

A247556


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 <= n1, and let d be the smallest positive integer not in S. If, for every i in 1..n1, a(n1) + d  a(i) is not in S, then a(n) = a(n1) + d. Otherwise, let r be the smallest positive integer such that, for every i in 1..n1, neither a(n1) + r  a(i) nor a(n1) + r + d  a(i) is in S; then a(n) = a(n1) + r and a(n+1) = a(n) + d.


2



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

0,3


COMMENTS

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(n1), we assign a(n) = a(n1) + d, provided that this would not cause any difference to be repeated; otherwise, we assign a(n) = a(n1) + 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(n1)) 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 MianChowla sequence A005282, which omits differences 33, 88, 98, 99, ...; see A080200).
The set is an optimization of Browkin's base, where r = a(n1) + 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.


REFERENCES

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).


LINKS

Jon E. Schoenfield, Table of n, a(n) for n = 0..6039
Andrew Pollington and Charles Vanden Eynden, The integers as differences of a sequence, Canad. Math. Bull. Vol. 24 (4), 1981 (497499)
Jon E. Schoenfield, Magma program


FORMULA

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 MianChowla sequence.


EXAMPLE

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 120=12, 121=11, 123=9, 127=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: 5930 = 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.


CROSSREFS

Cf. A001856, where a(1)=1, a(2)=2, a(2n+1)=2*a(2n), a(2n+2) = a(2n+1) + d.
Cf. A005282 (MianChowla sequence), A025582.
Cf. A080200.
Sequence in context: A330285 A002049 A025582 * A337656 A029452 A294422
Adjacent sequences: A247553 A247554 A247555 * A247557 A247558 A247559


KEYWORD

nonn,nice


AUTHOR

Thomas Ordowski, Sep 19 2014


EXTENSIONS

More terms from Jon E. Schoenfield, Jan 18 2015
Edited by Jon E. Schoenfield, Jan 22 2015


STATUS

approved



