cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A006575 Number of primitive (aperiodic, or Lyndon) asymmetric rhythm cycles: ones having no nontrivial shift automorphism.

Original entry on oeis.org

1, 2, 4, 10, 24, 60, 156, 410, 1092, 2952, 8052, 22140, 61320, 170820, 478288, 1345210, 3798240, 10761660, 30585828, 87169608, 249055976, 713205900, 2046590844, 5883948540, 16945772184, 48882035160, 141214767876
Offset: 1

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Comments

Asymmetric rhythm cycles (A115114): binary necklaces of length 2n subject to the restriction that for any k if the k-th bead is of color 1 then the (k+n)-th bead (modulo 2n) is of color 0. - Valery A. Liskovets, Jan 17 2006
This sequence differs from the Moebius transform of A115114 (for even n). Coincides with the second row (q=3) of array A098691. - Valery A. Liskovets, Jan 17 2006
This sequence is the number of Lyndon words on {1, 2, 3} with an odd number of 1's. Also, for even n, this sequence represents the differences between the number of Lyndon words on {1, 2, 3} with an odd number of 1's and the number of Lyndon words on {1, 2, 3} with an even number of 1's. - Jennifer Woodcock (jennifer.woodcock(AT)ugdsb.on.ca), Jan 03 2008

Examples

			Example. For n=3, out of 6=A115114(3) admissible rhythm cycles (necklaces) 000000, 100000, 110000, 101000, 111000 and 101010, only the first and the last ones are imprimitive. Thus a(3)=4.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A133267. Row q=3 of A098691.

Programs

  • Mathematica
    a[n_] := DivisorSum[n, If[BitAnd[#, 1]==1, MoebiusMu[#]*(3^(n/#)-1), 0]&] / (2n); Array[a, 30] (* Jean-François Alcover, Dec 01 2015, after Joerg Arndt *)
  • PARI
    a(n) = sumdiv( n, d, if ( bitand(d,1), moebius(d) * (3^(n/d)-1) , 0 ) ) / (2*n); /* Joerg Arndt, Dec 30 2012 */

Formula

From Valery A. Liskovets, Jan 17 2006: (Start)
a(n) = (Sum_{d|n, d odd} mu(d)*(3^(n/d)-1))/(2*n).
a(n) = (3^n-1)/(2*n) for n=2^k and a(n) = (Sum_{d|n, d odd} mu(d)*3^(n/d))/(2*n) otherwise. (End)

Extensions

Edited and extended by Valery A. Liskovets, Jan 17 2006