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A247556
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Exact differential base (a B_2 sequence) constructed as follows: Start with a(0)=0. For n>=1, let S be the set of all differences a(j)-a(i) for 0 <= i < j <= n-1, and let d be the smallest positive integer not in S. If, for every i in 1..n-1, a(n-1) + d - a(i) is not in S, then a(n) = a(n-1) + d. Otherwise, let r be the smallest positive integer such that, for every i in 1..n-1, neither a(n-1) + r - a(i) nor a(n-1) + r + d - a(i) is in S; then a(n) = a(n-1) + r and a(n+1) = a(n) + d.
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2
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0, 1, 3, 7, 12, 20, 30, 44, 65, 80, 96, 143, 165, 199, 224, 306, 332, 415, 443, 591, 624, 678, 716, 934, 973, 1134, 1174, 1449, 1491, 1674, 1720, 2113, 2161, 2468, 2517, 2855, 2906, 2961, 3245, 3302, 3711, 3772, 4081, 4148, 4603, 4673, 5557, 5628, 5917, 5989
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OFFSET
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0,3
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COMMENTS
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Every positive integer is uniquely represented as a difference of two distinct elements of the base set. This is a B_2 sequence.
By the definition of this sequence, with d as the smallest unused difference among terms a(0)..a(n-1), we assign a(n) = a(n-1) + d, provided that this would not cause any difference to be repeated; otherwise, we assign a(n) = a(n-1) + r and a(n+1) = a(n) + d, where r is the smallest integer that allows this assignment of a(n) and a(n+1) without causing any difference to be repeated. Thus, at each step, the smallest unused difference d is either used immediately (as a(n) - a(n-1)) or delayed by one step (and used as a(n+1) - a(n)). In this way, the sequence includes every positive integer as a difference (unlike the Mian-Chowla sequence A005282, which omits differences 33, 88, 98, 99, ...; see A080200).
The set is an optimization of Browkin's base, where r = a(n-1) + 1.
The series Sum_{n>=0} 1/(a(n+1) - a(n)) is divergent.
Conjecture: lim inf_{n->oo} (a(n+1) - a(n))/n = 1/2.
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REFERENCES
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Jerzy Browkin, Rozwiązanie pewnego zagadnienia A. Schinzla (Polish) [The solution of a certain problem of A. Schinzel], Roczniki Polskiego Towarzystwa Matematycznego [Annals Polish Mathematical Society], Seria I, Prace Matematyczne III (1959).
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LINKS
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FORMULA
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a(n) >= A025582(n+1) and for n <= 10 is here equality.
Conjecture: a(n) ~ log(log(n))*A025582(n+1), where A025582(m)+1 = A005282(m) is the Mian-Chowla sequence.
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EXAMPLE
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Given a(0)=0, a(1)=1, a(2)=3, a(3)=7, the differences used are 1,2,3,4,6,7, so d=5, and we can use a(4) = a(3)+d = 7+5 = 12 because appending a(4)=12 to the sequence will result in the differences 12-0=12, 12-1=11, 12-3=9, 12-7=5, none of which had already been used.
Similarly, given a(0)..a(4) = 0,1,3,7,12, the differences used are 1..7,9,11,12, so d=8, and we can use a(5) = a(4)+d = 12+8 = 20 because the resulting differences will be 20, 19, 17, 13, 8, none of which had already been used.
Proceeding as above, we get a(6)=30 and a(7)=44.
Given a(0)..a(7) = 0,1,3,7,12,20,30,44, the differences used are 1..14,17..20,23..24,27,29..30,32,37,41,43..44, so d=15, but we cannot use a(8) = a(7)+d = 44+15 = 59 because the difference 29 would be repeated: 59-30 = 30-1. Thus, we must find the smallest r such that using both a(8) = a(7)+r and a(9) = a(8)+d will not repeat any differences. The smallest such r is 21, so a(8) = a(7)+r = 44+21 = 65 and a(9) = a(8)+d = 65+15 = 80.
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CROSSREFS
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Cf. A001856, where a(1)=1, a(2)=2, a(2n+1)=2*a(2n), a(2n+2) = a(2n+1) + d.
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KEYWORD
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nonn,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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