900848, 1801696, 3603393, 7206787, 14413574, 28827148, 57654296, 115308593, 230617186, 461234373
Offset: 2

Views

Author

Jianqiang Zhao (jqz(AT)math.upenn.edu), Jan 06 2003

Keywords

Comments

The n-th partial sum of double harmonic series is defined to be Sum_{1 <= k < l <= n} 1/(kl).

Examples

			a(3)=6 because Sum_{1 <= k < l <= 6} 1/(kl) = 203/90, 4 does not divide 90, while 4 divides the denominators of both Sum_{1 <= k < l <= 5} 1/(kl) = 15/8 and Sum_{1 <= k < l <= 7} 1/(kl) = 469/180.

Links

Crossrefs

Cf. A079404.

Programs

  • Maple
    sequ := proc(T) local A,i,n,t,psum,innersum; psum := 0; innersum := 0; A := {}; for t to T-1 do for n from 2^t to 2^(t+1)-1 do innersum := innersum+2^T/(n-1) mod 2^(2*T); psum := psum+2^T*innersum/n mod 2^(2*T); if psum mod 2^(2*T-t+1)=0 then A := A union {n}; end if; od; od; RETURN(A); end:
  • Mathematica
    nmax = 15; dhs = Array[HarmonicNumber[# - 1]/# &, 2^nmax] // Accumulate; Print["dhs finished"];
f[s_] := IntegerExponent[s // Denominator, 2];
a[2] = 3; a[n_] := a[n] = For[k = 2*a[n - 1], k <= 2^n - 1, k++, fk = f[dhs[[k]]]; If[f[dhs[[k-1]]] > fk && f[dhs[[k+1]]] > fk, Return[k]]];
Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 2, nmax}] (* Jean-François Alcover, Jan 22 2018 *)

Formula

From Benoit Cloitre, Jan 24 2003: (Start)
a(n+1) - 2*a(n) = (a(n+1) mod 2);
a(n) = floor(c*2^n) where c = 1.718232... = 3/2 + Sum_{k>=2} (a(k+1) - 2*a(k))/2^k. (End)

Extensions

a(23)-a(29) from Sean A. Irvine, Aug 12 2025

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