A349685 Irregular triangle read by rows: the n-th row contains the elements in the continued fraction of the abundancy index of n.

1, 1, 2, 1, 3, 1, 1, 3, 1, 5, 2, 1, 7, 1, 1, 7, 1, 2, 4, 1, 1, 4, 1, 11, 2, 3, 1, 13, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 15, 1, 17, 2, 6, 1, 19, 2, 10, 1, 1, 1, 10, 1, 1, 1, 1, 3, 1, 23, 2, 2, 1, 4, 6, 1, 1, 1, 1, 1, 2, 1, 2, 13, 2, 1, 29, 2, 2, 2, 1, 31, 1, 1, 31

Offset

1,3

Comments

The abundancy index of n is sigma(n)/n = A000203(n)/n = A017665(n)/A017666(n).

For a prime p, the p-th row has a length 2 with a(p, 1) = 1 and a(p, 2) = p.

For multiply-perfect numbers m (A007691), the m-th row has a length 1, since their abundancy index is an integer. In particular, for a perfect number m (A000396), the m-th row has a length 1 with a(m, 1) = 2.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10489 (rows 1..2000)

Example

The first ten rows of the triangle are:
1,
1, 2,
1, 3,
1, 1, 3,
1, 5,
2,
1, 7,
1, 1, 7,
1, 2, 4,
1, 1, 4,
...

Mathematica

row[n_] := ContinuedFraction[DivisorSigma[1, n]/n]; Table[row[k], {k, 1, 32}] // Flatten

Prog

(PARI) row(n) = contfrac(sigma(n)/n); \\ Michel Marcus, Nov 25 2021

Crossrefs

Cf. A000203, A000396, A007691, A017665, A017666, A071862 (row lengths), A071865, A349473.

Sequence in context: A186007 A212623 A229214 * A218578 A006346 A244740

Adjacent sequences: A349682 A349683 A349684 * A349686 A349687 A349688

Keyword

nonn,tabf

Author

Amiram Eldar, Nov 25 2021

Status

approved