A308483 - OEIS

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).

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, 15, 10, 10, 15, 12, 20, 30, 6, 7, 42, 30, 20, 28, 21, 15, 35, 14, 14, 35, 15, 21, 28, 20, 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

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OFFSET

1,2

COMMENTS

This is also the product of the denominators of pairs of consecutive terms in the Farey sequence.

Each term of this sequence is an integer: (Proof by induction)

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).

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

LINKS

Table of n, a(n) for n=1..75.

FORMULA

T(n,k) = Farey(n,k+1) - Farey(n,k) with Farey(n,k) = A006842(n,k)/A006843(n,k).

T(n,k) = A006843(n,k)*A006843(n,k+1).

EXAMPLE

T(1,1) = 1/(1 - 0);

T(2,1) = 1/(1/2 - 0);

T(2,2) = 1/(1 - 1/2);

T(3,1) = 1/(1/3 - 0);

T(3,2) = 1/(1/2 - 1/3);

T(3,3) = 1/(2/3 - 1/2);

T(3,4) = 1/(1 - 2/3);

...

If written as an array:

2, 2;

3, 6, 6, 3;

4, 12, 6, 6, 12, 4;

5, 20, 12, 15, 10, 10, 15, 12, 20, 5;

...

PROG

(PARI) rowf(n) = {my(vf = [0]); for (k=1, n, for (m=1, k, vf = concat(vf, m/k); ); ); vecsort(Set(vf)); } \\ A006842/A006843

row(n) = my(vf = rowf(n)); vector(#vf-1, k, 1/(vf[k+1] - vf[k])); \\ Michel Marcus, Jun 07 2019

CROSSREFS

Cf. A006842, A006843.

Sequence in context: A345706 A132886 A119272 * A070871 A096115 A289838

Adjacent sequences: A308480 A308481 A308482 * A308484 A308485 A308486

KEYWORD

nonn,tabf

AUTHOR

Isaac Kaufmann, May 30 2019

EXTENSIONS

More terms from Michel Marcus, Jun 07 2019

STATUS

approved