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Inverse EULER transform of A064831 (with its initial 1 omitted).
3

%I #49 Aug 19 2018 10:45:22

%S 1,3,3,5,10,24,50,120,270,640,1500,3600,8610,20880,50700,124024,

%T 304290,750120,1854400,4600200,11440548,28527320,71289000,178526880,

%U 447910470,1125750120,2833885800,7144449920,18036373140,45591631800,115381697740,292329067800

%N Inverse EULER transform of A064831 (with its initial 1 omitted).

%C Also, dimensions of the graded Yang-Mills algebra ym(3) [Herscovich and Solotar]. - _N. J. A. Sloane_, Jan 02 2013

%C Apparently the same as A032170 for n > 2. - _Ralf Stephan_, Feb 01 2004

%C From _Petros Hadjicostas_, Dec 03 2017: (Start)

%C We clarify the name of the sequence and provide a sketch of a proof of R. Stephan's statement above.

%C For n>=0, let b(n) = A064831(n+1). Then B(x) = Sum_{n>=0} b(n)*x^n = Sum_{n>=0} A064831(n+1)*x^n = 1/((1-x^2)*(1-3*x+x^2)) (see the formula section for sequence A064831). Define a(0) = 1, and (a(n): n>=1) = invEULER(b(n): n>=0), i.e., we give a new (more correct) definition of the current sequence. Using Bernstein and Sloane (1995), we define the sequence (d(n): n>=1) via Sum_{n>=1} (d(n)/n)*x^n = log B(x). Since EULER(a(n): n>=1) = (b(n): n>=0), we have a(n) = (1/n)*Sum_{s|n} mu(s)*d(n/s). Using the change of indexes m=n/s, we get A(x) = 1 + Sum_{n>=1} a(n)*x^n = 1 + Sum_{s>=1} (mu(s)/s)*Sum_{m>=1} (d(m)/m)*(x^s)^m = 1 + Sum_{s>=1} (mu(s)/s)*log B(x^s).

%C From the documentation of sequence A032170, we see that its g.f. is Sum_{s>=1} (mu(s)/s)*log C(x^s), where C(x) = (1-x)^2/(1-3*x+x^2). To prove R. Stephan's claim above, we need to prove that Sum_{s>=1} (mu(s)/s)*log C(x^s) - x - 2*x^2 = Sum_{s>=1} (mu(s)/s)*log B(x^s) - 3*x - 3*x^2. The last equality is equivalent to Sum_{s>=1} (mu(s)/s)*log(B(x^s)/C(x^s)) = 2*x + x^2, which in turn is equivalent to -Sum_{s>=1} (mu(s)/s) log((1-x^s)^2*(1-x^{2*s})) = 2*x + x^2. The last equality follows from the identity -Sum_{s>=1} (mu(s)/s)*log(1-y^s) = y, which follows from the Lambert series Sum_{s>=1} mu(s)*y^s/(1-y^s) = y.

%C (End)

%H Alois P. Heinz, <a href="/A072337/b072337.txt">Table of n, a(n) for n = 0..1000</a>

%H M. Bernstein and N. J. A. Sloane, <a href="https://doi.org/10.1016/0024-3795(94)00245-9">Some canonical sequences of integers</a>, Linear Algebra and its Applications, Vol. 226-228 (1995), 57-72.

%H A. Connes and M. Dubois-Violette, <a href="https://arxiv.org/abs/math/0206205">Yang-Mills Algebra</a>, arXiv:math/0206205 [math.QA], 2002.

%H Herscovich, Estanislao, and Solotar, Andrea, <a href="http://dx.doi.org/10.4007/annals.2011.173.2.12">Representations of Yang-Mills algebras</a>, Ann. of Math. (2) 173 (2011), no. 2, 1043-1080. - From _N. J. A. Sloane_, Jan 02 2013

%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>

%F G.f.: 1 + Sum_{s>=1} (mu(s)/s)*log B(x^s), where B(x) = 1/((1-x^2) * (1-3*x+x^2)). - _Petros Hadjicostas_, Dec 03 2017

%F a(n) ~ ((1 + sqrt(5))/2)^(2*n) / n. - _Vaclav Kotesovec_, Aug 19 2018

%t mob[m_, n_] := If[Mod[m, n] == 0, MoebiusMu[m/n], 0];

%t EULERi[b_] := Module[{a, c, i, d}, c = {}; For[i = 1, i <= Length[b], i++, c = Append[c, i*b[[i]] - Sum[c[[d]]*b[[i - d]], {d, 1, i - 1}]]]; a = {}; For[i = 1, i <= Length[b], i++, a = Append[a, (1/i)*Sum[mob[i, d]*c[[d]], {d, 1, i}]]]; Return[a]];

%t Join[{1}, EULERi[LinearRecurrence[{3, 0, -3, 1}, {3, 9, 24, 64}, 31]]] (* _Jean-François Alcover_, Aug 19 2018 *)

%Y Cf. A032170, A064831.

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Jul 16 2002