A119590 a(n) = position of n in the lexicographical ordering A119589 of natural numbers from 1 to 100.

1, 13, 24, 35, 46, 57, 68, 79, 90, 2, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 3

Offset

1,2

Comments

Inverse of the permutation A119589. - M. F. Hasler, Oct 26 2019

Links

Table of n, a(n) for n=1..100.

Formula

a(n) = a(n-1) + a(n-10) - a(n-11) for 21 < n < 100. - M. F. Hasler, Sep 03 2018

a(n) = k such that A119589(k) = n. - M. F. Hasler, Oct 26 2019

Example

a(1) = 1
a(10) = 2 because "10" comes after "1"
a(100) = 3 because "100" comes after "10", but before "11"

Prog

(PARI) vecsort(vecsort(vector(100, n, Str(n)), , 1), , 1) \\ M. F. Hasler, Sep 03 2018, simplified Oct 25 2019

Crossrefs

Cf. A119589 (integers 1..100 in lexicographical order).

Cf. A190016, A190017 (integers 1..10^4 in lexicographical order, and inverse).

Sequence in context: A136316 A063315 A104342 * A266912 A363823 A032607

Adjacent sequences: A119587 A119588 A119589 * A119591 A119592 A119593

Keyword

nonn,base,fini,full

Author

Dmitry Kamenetsky, Jun 01 2006, Jun 03 2006

Status

approved