A334523 Irregular triangle read by rows: row n contains the sequence n, p(n), s(p(n)), p(s(p(n))), p(p(s(p(n)))), s(p(p(s(p(n))))), ..., repeatedy applying (p,s,p) to n, where p = phi (A000010), s = sigma = (A000203), stopping after the first 1 is reached. If 1 is never reached, row n contains n, -1.

1, 1, 2, 1, 3, 2, 3, 2, 1, 4, 2, 3, 2, 1, 5, 4, 7, 6, 2, 3, 2, 1, 6, 2, 3, 2, 1, 7, 6, 12, 4, 2, 3, 2, 1, 8, 4, 7, 6, 2, 3, 2, 1, 9, 6, 12, 4, 2, 3, 2, 1, 10, 4, 7, 6, 2, 3, 2, 1, 11, 10, 18, 6, 2, 3, 2, 1, 12, 4, 7, 6, 2, 3, 2, 1, 13, 12, 28, 12, 4, 7, 6, 2, 3, 2, 1, 14, 6, 12, 4, 2, 3, 2, 1

Offset

1,3

Links

Paolo Xausa, Table of n, a(n) for n = 1..9094 (rows 1..500 of the triangle, flattened)

L. Alaoglu and P. Erdős, A conjecture in elementary number theory, Bull. Amer. Math. Soc. 50 (1944), 881-882.

Example

Triangle begins:
1, 1,
2, 1,
3, 2, 3, 2, 1,
4, 2, 3, 2, 1,
5, 4, 7, 6, 2, 3, 2, 1,
6, 2, 3, 2, 1,
7, 6, 12, 4, 2, 3, 2, 1,
8, 4, 7, 6, 2, 3, 2, 1,
9, 6, 12, 4, 2, 3, 2, 1,
10, 4, 7, 6, 2, 3, 2, 1,
...
For row n=5, for example, we get 5 -> phi(5) = 4 -> sigma(4) = 7 -> phi(7) = 6 -> phi(6) = 2 -> sigma(2) = 3 -> phi(3) = 2 -> phi(2) = 1 (stop).

Mathematica

A334523row[n_]:=Module[{i=0}, NestWhileList[If[Mod[i++, 3]==1, DivisorSigma[1, #], EulerPhi[#]]&, n, i==0||#>1&]]; Array[A334523row, 15] (* Paolo Xausa, Nov 16 2023 *)

Crossrefs

See A032452, A334725 for other versions.

Cf. A000010, A000203, A334685, A334686.

Sequence in context: A026730 A318691 A075256 * A001480 A308117 A270755

Adjacent sequences: A334520 A334521 A334522 * A334524 A334525 A334526

Keyword

nonn,tabf

Author

N. J. A. Sloane, May 09 2020, following a suggestion from Amiram Eldar.

Status

approved