login
This site is supported by donations to The OEIS Foundation.

 

Logo


Hints
(Greetings from The On-Line Encyclopedia of Integer Sequences!)
A255315 Lower triangular matrix describing the shape of a half hyperbola in the Dirichlet divisor problem. 1
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,8

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)

Compare the formula for this table to the formula in A094820.

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)

LINKS

Table of n, a(n) for n=1..78.

Eric Weisstein's World of Mathematics, Dirichlet Divisor Problem

FORMULA

T(n,a) = Sum_{b=1..n} [a*b <= n and ((a + 1)*(b + 1) > n or a < b)]. - Mats Granvik, Oct 06 2018

EXAMPLE

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;

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) *)

CROSSREFS

Cf. A006218, A094820.

Sequence in context: A282380 A083661 A029369 * A125072 A162642 A139146

Adjacent sequences:  A255312 A255313 A255314 * A255316 A255317 A255318

KEYWORD

tabl,nonn

AUTHOR

Mats Granvik, May 31 2015

STATUS

approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

License Agreements, Terms of Use, Privacy Policy. .

Last modified September 16 08:36 EDT 2019. Contains 327091 sequences. (Running on oeis4.)