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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.

%I A347687 #34 Mar 12 2022 14:09:00
%S A347687 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,
%T A347687 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,
%U A347687 0,0,5,0,5,0,0,0,1,0,1,6,4,3,5,2,2,5,3,4,6,1,0
%N 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.
%C A347687 Stated another way, T(n,k) = (1/(2*n)) * (A347684(n,k+n) - A347684(n,k)) for 1 <= k <= n.
%C A347687 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.
%C A347687 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.
%C A347687 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.
%H A347687 N. J. A. Sloane, <a href="/A347687/b347687.txt">Table of n, a(n) for n = 1..5050</a> [Rows 1 through 100, flattened]
%H A347687 N. J. A. Sloane, <a href="/A347687/a347687.txt">Rows 1 through 40</a>
%e A347687 The initial rows are:
%e A347687 1,  [0],
%e A347687 2,  [1, 0],
%e A347687 3,  [1, 1, 0],
%e A347687 4,  [1, 0, 1, 0],
%e A347687 5,  [1, 2, 2, 1, 0],
%e A347687 6,  [1, 0, 0, 0, 1, 0],
%e A347687 7,  [1, 3, 2, 2, 3, 1, 0],
%e A347687 8,  [1, 0, 3, 0, 3, 0, 1, 0],
%e A347687 9,  [1, 4, 0, 2, 2, 0, 4, 1, 0],
%e A347687 10, [1, 0, 3, 0, 0, 0, 3, 0, 1, 0],
%e A347687 11, [1, 5, 4, 3, 2, 2, 3, 4, 5, 1, 0],
%e A347687 12, [1, 0, 0, 0, 5, 0, 5, 0, 0, 0, 1, 0],
%e A347687 ...
%e A347687 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.
%p A347687 myfun1 := proc(A,B) local Ar,Br;
%p A347687 if igcd(A,B) > 1 then return(0); fi;
%p A347687   Ar:=(A)^(-1) mod B;
%p A347687    if 2*Ar > B then Ar:=B-Ar; fi;
%p A347687   Br:=(B)^(-1) mod A;
%p A347687    if 2*Br > A then Br:=A-Br; fi;
%p A347687 A*Ar+B*Br;
%p A347687 end;
%p A347687 T87:=(n,k) -> (myfun1(n,n+k)-myfun1(n,k))/(2*n);
%p A347687 for n from 1 to 20 do lprint([seq(T87(n,k),k=1..n)]); od:
%Y A347687 Cf. A347681, A347682, A347683, A347684.
%K A347687 nonn,tabl
%O A347687 1,12
%A A347687 _N. J. A. Sloane_, Sep 19 2021