|
|
A115116
|
|
Number of imprimitive (periodic) asymmetric rhythm cycles: ones having nontrivial shift automorphisms. 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.
|
|
1
|
|
|
1, 1, 2, 1, 2, 3, 2, 1, 6, 3, 2, 11, 2, 3, 30, 1, 2, 63, 2, 11, 162, 3, 2, 411, 26, 3, 1098, 11, 2, 3015, 2, 1, 8058, 3, 182, 22151, 2, 3, 61326, 411, 2, 170883, 2, 11, 479410, 3, 2, 1345211, 158, 2955, 3798246, 11, 2, 10761723, 8078, 411, 30585834, 3, 2, 87191759, 2, 3, 249057230, 1, 61346, 713205963, 2, 11, 2046590850, 173775, 2
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,3
|
|
COMMENTS
|
a(2^k)=1 for all k including k=0.
a(p)=2, a(2p)=3, a(4p)=11, etc. for an odd prime p.
|
|
LINKS
|
|
|
FORMULA
|
|
|
MATHEMATICA
|
A006575[n_] := DivisorSum[n, If[BitAnd[#, 1] == 1, MoebiusMu[#] (3^(n/#) - 1), 0]&]/(2n);
A115114[n_] := Sum[EulerPhi[2d] + Boole[OddQ[d]] EulerPhi[d] 3^(n/d), {d, Divisors[n]}]/(2n);
|
|
PROG
|
(PARI)
A006575(n) = (sumdiv(n, d, bitand(d, 1)*moebius(d)*(3^(n/d)-1)) / (2*n)); \\ From A006575.
A115114(n) = (1/(2*n))*(sumdiv(n, d, eulerphi(2*d)+(bitand(d, 1)*eulerphi(d)*(3^(n/d)))));
|
|
CROSSREFS
|
|
|
KEYWORD
|
easy,nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|