A261729 Putative inverse of conjectured permutation in A261725.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 60, 61, 62, 63, 64, 65, 66, 67

Offset: 0

Author

Paul Tek, Aug 30 2015

Keywords

• nonn
• base

Comments

a(n) = A003100(n) for n < 100, but a(100) = 109, A003100(100) = 190.

a(n) = A118757(n) for n < 100, but a(100) = 109, A118757(100) = 190.

a(n) = A118758(n) for n < 201, but a(201) = 209, A118758(201) = 211.

a(n) = A174025(n) for n < 100, but a(100) = 109, A174025(100) = 199.

a(n) = A261725(n) for n < 100, but a(100) = 109, A261725(100) = 190.

Links

• Paul Tek, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A261725.