|
| |
|
|
A096098
|
|
a(1) = 2, a(2) = 1; for n >= 3, a(n) = least number not included earlier that divides the concatenation of all previous terms.
|
|
5
|
|
|
|
2, 1, 3, 71, 7, 21, 599, 173, 11, 23, 161, 49, 13, 9, 131, 19, 33, 17, 1489, 331, 3989, 69, 3097350956401900335673788279883089441874368101, 349387, 5651, 443, 29, 51, 479470832244949, 661, 1129, 1873, 181, 1544577973887516219070997863, 521
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
Conjecture (1) Every concatenation is squarefree.
Conjecture (2) This is a rearrangement of the squarefree numbers not divisible by 5. False! (The a(n) are not always squarefree, since a(12)=49 and a(14)=9.)
Fact: All a(n) for n >= 2 are odd, since a(2) = 1 and odd a(n) => odd concatenation => odd a(n+1). - Wolfdieter Lang, May 08 2014 (editing an earlier statement).
Conjecture (3) the sequence for n>=2 is a permutation of the positive integers not divisible by 2 or 5.
a(29) is probably 479470832244949, in which case the sequence continues 479470832244949, 661, 1129, 1873, 181. - Martin Fuller, Nov 21 2007
Factorization for a(29): 479470832244949*3*17*43217123024009614997922599713504735424547343*P51. - Sean A. Irvine, May 25 2010
Assuming Conjecture (3), the smallest number yet to appear is 89. - Sean A. Irvine, May 11 2014
The factorization given by Sean A. Irvine above is not for the prime a(29) = 479470832244949 but for the concatenation of a(1), a(2), ..., a(29), and P51 means a prime with 51 digits, namely 202232656574589264871780464738430216507933940172343. - Wolfdieter Lang, May 11 2014
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(6) = 21 as 213717 = 3*7*10177, and 3 = a(3) and 7 = a(4), hence 3*7 = 21 is the least number dividing 213717 not included earlier in the sequence.
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
base,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
