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A308483 Irregular triangle read by rows: T(n,k) = Farey(n,k+1) - Farey(n,k) where Farey(n,k) = A006842(n,k)/A006843(n,k).

%I A308483 #24 Jun 07 2019 11:05:00
%S A308483 1,2,2,3,6,6,3,4,12,6,6,12,4,5,20,12,15,10,10,15,12,20,5,6,30,20,12,
%T A308483 15,10,10,15,12,20,30,6,7,42,30,20,28,21,15,35,14,14,35,15,21,28,20,
%U A308483 30,42,7,8,56,42,30,20,28,21,24,40,35,14,14,35,40,24,21,28,20,30,42,56,8
%N A308483 Irregular triangle read by rows: T(n,k) = Farey(n,k+1) - Farey(n,k) where Farey(n,k) = A006842(n,k)/A006843(n,k).
%C A308483 This is also the product of the denominators of pairs of consecutive terms in the Farey sequence.
%C A308483 Each term of this sequence is an integer: (Proof by induction)
%C A308483 Assume that the reciprocal of Farey differences of order n are the product of the consecutive denominators, i.e., if x/y and c/d are adjacent, then |x/y - c/d| = 1/dy. Let a/b and p/q be adjacent in Farey sequence up to n, such that n+1 = b+q (so only their mediant is in the middle).
%C A308483 As |a/b - p/q| = 1/bq, |aq - bp| = 1, so |aq - bp + ab - ab| = 1, so |a/b - (a+p)/(b+q)| = 1. The base case is trivial. QED
%F A308483 T(n,k) = Farey(n,k+1) - Farey(n,k) with Farey(n,k) = A006842(n,k)/A006843(n,k).
%F A308483 T(n,k) = A006843(n,k)*A006843(n,k+1).
%e A308483 T(1,1) = 1/(1 - 0);
%e A308483 T(2,1) = 1/(1/2 - 0);
%e A308483 T(2,2) = 1/(1 - 1/2);
%e A308483 T(3,1) = 1/(1/3 - 0);
%e A308483 T(3,2) = 1/(1/2 - 1/3);
%e A308483 T(3,3) = 1/(2/3 - 1/2);
%e A308483 T(3,4) = 1/(1 - 2/3);
%e A308483 ...
%e A308483 If written as an array:
%e A308483   1;
%e A308483   2,  2;
%e A308483   3,  6,  6,  3;
%e A308483   4, 12,  6,  6, 12,  4;
%e A308483   5, 20, 12, 15, 10, 10, 15, 12, 20, 5;
%e A308483   ...
%o A308483 (PARI) rowf(n) = {my(vf = [0]); for (k=1, n, for (m=1, k, vf = concat(vf, m/k); ); ); vecsort(Set(vf));} \\ A006842/A006843
%o A308483 row(n) = my(vf = rowf(n)); vector(#vf-1, k, 1/(vf[k+1] - vf[k])); \\ _Michel Marcus_, Jun 07 2019
%Y A308483 Cf. A006842, A006843.
%K A308483 nonn,tabf
%O A308483 1,2
%A A308483 _Isaac Kaufmann_, May 30 2019
%E A308483 More terms from _Michel Marcus_, Jun 07 2019