A072337 Inverse EULER transform of A064831 (with its initial 1 omitted).

1, 3, 3, 5, 10, 24, 50, 120, 270, 640, 1500, 3600, 8610, 20880, 50700, 124024, 304290, 750120, 1854400, 4600200, 11440548, 28527320, 71289000, 178526880, 447910470, 1125750120, 2833885800, 7144449920, 18036373140, 45591631800, 115381697740, 292329067800

Offset: 0

Author

N. J. A. Sloane, Jul 16 2002

Keywords

• nonn

Comments

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

Apparently the same as A032170 for n > 2. - Ralf Stephan, Feb 01 2004

From Petros Hadjicostas, Dec 03 2017: (Start)

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

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).

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.

(End)

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..1000
• M. Bernstein and N. J. A. Sloane, Some canonical sequences of integers, Linear Algebra and its Applications, Vol. 226-228 (1995), 57-72.
• A. Connes and M. Dubois-Violette, Yang-Mills Algebra, arXiv:math/0206205 [math.QA], 2002.
• Herscovich, Estanislao, and Solotar, Andrea, Representations of Yang-Mills algebras, Ann. of Math. (2) 173 (2011), no. 2, 1043-1080. - From _N. J. A. Sloane_, Jan 02 2013
• N. J. A. Sloane, Transforms

Crossrefs

Cf. A032170, A064831.

Programs

Mathematica

mob[m_, n_] := If[Mod[m, n] == 0, MoebiusMu[m/n], 0];
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]];
Join[{1}, EULERi[LinearRecurrence[{3, 0, -3, 1}, {3, 9, 24, 64}, 31]]] (* Jean-François Alcover, Aug 19 2018 *)

Formula

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

a(n) ~ ((1 + sqrt(5))/2)^(2*n) / n. - Vaclav Kotesovec, Aug 19 2018