Permutations based on themselves
franktaw at netscape.net
franktaw at netscape.net
Sun Apr 23 19:00:39 CEST 2006
Sorry, I was misreading the definition. These aren't Beatty sequences,
and I don't see any errors.
Franklin T. Adams-Watters
-----Original Message-----
From: franktaw at netscape.net
The 17 in A118315 is wrong. Odd index members must increase by at
least 3. (Likewise for the 13 in A118317.)
The odd index members of this sequence, excluding a(1), look like they
might be a Beatty sequence. (This would make the even index members,
together with 1, another Beatty sequence.) Even if not, this really is
two related sequences more than a single sequence.
Franklin T. Adams-Watters
-----Original Message-----
From: Leroy Quet <qq-quet at mindspring.com>
I just submitted the following permutations of the positive integers:
>%S A118315 1,2,3,4,6,5,9,7,12,8,16,10,17,11
>%N A118315 a(1)=1. a(2n)= lowest positive integer not occurring among
the
>earlier terms of the sequence. a(2n+1) = the a(n)th positive integer
among
>those positive integers not occurring earlier in the sequence.
>%C A118315 Sequence is a permutation of the positive integers.
>%e A118315 For a(9) we want the a(4)th = 4th positive integer among
those
>not equal to any of the first 8 terms of the sequence (those positive
>integers not equal to 1,2,3,4,6,5,9, or 7). Among those positive
integers
>not equal to any the first 8 terms (which is the sequence
>8,10,11,12,13...), 12 is the 4th. So a(9) = 12.
>Now for a(10) we want the lowest positive integer that does not occur
>among the first 9 terms of the sequence. So a(10) = 8.
>%Y A118315 A118316,A118317,A118318
>%O A118315 1
>%K A118315 ,easy,more,nonn,
>%S A118316 1,2,3,4,6,5,8,10,7,12,14,9
>%N A118316 Inverse permutation of sequence A118315.
>%C A118316 Sequence is a permutation of the positive integers.
>%Y A118316 A118315,A118317,A118318
>%O A118316 1
>%K A118316 ,easy,more,nonn,
>%S A118317 1,2,3,5,4,8,6,12,7,13,9,19
>%N A118317 a(2n-1)= lowest positive integer not occurring among the
>earlier terms of the sequence. a(2n) = the a(n)th positive integer
among
>those positive integers not occurring earlier in the sequence.
>%C A118317 Sequence is a permutation of the positive integers.
>%e A118317 For a(8) we want the a(4)th = 5th positive integer among
those
>not equal to any of the first 7 terms of the sequence (those positive
>integers not equal to 1,2,3,5,4,8, or 6). Among those positive
integers
>not equal to any the first 7 terms (which is the sequence
>7,9,10,11,12,13...), 12 is the 5th. So a(8) = 12.
>Now for a(9) we want the lowest positive integer that does not occur
among
>the first 8 terms of the sequence. So a(9) = 7.
>%Y A118317 A118315,A118316,A118318
>%O A118317 1
>%K A118317 ,easy,more,nonn,
>%S A118318 1,2,3,5,4,7,9,6,11,13
>%N A118318 Inverse permutation of sequence A118317.
>%C A118318 Sequence is a permutation of the positive integers.
>%Y A118318 A118315,A118316,A118317
>%O A118318 1
>%K A118318 ,easy,more,nonn,
(It should be noted that A118318 is different than A102399.)
Is there a direct way to calculate the nth terms of any of these
sequences? (Something involving the highest power of 2 dividing n,
perhaps?)
thanks,
Leroy Quet
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