A255315 - OEIS

%I #95 May 06 2020 02:42:20

%S 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,

%T 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,

%U 1,2,1,2,1,1,1,1,1,1

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

%C The sum of terms of row n is n. Length of row n is n.

%C From _Mats Granvik_, Feb 21 2016: (Start)

%C 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))).

%C 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)))).

%C (End)

%C From _Mats Granvik_, May 28 2017: (Start)

%C 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))).

%C 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)))).

%C (End)

%C From _Mats Granvik_, Sep 07 2017: (Start)

%C It appears that:

%C 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.

%C The number of 1's in column k is given by A152948(k+2).

%C The number of 2's in column k is given by A000096(k-1).

%C The row index of the last nonzero entry in column k is given by A005563(k).

%C (End)

%C From _Mats Granvik_, Oct 06 2018: (Start)

%C The smallest k such that T(n,k)=2 is given by A079643(n) = floor(n/floor(sqrt(n))).

%C This gives the lower bound: A006218(n) >= A094761(n) + A079643(n)*2*(A000196(n)-1).

%C <=> A006218(n) >= 2*n - (floor(sqrt(n)))^2 + floor(n/floor(sqrt(n)))*2*floor(sqrt(n)-1).

%C The average of k:s such that T(n,k)=2, for n>3 is given by:

%C b(n) = Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2)))/floor(sqrt(n)-1).

%C 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.

%C The largest k such that T(n,k)=2 is given by A004526(n) = floor(n/2).

%C This gives the upper bound: A006218(n) <= A094761(n) + A004526(n)*2*(A000196(n)-1).

%C <=> A006218(n) <= 2*n - (floor(sqrt(n)))^2 + floor(n/2)*2*floor(sqrt(n)-1).

%C The lower bound starts:  1, 3, 5, 8, 10, 14, 16, 20, 21, 23, ...

%C Sequence A006218 starts: 1, 3, 5, 8, 10, 14, 16, 20, 23, 27, ...

%C The upper bound starts:  1, 3, 5, 8, 10, 14, 16, 20, 25, 31, ...

%C (End)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DirichletDivisorProblem.html">Dirichlet Divisor Problem</a>

%F See Mathematica program.

%e 1;

%e 1, 1;

%e 1, 1, 1;

%e 0, 2, 1, 1;

%e 0, 2, 1, 1, 1;

%e 0, 1, 2, 1, 1, 1;

%e 0, 1, 2, 1, 1, 1, 1;

%e 0, 1, 1, 2, 1, 1, 1, 1;

%e 0, 0, 2, 2, 1, 1, 1, 1, 1;

%e 0, 0, 2, 1, 2, 1, 1, 1, 1, 1;

%e 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1;

%e 0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1;

%t (* From _Mats Granvik_, Feb 21 2016: (Start) *)

%t nn = 12;

%t T = Table[

%t    Sum[Table[

%t      If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,

%t        If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,

%t        1, r}], {k, 1, r}], {r, 1, nn}];

%t Flatten[T]

%t A006218a = Table[(n^2 - (2*Sum[Sum[T[[n, k]], {k, 1, kk}], {kk, 1, n}] -

%t         n)) + 2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n,

%t      1, nn}];

%t A006218b = -Table[(n^2 - (2*

%t           Sum[Sum[T[[n, n - k + 1]], {k, 1, kk}], {kk, 1, n}] - n)) -

%t      2*n + Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];

%t (A006218b - A006218a);

%t (* (End) *)

%t (* From _Mats Granvik_, May 28 2017: (Start) *)

%t nn = 12;

%t T = Table[

%t    Sum[Table[

%t      If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,

%t        If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,

%t        1, r}], {k, 1, r}], {r, 1, nn}];

%t Flatten[T]

%t A006218a = Table[(n^2 - (2*Sum[T[[n, k]]*(n - k + 1), {k, 1, n}] - n)) +

%t     2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];

%t A006218b = Table[-((n^2 - (2*Sum[T[[n, n - k + 1]]*(n - k + 1), {k, 1, n}] -

%t           n)) - 2*n +

%t       Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])]), {n, 1, nn}];

%t (A006218b - A006218a);

%t (* (End) *)

%Y Cf. A000096, A000196, A004526, A005563, A006218, A079643, A094761, A094820, A152948.

%K tabl,nonn

%O 1,8

%A _Mats Granvik_, May 31 2015