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

Author

Amiram Eldar, Nov 25 2021

Keywords

• nonn
• tabf

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.

Examples

			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,
...
		

Links

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

Crossrefs

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

Programs

Mathematica

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

PARI

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