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%I #21 Dec 10 2023 17:15:25
%S 0,1,1,1,0,1,1,5,5,1,1,0,0,0,1,1,9,7,7,9,1,1,0,11,0,11,0,1,1,13,0,9,9,
%T 0,13,1,1,0,13,0,0,0,13,0,1,1,17,17,15,11,11,15,17,17,1,1,0,0,0,29,0,
%U 29,0,0,0,1,1,21,19,17,31,13,13,31,17,19,21,1,1,0,23,0,19,0,0,0,19,0,23,0,1
%N Array read by antidiagonals: T(n,k) (n>=1, k>=1) = f(n,k), where f(x,y) = x*red_inv(x,y) + y*red_inv(y,x) if gcd(x,y)=1, or 0 if gcd(x,y)>1, and red_inv is defined in the comments.
%C If u, v are positive integers with gcd(u,v) = 1, the "reduced inverse" red_inv(u,v) of u mod v is u^(-1) mod v if u^(-1) mod v <= v/2, otherwise it is v - u^(-1) mod v.
%C That is, we map u to whichever of +-u has a representative mod v in the range 0 to v/2. Stated another way, red_inv(u,v) is a number r in the range 0 to v/2 such that r*u == +-1 mod v.
%C For example, red_inv(3,11) = 4, since 3^(-1) mod 11 = 4. But red_inv(2,11) = 5 = 11-6, since red_inv(2,11) = 6.
%C Arises in the study of A344005.
%C Conjecture: The entries of this array can be expressed in terms of A347683 and A347687 as follows. Write k = j*n+i, where j >= 0 and 1 <= i <= n. Then T(n,k) = A347683(n,i) + 2*n*j*A347687(n,i). For example, suppose n=3, k=11, so k = 3*3+2, with j=3, i=2. Then A347683(3,2) + 2*3*3*A347687(3,2) = 5 + 2*3*3*1 = 23, which is indeed T(3,11).
%H N. J. A. Sloane, <a href="/A347684/b347684.txt">Table of n, a(n) for n = 1..5050</a> [First 100 antidiagonals, flattened]
%e The array begins:
%e 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,...
%e 1, 0, 5, 0, 9, 0, 13, 0, 17, 0, 21, 0, 25, 0, 29, 0,...
%e 1, 5, 0, 7, 11, 0, 13, 17, 0, 19, 23, 0, 25, 29, 0, 31,...
%e 1, 0, 7, 0, 9, 0, 15, 0, 17, 0, 23, 0, 25, 0, 31, 0,...
%e 1, 9, 11, 9, 0, 11, 29, 31, 19, 0, 21, 49, 51, 29, 0, 31,...
%e 1, 0, 0, 0, 11, 0, 13, 0, 0, 0, 23, 0, 25, 0, 0, 0,...
%e 1, 13, 13, 15, 29, 13, 0, 15, 55, 41, 43, 71, 27, 0, 29, 97,...
%e 1, 0, 17, 0, 31, 0, 15, 0, 17, 0, 65, 0, 79, 0, 31, 0,...
%e 1, 17, 0, 17, 19, 0, 55, 17, 0, 19, 89, 0, 53, 55, 0, 127,...
%e ...
%e The first few antidiagonals are:
%e 0,
%e 1, 1,
%e 1, 0, 1,
%e 1, 5, 5, 1,
%e 1, 0, 0, 0, 1,
%e 1, 9, 7, 7, 9, 1,
%e 1, 0, 11, 0, 11, 0, 1,
%e 1, 13, 0, 9, 9, 0, 13, 1,
%e 1, 0, 13, 0, 0, 0, 13, 0, 1,
%e 1, 17, 17, 15, 11, 11, 15, 17, 17, 1,
%e ...
%p myfun1 := proc(A,B) local Ar,Br;
%p if igcd(A,B) > 1 then return(0); fi;
%p Ar:=(A)^(-1) mod B;
%p if 2*Ar > B then Ar:=B-Ar; fi;
%p Br:=(B)^(-1) mod A;
%p if 2*Br > A then Br:=A-Br; fi;
%p A*Ar+B*Br;
%p end;
%p for i from 1 to 14 do lprint([seq(myfun1(i-j+1,j),j=1..i)]); od:
%Y Cf. A344005, A347681, A347682, A347683, A347687.
%Y Row 3 is A147666.
%K nonn,tabl
%O 1,8
%A _N. J. A. Sloane_, Sep 18 2021
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