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A268492
Orbit of 2 under the map A268488: n -> least number k of the form k = n*(last digit of k) + (k without its last digit).
2
2, 19, 21, 209, 2089, 2321, 23209, 77363, 773629, 2578763, 25787629, 28652921, 286529209, 955097363, 9550973629, 10612192921, 35373976403, 353739764029, 1179132546763, 1310147274181, 4367157580603, 4852397311781, 48523973117809, 161746577059363
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
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 2, we have 2*(19 mod 10) + [19 / 10] = 2*9 + 1 = 19, then 19*(21 mod 10) + [21 / 10] = 19*1 + 2 = 21, etc.
a(4) = 2089: 2089*10-1 = 20889 == 0 (mod 9), so a(5) = 20889/9 = 2321; 2321*10-1 = 23209 == 1 (mod 3), so a(6) = 23209; 23209*10-1 = 232089 == 0 (mod 3) but not 0 (mod 9), so a(7) = 232089/3 = 77363. - Bob Selcoe, Feb 17 2016
PROG
(PARI) vector(30, n, t=if(n>1, A268488(t), 2))
CROSSREFS
Cf. A268493 for the orbit of 3.
Sequence in context: A022119 A042247 A041451 * A130112 A019348 A155025
KEYWORD
nonn,base
AUTHOR
Eric Angelini and M. F. Hasler, Feb 14 2016
STATUS
approved