A260616 Irregular triangle read by rows: continued fraction expansion of k/n, 1 <= k <= n.

1, 0, 2, 1, 0, 3, 0, 1, 2, 1, 0, 4, 0, 2, 0, 1, 3, 1, 0, 5, 0, 2, 2, 0, 1, 1, 2, 0, 1, 4, 1, 0, 6, 0, 3, 0, 2, 0, 1, 2, 0, 1, 5, 1, 0, 7, 0, 3, 2, 0, 2, 3, 0, 1, 1, 3, 0, 1, 2, 2, 0, 1, 6, 1, 0, 8, 0, 4, 0, 2, 1, 2, 0, 2, 0, 1, 1, 1, 2, 0, 1, 3, 0, 1, 7, 1, 0, 9, 0, 4, 2, 0, 3, 0, 2, 4, 0, 1, 1, 4, 0, 1, 2, 0, 1, 3, 2, 0, 1, 8, 1

Offset

1,3

Comments

This sequence contains the terms of the continued fraction expansion of all fractions involving integers between 0 and 1, excluding 0/n.

The number of distinct continued fractions in row n is equal to A000010(n).

Links

Matthew Campbell and Robert Israel, Table of n, a(n) for n = 1..22438 (rows 1 to 100, flattened; n = 1..1722 from Matthew Campbell)

Example

Triangle begins ({} included for fraction separation):
{1}
{0, 2}, {1}
{0, 3}, {0, 1, 2}, {1}
{0, 4}, {0, 2}, {0, 1, 3}, {1}
{0, 5}, {0, 2, 2}, {0, 1, 1, 2}, {0, 1, 4}, {1}
{0, 6}, {0, 3}, {0, 2}, {0, 1, 2}, {0, 1, 5}, {1}
{0, 7}, {0, 3, 2}, {0, 2, 3}, {0, 1, 1, 3}, {0, 1, 2, 2}, {0, 1, 6}, {1}
{0, 8}, {0, 4}, {0, 2, 1, 2}, {0, 2}, {0, 1, 1, 1, 2}, {0, 1, 3}, {0, 1, 7}, {1}
{0, 9}, {0, 4, 2}, {0, 3}, {0, 2, 4}, {0, 1, 1, 4}, {0, 1, 2}, {0, 1, 3, 2}, {0, 1, 8}, {1}
...

Maple

seq(seq(op(numtheory:-cfrac(k/n, 'quotients')), k=1..n), n=1..10); # Robert Israel, Sep 04 2015

Mathematica

Table[ContinuedFraction[k/n], {n, 9}, {k, n}] // Flatten (* Michael De Vlieger, Sep 04 2015 *)

Prog

(PARI) row(n) = {v = []; for (k=1, n, v = concat(v, contfrac(k/n)); ); v; }
tabf(nn) = for (n=1, nn, print(row(n), ", ")) \\ Michel Marcus, Sep 04 2015

Crossrefs

Denominator: A002024; Numerator: A002260.

Cf. A000010, A072193 (0<k<n).

Sequence in context: A200123 A187616 A217262 * A284950 A308298 A364106

Adjacent sequences: A260613 A260614 A260615 * A260617 A260618 A260619

Keyword

nonn,cofr,tabf

Author

Matthew Campbell, Sep 03 2015

Status

approved