A347687 Triangle read by rows: T(n,k) (1<=k<=n) = (r(n+k)-r(k))/(2*n), where {r(i), i>=1} is the n-th row of array A347684.

0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 2, 2, 1, 0, 1, 0, 0, 0, 1, 0, 1, 3, 2, 2, 3, 1, 0, 1, 0, 3, 0, 3, 0, 1, 0, 1, 4, 0, 2, 2, 0, 4, 1, 0, 1, 0, 3, 0, 0, 0, 3, 0, 1, 0, 1, 5, 4, 3, 2, 2, 3, 4, 5, 1, 0, 1, 0, 0, 0, 5, 0, 5, 0, 0, 0, 1, 0, 1, 6, 4, 3, 5, 2, 2, 5, 3, 4, 6, 1, 0

Offset: 1

Author

N. J. A. Sloane, Sep 19 2021

Keywords

• nonn
• tabl

Comments

Stated another way, T(n,k) = (1/(2*n)) * (A347684(n,k+n) - A347684(n,k)) for 1 <= k <= n.

Conjecture: Row n is not just the rescaled differences between the first two blocks of n terms of row n of A347684, but is in fact the rescaled differences between any two successive blocks of n terms that start at positions == 1 (mod n). See also the comments in A347684.

It would be nice to have an independent characterization of this triangle. There is a lot of structure - see the link for the first 40 rows.

Except for the final 0 in each row, the rows begin with 1 and are palindromic. T(n,k) = 0 iff gcd(n,k) > 1.

Examples

			The initial rows are:
1,  [0],
2,  [1, 0],
3,  [1, 1, 0],
4,  [1, 0, 1, 0],
5,  [1, 2, 2, 1, 0],
6,  [1, 0, 0, 0, 1, 0],
7,  [1, 3, 2, 2, 3, 1, 0],
8,  [1, 0, 3, 0, 3, 0, 1, 0],
9,  [1, 4, 0, 2, 2, 0, 4, 1, 0],
10, [1, 0, 3, 0, 0, 0, 3, 0, 1, 0],
11, [1, 5, 4, 3, 2, 2, 3, 4, 5, 1, 0],
12, [1, 0, 0, 0, 5, 0, 5, 0, 0, 0, 1, 0],
...
Row 5 of A347684 starts 1, 9, 11, 9, 0, 11, 29, 31, 19, 0, 21, 49, 51, 29, 0, 31, ... Subtracting the first block of five terms from the second block of five terms we get [11, 29, 31, 19, 0] - [1, 9, 11, 9, 0] = [10, 20, 20, 10, 0], and after dividing by 10 we get [1, 2, 2, 1, 0], which is row 5 of the present triangle.
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..5050 [Rows 1 through 100, flattened]
• N. J. A. Sloane, Rows 1 through 40

Crossrefs

Cf. A347681, A347682, A347683, A347684.

Programs

Maple

myfun1 := proc(A,B) local Ar,Br;
if igcd(A,B) > 1 then return(0); fi;
  Ar:=(A)^(-1) mod B;
   if 2*Ar > B then Ar:=B-Ar; fi;
  Br:=(B)^(-1) mod A;
   if 2*Br > A then Br:=A-Br; fi;
A*Ar+B*Br;
end;
T87:=(n,k) -> (myfun1(n,n+k)-myfun1(n,k))/(2*n);
for n from 1 to 20 do lprint([seq(T87(n,k),k=1..n)]); od: