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A082936
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a(n) = (1/(3*n))*Sum_{d|n, d even} phi(2*n/d)*binomial(3d/2,d).
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4
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1, 1, 3, 10, 43, 201, 1038, 5538, 30667, 173593, 1001603, 5864750, 34769374, 208267320, 1258579654, 7663720710, 46976034379, 289628805623, 1794932468571, 11175157356522, 69864075597643, 438403736549145, 2760351032959050, 17433869214973754, 110420300879752990
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OFFSET
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0,3
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COMMENTS
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a(n) = number of necklaces of n white beads and 2n black beads. - David Callan, Mar 28 2004
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LINKS
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FORMULA
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a(n) = sum( d divides n, phi(n/d) * C(3*d,d) ) / (3*n) for n>=1, a(0)=1.
a(n) = sum( d divides n, phi(n/d) * C(3*d-1,d) ) / (2*n) for n>=1, a(0)=1.
(End)
a(n) ~ 3^(3*n) / (2^(2*n+1) * sqrt(3*Pi) * n^(3/2)). - Vaclav Kotesovec, Aug 22 2015
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MAPLE
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with(numtheory): f := proc(n) local t1, d; t1 := 0; for d from 1 to n do if n mod d = 0 then if d mod 2 = 0 then t1 := t1+phi(n/d)*binomial(3*d/2, d) fi; fi; od; 2*t1/(3*n); end; # use with n even
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MATHEMATICA
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a[n_] := DivisorSum[n, EulerPhi[n/#]*Binomial[3#, #]&]/(3n); a[0] = 1; Array[a, 30, 0] (* Jean-François Alcover, Dec 02 2015 *)
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PROG
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(PARI)
C(n, k)=binomial(n, k);
a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(3*d, d)) / (3*n) );
/* or, second formula: */
/* a(n) = if(n<=0, n==0, sumdiv(n, d, eulerphi(n/d) * C(3*d-1, d)) / (2*n) ); */
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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