[seqfan] Dividing a concatenation
Eric Angelini
Eric.Angelini at kntv.be
Mon May 4 19:03:03 CEST 2015
Hello SeqFans,
we want a(n) to be the smallest unused integer in S so far;
we want a(n) to divide the concatenation of [a(1),a(2)... a(n-1)]
S = 2,1,3,71,7,10177,2100001,101770000001,4603,13,107,4013,23,...
Explanations:
S cannot start with "1" -- thus a(1)=2;
2 (start)
1 is the smallest integer (not yet present in S) dividing 2
3 is the smallest integer (not yet present in S) dividing 21
71 is the smallest integer (not yet present in S) dividing 213
7 is ... dividing 21371
10177 is ... dividing 213717
2100001 is ... dividing 21371710177
101770000001 is ... dividing 213717101772100001
4603 is ... dividing 213717101772100001101770000001
13 is ... dividing 2137171017721000011017700000014603
107 is ... dividing 213717101772100001101770000001460313
4013 is ... dividing 213717101772100001101770000001460313107
23 is ... dividing 2137171017721000011017700000014603131074013
I guess the next terms are (not sure):
3097349301044927552199565217412468305904367
1847
37
57761378857324324622100000000394679218216573867495927055268312221609870740877521781199113101
73
9
27
243
11
201010661
19
49
21
...
Is S a permutation of the positive integers?
Best,
É.
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