Walking base sequence

[Alexandre Wajnberg]

Hi all,

A "base" sequence of another kind, specially dealing with bases, going
through all bases, but in the scope of base 10.


1 1 2 3 4 5 6 7 8 9 10 2 4 6 8 10 3 5 7 9 11 2 5 8 11 3 6 9 12 3 7 10 4 7 11
4 8 12 4 9 13 4 10 5 9 14 5 10 6 10 7 12 5 11 5 12 6 11 6 12 7 13 5 13 6 13
7 14 6 14 7 15 6 15 7 16 7 17 8 13 8 14 8 15 8 16 8 17 9 15 9 16 9 17 10 8
18 9 18 10 9 19 10 10 11 7 18 11 8 19 11 9 20 3 8 20 4 11 10 12 8 21 3 9 21
4 12 9 22 3 10 13 9 23 4 13 10 14 9 24
 
%N A000001 
"Walking base" sequence: the number becomes the least base in which it could
be read, once; written in base 10.

%C A000001 
The number do not become its value in the new base, but becomes the base
itself. Each term has a double status according to its preceding or
following neighbor: a *base* (the least one not used so far) which read the
preceding one, and a *number* to be read in the base expressed by the next
term. The sequence walks through all bases but is written in base 10. The
first break, specific of this sequence, occurs after a(11)=10.

%e A000001 
The beginning is 1,1,2,3 but could also be 0,2,3.
‹ a(1)=1. The least base in which 1 has a meaning is the unary base, so next
term, a(2)=1.
‹ The least base in which this second 1 could now be found is the base 2
(since base 1 has already been used), so next term, a(3)=2.
‹ The least base in which 2 makes sense is 3, so next term, a(4)=3.
‹ The least base in which "10" makes sense is not base 11 but base 2, so
next term, a(12)=2 (although 2 was already used to read 1, it has not yet
been used to read "10").
‹ The least base in which this second 2 makes sense now is not 3 (because 3
has already been used to read a(3)=2), but 4, so next term a(13)=4.
‹ a(103)=10: the least base not used so far to read "10" is base 10, so
a(104)=10; then a(105)=11 (and although the value a(104)="10" in base 11
should be written "A", which is impossible in OEIS, this does not affect the
next term a(105); anyway, this walking base is written all along in base 10,
so a(104)=10).

%O A000001 1,3



It's quite tricky to do by hand and I didn't submit it yet. Could someone
verify it? And may be compute it in the scope of other bases [For instance,
in the scope of base 8, the beginning would be: 1,1,2,3,4,5,6,7,10,2,4...
the first break occurring after a(9) instead of a(11). ]

Thanx

Alexandre







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