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A255315
Lower triangular matrix describing the shape of a half hyperbola in the Dirichlet divisor problem.
0
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
1,8
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)
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
Eric Weisstein's World of Mathematics, Dirichlet Divisor Problem
FORMULA
See Mathematica program.
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) *)
KEYWORD
tabl,nonn
AUTHOR
Mats Granvik, May 31 2015
STATUS
approved