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A032275
Number of bracelets (turnover necklaces) of n beads of 4 colors.
12
4, 10, 20, 55, 136, 430, 1300, 4435, 15084, 53764, 192700, 704370, 2589304, 9608050, 35824240, 134301715, 505421344, 1909209550, 7234153420, 27489127708, 104717491064, 399827748310, 1529763696820
OFFSET
1,1
FORMULA
"DIK" (bracelet, indistinct, unlabeled) transform of 4, 0, 0, 0, ...
Equals (A001868(n) + A056486(n)) / 2 = A001868(n) - A278640(n) = A278640(n) + A056486(n), for n>=1.
a(n) = A081720(n,4), n >= 4. - Wolfdieter Lang, Jun 03 2012
G.f.: (1 - Sum_{n>=1} phi(n)*log(1 - 4*x^n)/n + (1+4*x+6*x^2)/(1-4*x^2))/2. - Herbert Kociemba, Nov 02 2016
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))* Sum_{d|n} phi(d)*k^(n/d), where k=4 is the maximum number of colors. - Robert A. Russell, Sep 24 2018
a(n) = (k^floor((n+1)/2) + k^ceiling((n+1)/2))/4 + (1/(2*n))*Sum_{i=1..n} k^gcd(n,i), where k=4 is the maximum number of colors. (See A075195 formulas.) - Richard L. Ollerton, May 04 2021
EXAMPLE
For n=2, the ten bracelets are AA, AB, AC, AD, BB, BC, BD, CC, CD, and DD. - Robert A. Russell, Sep 24 2018
MATHEMATICA
mx=40; CoefficientList[Series[(1-Sum[ EulerPhi[n]*Log[1-4*x^n]/n, {n, mx}]+(1+4 x+6 x^2)/(1-4 x^2))/2, {x, 0, mx}], x] (* Herbert Kociemba, Nov 02 2016 *)
k=4; Table[DivisorSum[n, EulerPhi[#] k^(n/#) &]/(2n) + (k^Floor[(n+1)/2] + k^Ceiling[(n+1)/2])/4, {n, 1, 30}] (* Robert A. Russell, Sep 24 2018 *)
CROSSREFS
Column 4 of A051137.
Sequence in context: A237626 A020149 A056412 * A220828 A015220 A047199
KEYWORD
nonn
STATUS
approved