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%I #26 Jul 27 2019 16:45:06
%S 0,0,1,-1,1,2,-2,2,2,1,5,-5,5,2,5,2,2,1,8,-8,8,2,5,8,3,-3,3,2,6,-6,6,
%T 2,4,-4,4,2,4,2,2,1,22,-22,22,2,5,22,3,20,-20,20,2,7,-7,7,2,4,19,-19,
%U 19,2,5,16,-16,16,2,5,5,1,28,-28,28,2,7,19,15,-15,15,2,6,48,-48,48,2,5,17,-17,17,2,5,5,1,23,-23,23,2,7,23,3,51,-51
%N Negative van Eck's sequence: For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = n-m, otherwise a(n+1) = -a(n). Start with a(1)=0.
%C Similar logic as van Eck's sequence except that if a(n) hasn't appeared before, then a(n+1) = -a(n) instead of being 0.
%F If a(n) < 0, then a(n+1) = -a(n) and a(n+2) = 2. - _Rémy Sigrist_, Jul 14 2019
%e We start with a(1) = 0.
%e 0 has not occurred before, so our rule says that a(2) = -a(1) = 0.
%e Now 0 has occurred before, at a(1), which is 1 term before, so a(3) = 1.
%e 1 has not occurred before, so a(4) = -a(3) = -1.
%e -1 has not occurred before, so a(5) = -a(4) = 1.
%e Now 1 has occurred before, at a(3), which is 2 term before, so a(6) = 2.
%e 2 has not occurred before, so a(6) = -a(5) = -2.
%e And so on...
%o (Python)
%o import numpy as np
%o def A308619(n):
%o NegVanEck = np.array([0])
%o #first value is 0
%o i=NegVanEck[0]
%o while i < n:
%o last = NegVanEck[-1]
%o pos = np.where(NegVanEck == last)
%o if pos[0].size == 1:
%o NegVanEck = np.append(NegVanEck, -last)
%o else:
%o NegVanEck = np.append(NegVanEck, pos[0][-1] - pos[0][-2])
%o i += 1
%o return NegVanEck
%o print(A308619(1000)) #prints first 1000 integers of the sequence
%Y Cf. A181391.
%K sign
%O 1,6
%A _João D. R. Camacho_, Jun 11 2019
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