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A166711
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Permutation of the integers: two positives, one negative.
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6
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0, 1, 2, -1, 3, 4, -2, 5, 6, -3, 7, 8, -4, 9, 10, -5, 11, 12, -6, 13, 14, -7, 15, 16, -8, 17, 18, -9, 19, 20, -10, 21, 22, -11, 23, 24, -12, 25, 26, -13, 27, 28, -14, 29, 30, -15, 31, 32, -16, 33, 34, -17, 35, 36, -18, 37, 38, -19, 39, 40, -20, 41, 42, -21, 43, 44, -22, 45, 46
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OFFSET
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0,3
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COMMENTS
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Setting m=2 in
log(m) = Sum_{n>0} (n mod m - (n-1) mod m)/n [1]
yields the sum
log(2) = (1 -1/2) +(1/3 -1/4) +(1/5 -1/6)+...
Substituting every -1/d by 1/d - 2/d we obtain
log(2) = (1+1/2-1)+(1/3+1/4-1/2)+(1/5+1/6-1/3)+...
a(n) is the sequence of denominators of this modified sum with unit numerators, so
Sum_{k>0} 1/a(k) = log(2)
Substituting -1/d by -2/d + 1/d would yield another permutation (one positive, one negative, one positive) with the same sum of inverses.
Similar sequences (m positives, one negative) may be obtained for the logarithm of any integer m>0. A001057 is the case m=1, with sum of inverses log(1).
Equation [1] is a result of expanding log( Sum_{0<=k<=m-1} x^k ) at x=1 (see comment to A061347.)
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LINKS
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FORMULA
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G.f.: (x*(1+2*x-x^2+x^3)/((1-x)^2*(1+x+x^2)^2)).
a(0)=0, a(1)=1, a(2)=2, a(3)=-1, a(4)=3, a(5)=4, a(n)=2*a(n-3)-a(n-6), n>=6.
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MATHEMATICA
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LinearRecurrence[{0, 0, 2, 0, 0, -1}, {0, 1, 2, -1, 3, 4}, 100] (* G. C. Greubel, May 24 2016 *)
Join[{0}, With[{nn=50}, Riffle[Range[nn], Range[-1, -nn/2, -1], 3]]] (* Harvey P. Dale, May 15 2023 *)
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PROG
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(PARI) a(n)=(2*(n+1)\3)*(1-3/2*!(n%3))
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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