A274647 A variation on Recamán's sequence (A005132): to get a(n), we first try to subtract n from a(n-1): a(n) = a(n-1)-n if positive and not already in the sequence; if not then we try to add n: a(n) = a(n-1)+n if not already in the sequence; if this fails we try to subtract 2n from a(n-1), or to add 2n to a(n-1), or to subtract 3n, or to add 3n, etc., until one of these produces a positive number not already in the sequence.
0, 1, 3, 6, 2, 7, 13, 20, 12, 21, 11, 22, 10, 23, 9, 24, 8, 25, 43, 62, 42, 63, 41, 18, 66, 91, 65, 38, 94, 123, 93, 124, 92, 59, 127, 162, 126, 89, 51, 90, 50, 132, 174, 131, 87, 177, 223, 176, 128, 79, 29, 80, 28, 81, 27, 82, 26, 83, 141, 200, 140, 201, 139
Offset: 0
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Programs
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Mathematica
f[s_List] := Block[{a = b = 0, k = 1, l = s[[-1]], n = Length@ s}, While[ If[l > k*n && !MemberQ[s, l - k*n], a = l - k*n]; If[ !MemberQ[s, l + k*n], b = l + k*n; Break[]]; a == b == 0, k++]; Append[s, If[a > 0, a, b]]]; Nest[f, {0}, 70] (* Robert G. Wilson v, Sep 09 2016 *)
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Python
l=[0] for n in range(1, 101): i=1 while True: a=l[n - 1] x=a - i*n if x>0 and x not in l: l.append(x) break y=a + i*n if y>0 and not y in l: l.append(y) break else : i+=1 print(l) # Indranil Ghosh, Jun 03 2017
Formula
A276342(a(n)) = n for all n.
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