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%I #16 May 15 2023 15:50:07
%S 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,
%T 18,-9,19,20,-10,21,22,-11,23,24,-12,25,26,-13,27,28,-14,29,30,-15,31,
%U 32,-16,33,34,-17,35,36,-18,37,38,-19,39,40,-20,41,42,-21,43,44,-22,45,46
%N Permutation of the integers: two positives, one negative.
%C Setting m=2 in
%C log(m) = Sum_{n>0} (n mod m - (n-1) mod m)/n [1]
%C yields the sum
%C log(2) = (1 -1/2) +(1/3 -1/4) +(1/5 -1/6)+...
%C Substituting every -1/d by 1/d - 2/d we obtain
%C log(2) = (1+1/2-1)+(1/3+1/4-1/2)+(1/5+1/6-1/3)+...
%C a(n) is the sequence of denominators of this modified sum with unit numerators, so
%C Sum_{k>0} 1/a(k) = log(2)
%C Substituting -1/d by -2/d + 1/d would yield another permutation (one positive, one negative, one positive) with the same sum of inverses.
%C 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).
%C Equation [1] is a result of expanding log( Sum_{0<=k<=m-1} x^k ) at x=1 (see comment to A061347.)
%H G. C. Greubel, <a href="/A166711/b166711.txt">Table of n, a(n) for n = 0..10000</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Riemann_series_theorem">Riemann series theorem</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,2,0,0,-1).
%F G.f.: (x*(1+2*x-x^2+x^3)/((1-x)^2*(1+x+x^2)^2)).
%F 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.
%F a(n) = (n+1)/3 +2*A049347(n)/3 -(-1)^n*A076118(n+1). - _R. J. Mathar_, Oct 30 2009
%t LinearRecurrence[{0, 0, 2, 0, 0, -1}, {0, 1, 2, -1, 3, 4}, 100] (* _G. C. Greubel_, May 24 2016 *)
%t Join[{0},With[{nn=50},Riffle[Range[nn],Range[-1,-nn/2,-1],3]]] (* _Harvey P. Dale_, May 15 2023 *)
%o (PARI) a(n)=(2*(n+1)\3)*(1-3/2*!(n%3))
%o (PARI) a(n)=if(n>=0,[ -n\3, 2*(n\3)+1, 2*(n\3)+2][n%3+1]) \\ _Jaume Oliver Lafont_, Nov 14 2009
%Y Cf. A001057, A002162, A038608. Signed and shifted version of A009947.
%K sign,easy
%O 0,3
%A _Jaume Oliver Lafont_, Oct 18 2009
%E Corrected by _Jaume Oliver Lafont_, Oct 22 2009
%E frac keyword removed by _Jaume Oliver Lafont_, Nov 02 2009
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