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.
%I A255315 #96 Feb 16 2025 08:33:25
%S A255315 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 A255315 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 A255315 1,2,1,2,1,1,1,1,1,1
%N A255315 Lower triangular matrix describing the shape of a half hyperbola in the Dirichlet divisor problem.
%C A255315 The sum of terms of row n is n. Length of row n is n.
%C A255315 From _Mats Granvik_, Feb 21 2016: (Start)
%C A255315 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 A255315 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 A255315 (End)
%C A255315 From _Mats Granvik_, May 28 2017: (Start)
%C A255315 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 A255315 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 A255315 (End)
%C A255315 From _Mats Granvik_, Sep 07 2017: (Start)
%C A255315 It appears that:
%C A255315 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 A255315 The number of 1's in column k is given by A152948(k+2).
%C A255315 The number of 2's in column k is given by A000096(k-1).
%C A255315 The row index of the last nonzero entry in column k is given by A005563(k).
%C A255315 (End)
%C A255315 From _Mats Granvik_, Oct 06 2018: (Start)
%C A255315 The smallest k such that T(n,k)=2 is given by A079643(n) = floor(n/floor(sqrt(n))).
%C A255315 This gives the lower bound: A006218(n) >= A094761(n) + A079643(n)*2*(A000196(n)-1).
%C A255315 <=> A006218(n) >= 2*n - (floor(sqrt(n)))^2 + floor(n/floor(sqrt(n)))*2*floor(sqrt(n)-1).
%C A255315 The average of k:s such that T(n,k)=2, for n>3 is given by:
%C A255315 b(n) = Sum_{k=1..n} (k*floor(abs(T(n, k)-1/2)))/floor(sqrt(n)-1).
%C A255315 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 A255315 The largest k such that T(n,k)=2 is given by A004526(n) = floor(n/2).
%C A255315 This gives the upper bound: A006218(n) <= A094761(n) + A004526(n)*2*(A000196(n)-1).
%C A255315 <=> A006218(n) <= 2*n - (floor(sqrt(n)))^2 + floor(n/2)*2*floor(sqrt(n)-1).
%C A255315 The lower bound starts: 1, 3, 5, 8, 10, 14, 16, 20, 21, 23, ...
%C A255315 Sequence A006218 starts: 1, 3, 5, 8, 10, 14, 16, 20, 23, 27, ...
%C A255315 The upper bound starts: 1, 3, 5, 8, 10, 14, 16, 20, 25, 31, ...
%C A255315 (End)
%H A255315 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DirichletDivisorProblem.html">Dirichlet Divisor Problem</a>
%F A255315 See Mathematica program.
%e A255315 1;
%e A255315 1, 1;
%e A255315 1, 1, 1;
%e A255315 0, 2, 1, 1;
%e A255315 0, 2, 1, 1, 1;
%e A255315 0, 1, 2, 1, 1, 1;
%e A255315 0, 1, 2, 1, 1, 1, 1;
%e A255315 0, 1, 1, 2, 1, 1, 1, 1;
%e A255315 0, 0, 2, 2, 1, 1, 1, 1, 1;
%e A255315 0, 0, 2, 1, 2, 1, 1, 1, 1, 1;
%e A255315 0, 0, 2, 1, 2, 1, 1, 1, 1, 1, 1;
%e A255315 0, 0, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1;
%t A255315 (* From _Mats Granvik_, Feb 21 2016: (Start) *)
%t A255315 nn = 12;
%t A255315 T = Table[
%t A255315 Sum[Table[
%t A255315 If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
%t A255315 If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
%t A255315 1, r}], {k, 1, r}], {r, 1, nn}];
%t A255315 Flatten[T]
%t A255315 A006218a = Table[(n^2 - (2*Sum[Sum[T[[n, k]], {k, 1, kk}], {kk, 1, n}] -
%t A255315 n)) + 2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n,
%t A255315 1, nn}];
%t A255315 A006218b = -Table[(n^2 - (2*
%t A255315 Sum[Sum[T[[n, n - k + 1]], {k, 1, kk}], {kk, 1, n}] - n)) -
%t A255315 2*n + Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
%t A255315 (A006218b - A006218a);
%t A255315 (* (End) *)
%t A255315 (* From _Mats Granvik_, May 28 2017: (Start) *)
%t A255315 nn = 12;
%t A255315 T = Table[
%t A255315 Sum[Table[
%t A255315 If[And[If[n*k <= r, If[n >= k, 1, 0], 0] == 1,
%t A255315 If[(n + 1)*(k + 1) <= r, If[n >= k, 1, 0], 0] == 0], 1, 0], {n,
%t A255315 1, r}], {k, 1, r}], {r, 1, nn}];
%t A255315 Flatten[T]
%t A255315 A006218a = Table[(n^2 - (2*Sum[T[[n, k]]*(n - k + 1), {k, 1, n}] - n)) +
%t A255315 2*n - Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])], {n, 1, nn}];
%t A255315 A006218b = Table[-((n^2 - (2*Sum[T[[n, n - k + 1]]*(n - k + 1), {k, 1, n}] -
%t A255315 n)) - 2*n +
%t A255315 Round[1 + (1/2)*(-3 + Sqrt[n] + Sqrt[1 + n])]), {n, 1, nn}];
%t A255315 (A006218b - A006218a);
%t A255315 (* (End) *)
%Y A255315 Cf. A000096, A000196, A004526, A005563, A006218, A079643, A094761, A094820, A152948.
%K A255315 tabl,nonn
%O A255315 1,8
%A A255315 _Mats Granvik_, May 31 2015