A128382 Inverse Moebius transform operation performed 24 times on A000594: A051731^24 * A000594.

1, 0, 276, -1748, 4854, 0, -16720, 44552, -107295, 0, 534636, -482448, -577714, 0, 1339704, 2528206, -6905910, 0, 10661444, -8484792, -4614720, 0, 18643296, 12296352, -25383005, 0, -75928312, 29226560, 128406654, 0, -52843144, -151821160, 147559536, 0, -81158880

Offset: 1

Author

Gary W. Adamson, Feb 28 2007

Keywords

• sign
• mult

Comments

Conjecture: given the inverse Moebius transform operation performed any k times (k=1,2,3,...); k=24 is the only such sequence with zeros. A weaker conjecture: "zero" occurs an infinite number of times in A128382.

Multiplicative because A000594 is. Each application of A051731 corresponds to an inverse Moebius transform. - Andrew Howroyd, Aug 03 2018

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A051731, A000594, A128378, A128379, A128380, A128381.

Programs

Mathematica

nmax = 40;
M = Table[If[Mod[n, k] == 0, 1, 0], {n, nmax}, {k, nmax}];
MatrixPower[M, 24].RamanujanTau[Range[nmax]] (* Jean-François Alcover, Sep 20 2019 *)

PARI

seq(n, k=24)={my(u=vector(n,n,1), v=vector(n,n,ramanujantau(n))); for(i=1, k, v=dirmul(u,v)); v} \\ Andrew Howroyd, Aug 03 2018

Formula

Dirichlet g.f.: Sum_{n>=1} a(n)/n^s = (Sum_{n>=1} A000594(n)/n^s)*zeta(s)^24. - Jianing Song, Aug 04 2018

Extensions

Terms a(11) and beyond from Andrew Howroyd, Aug 03 2018