A268493 Orbit of 3 under the map A268488: n -> least number k of the form k = n*(last digit of k) + (k without its last digit).

3, 29, 289, 321, 3209, 32089, 106963, 356543, 3565429, 11884763, 118847629, 132052921, 440176403, 4401764029, 4890848921, 48908489209, 163028297363, 1630282973629, 5434276578763, 18114255262543, 60380850875143, 201269502917143, 670898343057143

Offset

0,1

Comments

See A268488 and the link to the SeqFan list for further information.

There can be no more than three consecutive terms where a(n) = 10*a(n-1)-1 and a(n+1) = 10*a(n)-1. This occurs when a(n-1) == 0 (mod 3), which must be followed by a(n) == 2 (mod 3) and a(n+1) == 1 (mod 3). - Bob Selcoe, Feb 17 2016

Links

Table of n, a(n) for n=0..22.

Eric Angelini (and reply by M. Hasler), 3, 29, 289, 321, ..., SeqFan list, Feb. 13, 2016

Formula

a(n) = 10*a(n-1)-1 unless divisible by 3 or 9, then a(n) = (10*a(n-1)-1)/{3,9}, respectively. - Bob Selcoe, Feb 17 2016

Example

Starting from 3, we have 3*(29 mod 10) + [29 / 10] = 3*9+2 = 29, then 29*(289 mod 10) + [289 / 10] = 29*9 + 28 = 289, etc.
a(2) = 289: 289*10-1 = 2889 == 0 (mod 9), so a(3) = 2889/9 = 321; 321*10-1 = 3209 == 2 (mod 3), so a(4) = 3209; 3209*10-1 = 32089 == 1 (mod 3), so a(5) = 32089; 32089*10-1 = 320889 == 0 (mod 3) but not 0 (mod 9), so a(6) = 320889/3 = 106963. - Bob Selcoe, Feb 17 2016

Prog

(PARI) vector(30, n, t=if(n>1, A268488(t), 3))

Crossrefs

Cf. A268492 for the orbit of 2.

Sequence in context: A026159 A025186 A140780 * A002669 A340990 A112711

Adjacent sequences: A268490 A268491 A268492 * A268494 A268495 A268496

Keyword

nonn,base

Author

Eric Angelini and M. F. Hasler, Feb 14 2016

Status

approved