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Write 2*x/(1-x) in the form Sum_{j>=1} ((1-x^j)^a(j) - 1).
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%I #8 Aug 27 2017 21:47:42

%S -2,1,2,3,4,6,6,10,8,15,10,25,12,28,10,60,16,25,18,125,0,66,22,218,24,

%T 91,-30,420,28,-387,30,2011,-88,153,28,-1894,36,190,-182,8902,40,

%U -3234,42,2398,-132,276,46,2340,48,-2678,-510,4641,52,-1754,-198,108400

%N Write 2*x/(1-x) in the form Sum_{j>=1} ((1-x^j)^a(j) - 1).

%F For all n > 0 we have: 2 = Sum_{d|n} binomial(-a(d) + n/d - 1, n/d).

%e 2x/(1-x) = (1-x)^(-2) - 1 + (1-x^2)^1 - 1 + (1-x^3)^2 - 1 + (1-x^4)^3 - 1 + ...

%p a:= n-> add(binomial(n/d-1-a(d), n/d), d=

%p numtheory[divisors](n) minus {n})-2:

%p seq(a(n), n=1..60); # _Alois P. Heinz_, Aug 27 2017

%t nn=60;

%t rus=SolveAlways[Normal[Series[2x/(1-x)==Sum[(1-x^n)^a[n]-1,{n,nn}],{x,0,nn}]],x];

%t Array[a,nn]/.First[rus]

%Y Cf. A048272, A220418, A260685, A281145, A289078, A289501, A290261, A290971.

%K sign

%O 1,1

%A _Gus Wiseman_, Aug 16 2017