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.

A255315 Lower triangular matrix describing the shape of a half hyperbola in the Dirichlet divisor problem.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 0, 2, 1, 1, 1, 0, 1, 2, 1, 1, 1, 0, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 0, 2, 2, 1, 1, 1, 1, 1, 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1
Offset: 1

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Author

Mats Granvik, May 31 2015

Keywords

Comments

The sum of terms of row n is n. Length of row n is n.
From Mats Granvik, Feb 21 2016: (Start)
A006218(n) = (n^2 - ((2*Sum_{kk=1..n} Sum_{k=1..kk} T(n,k)) - n)) + 2*n - round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n))).
A006218(n) = -((n^2 - ((2*Sum_{kk=1..n} Sum_{k=1..kk} T(n,n-k+1)) - n)) - 2*n + round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n)))).
(End)
From Mats Granvik, May 28 2017: (Start)
A006218(n) = (n^2 - (2*(Sum_{k=1..n} T(n, k)*(n - k + 1)) - n)) + 2*n - round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n))).
A006218(n) = -((n^2 - (2*(Sum_{k=1..n} T(n, n - k + 1)*(n - k + 1)) - n)) - 2*n + round(1 + (1/2)*(-3 + sqrt(n) + sqrt(1 + n)))).
(End)
From Mats Granvik, Sep 07 2017: (Start)
It appears that:
The number of 0's in row n is equal to the number of 2's in row n and their number is given by A000196(n) - 1.
The number of 1's in column k is given by A152948(k+2).
The number of 2's in column k is given by A000096(k-1).
The row index of the last nonzero entry in column k is given by A005563(k).
(End)
From Mats Granvik, Oct 06 2018: (Start)
The smallest k such that T(n,k)=2 is given by A079643(n) = floor(n/floor(sqrt(n))).
This gives the lower bound: A006218(n) >= A094761(n) + A079643(n)*2*(A000196(n)-1).
<=> A006218(n) >= 2*n - (floor(sqrt(n)))^2 + floor(n/floor(sqrt(n)))*2*floor(sqrt(n)-1).
The average of k:s such that T(n,k)=2, for n>3 is given by:
b(n) = Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2)))/floor(sqrt(n)-1).
This gives A006218(n) = 2*n - (floor(sqrt(n)))^2 + b(n)*2*floor(sqrt(n)-1) = 2*n - (floor(sqrt(n)))^2 + (Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2))))*2, for n>3.
The largest k such that T(n,k)=2 is given by A004526(n) = floor(n/2).
This gives the upper bound: A006218(n) <= A094761(n) + A004526(n)*2*(A000196(n)-1).
<=> A006218(n) <= 2*n - (floor(sqrt(n)))^2 + floor(n/2)*2*floor(sqrt(n)-1).
The lower bound starts: 1, 3, 5, 8, 10, 14, 16, 20, 21, 23, ...
Sequence A006218 starts: 1, 3, 5, 8, 10, 14, 16, 20, 23, 27, ...
The upper bound starts: 1, 3, 5, 8, 10, 14, 16, 20, 25, 31, ...
(End)

Examples

			1;
1, 1;
1, 1, 1;
0, 2, 1, 1;
0, 2, 1, 1, 1;
0, 1, 2, 1, 1, 1;
0, 1, 2, 1, 1, 1, 1;
0, 1, 1, 2, 1, 1, 1, 1;
0, 0, 2, 2, 1, 1, 1, 1, 1;
0, 0, 2, 1, 2, 1, 1, 1, 1, 1;
0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1;
0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1;
		

Crossrefs

Programs

  • Mathematica
    (* From Mats Granvik, Feb 21 2016: (Start) *)
    nn = 12;
    T = Table[
       Sum[Table[
         If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
           If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
           1, r}], {k, 1, r}], {r, 1, nn}];
    Flatten[T]
    A006218a = Table[(n^2 - (2*Sum[Sum[T[[n, k]], {k, 1, kk}], {kk, 1, n}] -
            n)) + 2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n,
         1, nn}];
    A006218b = -Table[(n^2 - (2*
              Sum[Sum[T[[n, n - k + 1]], {k, 1, kk}], {kk, 1, n}] - n)) -
         2*n + Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
    (A006218b - A006218a);
    (* (End) *)
    (* From Mats Granvik, May 28 2017: (Start) *)
    nn = 12;
    T = Table[
       Sum[Table[
         If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
           If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
           1, r}], {k, 1, r}], {r, 1, nn}];
    Flatten[T]
    A006218a = Table[(n^2 - (2*Sum[T[[n, k]]*(n - k + 1), {k, 1, n}] - n)) +
        2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
    A006218b = Table[-((n^2 - (2*Sum[T[[n, n - k + 1]]*(n - k + 1), {k, 1, n}] -
              n)) - 2*n +
          Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])]), {n, 1, nn}];
    (A006218b - A006218a);
    (* (End) *)

Formula

See Mathematica program.