A362181 Number of numbers k such that A323410(k) = n.

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

Offset: 2

Author

Amiram Eldar, Apr 10 2023

Keywords

• nonn

Comments

The offset is 2 since A323410(p) = 1 for all prime powers p (A246655).

a(0) = 1, since there is only one solution, x = 1, to A323410(x) = 0.

Links

• Amiram Eldar, Table of n, a(n) for n = 2..10000

Crossrefs

Row lengths of A362180.

The unitary version of A063740.

Cf. A246655, A323410, A362182 (positions of 0's), A362183 (indices of records), A362184, A362185 (positions of 1's), A362186.

Similar sequences: A014197, A361967.

Programs

Mathematica

ucototient[n_] := n - Times @@ (Power @@@ FactorInteger[n] - 1); ucototient[1] = 0; With[{max = 100}, ucot = Table[ucototient[n], {n, 1, max^2}]; Table[Length[Position[ucot, n]], {n, 2, max}] // Flatten]

Formula

a(A362182(n)) = 0.

a(A362185(n)) = 1.

a(A362186(n)) = n.