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A308021
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Start at n=1. Fill in a(n) with a value d > 0 not used earlier such that n-d or n+d is the smallest possible index not visited earlier, then continue with that index.
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5
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1, 2, 5, 3, 7, 10, 4, 6, 12, 15, 17, 8, 20, 9, 22, 11, 25, 27, 29, 14, 13, 33, 35, 37, 16, 40, 18, 19, 42, 45, 47, 49, 21, 24, 52, 55, 23, 58, 61, 62, 30, 26, 65, 66, 28, 68, 34, 31, 71, 74, 76, 79, 81, 39, 32, 83, 86, 36, 89, 43, 38, 91, 93, 96, 99, 41, 101, 50, 103, 106, 44, 108, 54, 111, 46, 113, 117, 48, 57, 118, 51, 120, 123
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OFFSET
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1,2
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COMMENTS
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A variant of Recamán's sequence: start at n=1, a(1)=1, then iterate: let n' = n - a(n) if this n' > 0 and was not visited earlier, otherwise n' = n + a(n). Then let n'' <> n' be the smallest index not visited earlier (a(n'') not yet defined) such that the value |n'-n''| was not yet used (meaning not yet a value of any a(i)). Set a(n') = |n'-n''| and continue with n = n'. [Name and Comment suggested by M. F. Hasler at the request of the authors.]
Conjectured to be a permutation of the positive integers. The conjectured inverse permutation is given in A308049. - M. F. Hasler, May 10 2019
The sequence is given by the following formula. Let R(t) = A081145(t). Then for all t >= 1, a(R(t)) = |R(t+1)-R(t)|.
For example, for t=10, R(10)=20, R(11)=6, and a(R(10)) = a(20) = |6-20| = 14.
Since it is known that {R(t): t>=1} is a permutation of the positive integers (it is the "Slater-Velez permutation of the first kind"), this specifies a(n) for all n.
The connection with the definition as interpreted above by M. F. Hasler is that at step t of the procedure, n' is R(t) = A081145(t), n'' is R(t+1) = A081145(t+1), and we calculate a(R(t)) = |n'-n''| = |R(t)-R(t+1)|.
The conjecture that {a(n)} is a permutation of the positive integers is equivalent to Slater and Velez's conjecture (see references) that the absolute values of the first differences of A081145 are also a permutation of the positive integers. This problem appears to be still unsolved. (End)
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REFERENCES
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P. J. Slater and W. Y. Velez, Permutations of the Positive Integers with Restrictions on the Sequence of Differences, II, Pacific Journal of Mathematics, Vol. 82, No. 2, 1979, 527-531.
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LINKS
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EXAMPLE
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a(1) = 1 drives us to the empty cell a(2) since we can't go further to the left. We fill this cell with the number 2 which is the smallest integer not used before and thus allows us to go to the leftmost possible empty cell, 2 + 2 = 4. (There are no empty places to the left and we can't go to 3 = 2 + 1 since a step 1 has already been used.) So we have a(2) = 2.
a(2) = 2 drives us to the empty cell a(4). We see that the leftmost empty cell a(3) cannot be reached from a(4) since a step of 1 has already been used. We thus fill the cell a(4) with the smallest integer not used before, a(4) = 3.
a(4) = 3 drives us to the empty cell a(7). We see that the leftmost empty cell a(3) can now be reached from a(7) if we fill a(7) with 4; we have thus a(7) = 4.
a(7) = 4 drives us to the empty cell a(3), which is the one we wanted to fill. We fill a(3) with 5 which is the smallest integer not leading to a contradiction, whence a(3) = 5.
a(3) = 5 drives us to the empty cell a(8). We would like to fill this cell with 3, as this 3 would allow us to fill the leftmost empty cell of the sequence - but 3 has been used before; thus we'll have a(8) = 6.
a(8) = 6 drives us to the empty cell a(14). We fill a(14) with 9 as this will allow us to reach the leftmost empty cell of the sequence, whence a(14) = 9.
a(14) = 9 drives us to the empty cell a(5). We fill a(5) with 7 as this is the smallest integer not leading to a contradiction, so we have a(5) = 7, etc.
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PROG
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(PARI) {A=vector(N=199); n=1; while (n<=N, S=Set(A); Z=select(t->!t, A, 1); for (i=1, #Z, Z[i]!=n||next; setsearch(S, abs(n-z=Z[i]))&& next; A[n]=abs(n-z); n=z; next(2)); break); if(#Z, A[1..Z[1]-!A[Z[1]]], A)} \\ M. F. Hasler, May 09 2019
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CROSSREFS
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Cf. A308049 (conjectured inverse permutation).
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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