OFFSET
1,2
COMMENTS
Proof that this is the same as the positions of 0 in A284939, from Joe Slater, Apr 26 2017: (Start)
The sequence A284939 consists of a concatenation of the words 01 and 1101, each word representing a single term from the previous iteration of the transformation taken in order.
Because the sequence has reached its fixed point we don't need to compare these words to the terms of the previous iteration: we can actually relate the present iteration to itself.
Thus the following table relates terms of A284939 to its words. We would end up with the same sequence by concatenating either the terms or the words:
0 -> 01
1 -> 1101
1 -> 1101
1 -> 1101
0 -> 01
1 -> 1101
... ...
Consider any arbitrary term A284939(n) and its successor A284939(n+1) . They relate to the n-th and (n+1)-st words, and since neither of the possible words have sequential zeros there are only three possibilities for the two terms (0,1; 1,0; and 1,1) and therefore three possibilities for the words: 01 1101, 1101 01, or 1101 1101.
We can see that the only time there is a gap of three 1's between the first and second zeros will be when the two terms A284939(n) and A284939(n+1) are (1,0). Therefore, if the n-th zero of the sequence (corresponding to the term A284939(n)) is at position k, it will be followed by a zero at position k+2 if the term A284939(n+1)=0, but otherwise it will be followed by a zero at position k+4.
Let's relate this to the present sequence A080580:
We know sequence A284939 starts with a zero in position 1 (i.e., A284939(1)=0), so we can make a sequence S (say) listing the position of the zeros with S(1)=1. From our earlier discussion we know that the n-th zero of A284939 relates to the n-th term of A284939. Therefore, if S(n)=k, the following term S(n+1) will be +4 if A284939(n+1)=1 and +2 if A284939(n+1)=0.
But we don't actually need to refer to A284939 at all! When we come to S(n+1) we can just see whether n+1 appears in our sequence. If the number n+1 already appears in S then we know that A284939(n+1)=0, so S(n+1)=S(n)+2. If the number n+1 does not appear in S then we know that A284939(n+1)=1, so S(n+1)=S(n)+4.
This rule is exactly the rule of A080580, which means that A080580 and S are identical, and A080580 manages to predict the positions of the zeros in A284939 without ever referring to the sequence itself. QED
(End)
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..10000
B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence, J. Integer Seqs., Vol. 6 (2003), #03.2.2.
B. Cloitre, N. J. A. Sloane and M. J. Vandermast, Numerical analogues of Aronson's sequence (math.NT/0305308)
MAPLE
A080580 := proc(n)
option remember;
if n = 1 then
1;
else
known := false ;
for i from 1 to n-1 do
if procname(i) = n then
known := true;
break;
end if;
end do:
if known then
procname(n-1)+2 ;
else
procname(n-1)+4 ;
end if;
end if;
end proc:
seq(A080580(n), n=1..100) ; # R. J. Mathar, Apr 25 2017
MATHEMATICA
a[1] = 1; a[n_] := a[n] = If[MemberQ[Array[a, n-1], n], a[n-1]+2, a[n-1]+4 ];
Array[a, 60] (* Jean-François Alcover, Nov 23 2017 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Mar 23 2003
EXTENSIONS
Edited by N. J. A. Sloane, Apr 27 2017
STATUS
approved