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A032275
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Number of bracelets (turnover necklaces) of n beads of 4 colors.
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12
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4, 10, 20, 55, 136, 430, 1300, 4435, 15084, 53764, 192700, 704370, 2589304, 9608050, 35824240, 134301715, 505421344, 1909209550, 7234153420, 27489127708, 104717491064, 399827748310, 1529763696820
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OFFSET
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1,1
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LINKS
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FORMULA
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"DIK" (bracelet, indistinct, unlabeled) transform of 4, 0, 0, 0, ...
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
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EXAMPLE
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For n=2, the ten bracelets are AA, AB, AC, AD, BB, BC, BD, CC, CD, and DD. - Robert A. Russell, Sep 24 2018
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MATHEMATICA
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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 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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