A269783 - OEIS

%I #33 May 21 2023 08:24:25

%S 0,0,1,1,0,-1,-1,0,2,2,1,2,3,3,2,0,-2,-2,-1,-2,-3,-3,-2,0,3,4,4,3,1,

%T -1,1,3,5,5,4,5,6,6,5,3,0,-3,-4,-4,-3,-1,-3,-5,-5,-4,-5,-6,-6,-5,-3,0,

%U 4,2,4,6,7,7,6,4,1,-2,-4,-2,1,4

%N Starting with a(0) = 0, the subsequence of immediate successors of the instances of each integer n in the sequence is n, n-1, n+1, n-2, n+2, ... for n > 0 and n, n+1, n-1, n+2, n-2, ... for n <= 0.

%C Note that "subsequence" refers to any sequence of values taken at increasing indices in the base sequence, not necessarily consecutive terms.

%C In more detail, for any integer n, let i(k) be the index of the k-th occurrence of n in this sequence, i.e., { i(1), i(2), i(3), ...} = { i | a(i) = n }. Then S(n) := (a(i(1)+1), a(i(2)+1), a(i(3)+1), ...) = (n, n-1, n+1, n-2, ...) if n > 0 and S(n) = (n, n+1, n-1, n+2, ...) if n <= 0. This property, together with the choice if the starting value a(0) = 0, uniquely defines the sequence, because it unambiguously defines the successor of any given term. - _M. F. Hasler_, Mar 07 2016

%C Every ordered pair of integers occurs exactly once in this sequence.

%C Indeed, given any pair (n,m), there is a unique k such that m-n = (-1)^(k-[n>0])*floor(k/2), and by definition the pair (n,m) occurs exactly at the k-th occurrence of n. It remains to show that all integers occur infinitely often. - _M. F. Hasler_, Mar 07 2016

%H Rémy Sigrist, <a href="/A269783/b269783.txt">Table of n, a(n) for n = 0..10000</a>

%F The immediate successor of any term  a(j) = n  is  a(j+1) = n + (-1)^(k-e) * floor(k/2), where k = # { i <= j | a(i) = n } (meaning that a(j) is the k-th occurrence of n in the sequence), and e := [n>0] = 1 if n > 0 and e = 0 if n <= 0). - _M. F. Hasler_, Mar 07 2016

%e a(0) is 0. As a(0) is the first 0 in the sequence, a(1) is the first term in the sequence 0, 1, -1, 2, -2 (incidently A001057, but ignoring the offset of that sequence), nameley 0. Now a(1) is the second 0, so we take the second term in A001057, and a(2) = 1. For the first 1, we take the first term associated with 1, which is 1, and a(3) = 1. Now we have the second 1, we take the second term of 1, 0, 2, -1, ..., which is again 0.

%e Table of indices where the pairs (i,j) start in the sequence:

%e     i \ j  -3   -2   -1   0    1    2    3

%e    ...    ...  ...  ...  ...  ...  ...  ...

%e    -3 ...  20   21   44   54  129  326  558  ...

%e    -2 ...  19   16   17   22   67  200  374  ...

%e    -1 ...  45   18    5    6   29   92  169  ...

%e     0 ...  40   15    4   (0)   1    7   23  ...

%e     1 ... 124   64   28    3    2   10   30  ...

%e     2 ... 325  199   93   14    9    8   11  ...

%e     3 ... 557  373  170   39   27   13   12  ...

%e    ...    ...  ...  ...  ...  ...  ...  ...

%e extending in all directions. The table always fills in faster on the top-left to bottom-right diagonal, so it grows faster on the other diagonal.

%o (PARI) A269783(n,show=0,a=0,C=[])={for(n=1,n,show&&print1(a",");i=setsearch(C,[a,0],1);(i>#C || C[i][1] != a) && C=setunion(C,[[a,0]]); a+=(-1)^((a>0)+C[i][2]+=1)*(C[i][2]\2)));a} \\ Set 2nd (optional) arg to 1 to print out all terms, 3rd arg to change starting value. Change "for(n=1...);a" to "vector(n...)" to return the vector of values. - _M. F. Hasler_, Mar 07 2016

%Y Cf. A269501, A001057.

%K sign

%O 0,9

%A _Franklin T. Adams-Watters_, Mar 05 2016