%I #63 Sep 08 2022 08:44:51
%S 2,4,10,16,28,42,64,90,126,168,224,288,370,462,576,704,858,1030,1232,
%T 1456,1716,2002,2330,2688,3094,3536,4032,4570,5168,5814,6528,7296,
%U 8140,9044,10032,11088,12236,13460,14784,16192,17710,19320,21050,22880,24840,26910
%N "DHK[ 5 ]" (bracelet, identity, unlabeled, 5 parts) transform of 1,1,1,1,...
%C a(n) is the number of bracelets with k = 5 black beads and n-k white beads which have no reflection symmetry. - _Herbert Kociemba_, Nov 27 2016
%C From _Petros Hadjicostas_, Feb 24 2019: (Start)
%C When k is odd >= 3, the DHK[k] transform of sequence c = (c(n): n >= 1), whose g.f. is C(x) = Sum_{n>=1} c(n)*x^n, has g.f. Sum_{n>=1} (DHK[k] c)_n*x^n = (1/2)*Sum_{d|k} mu(d)*((1/k)*C(x^d)^(k/d) - C(x^d)*C(x^(2*d))^((k/d) - 1)/2)).
%C For the current sequence we have k = 5 and c(n) = 1 for all n >= 1. Hence, C(x) = x/(1-x) and A(x) = Sum_{n>=1} a(n)*x^n = (x^k/2)*Sum_{d|k} mu(d)*((1/k)*(1-x^d)^(-k/d) - (1-x^d)^(-1)*(1-x^(2*d))^(-((k/d) - 1)/2)).
%C The latter g.f. agrees with Herbert Kociemba's formula found below only when k is an odd prime. The reason is that (DHK[k] c)_n, with c=(1,1,1,...), is the number of aperiodic bracelets without reflection symmetry with k black beads and n-k white beads, while Herbert Kociemba's formula counts all (periodic and aperiodic) bracelets without reflection symmetry with k black beads and n-k white beads. Hence, in the case k is an odd prime, the two formulas agree.
%C When k is even, the g.f. of the DHK[k] transform of sequence c = (c(n): n >= 1) is much more complicated.
%C Note that Herbert Kociemba's formula for counting all (periodic and aperiodic) bracelets with no reflection symmetry is still valid even when k is even; e.g., see sequence A008804 for the case k=4. For k = 4, all bracelets with 4 black beads and n-k = n-4 white beads that have no reflection symmetry are aperiodic, but this is not true anymore for k even >= 6.
%C (End)
%H G. C. Greubel, <a href="/A032246/b032246.txt">Table of n, a(n) for n = 8..1007</a>
%H C. G. Bower, <a href="/transforms2.html">Transforms (2)</a>
%H Petros Hadjicostas, <a href="/A032246/a032246.pdf">The aperiodic version of Herbert Kociemba's formula for bracelets</a>.
%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (2,1,-4,1,3,-3,-1,4,-1,-2,1).
%F G.f.: 2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)).
%F From _Herbert Kociemba_, Nov 27 2016: (Start)
%F More generally gf(k) is the g.f. for the number of bracelets without reflection symmetry with k black beads and n-k white beads.
%F gf(k): x^k/2 * ((1/k)*Sum_{n|k} phi(n)/(1 - x^n)^(k/n) - (1 + x)/(1-x^2)^floor(k/2 + 1)). (End)
%F a(n) = a(5-n) for all n in Z. - _Michael Somos_, Nov 28 2016
%F 0 = a(n) - 2*a(n+1) - a(n+2) + 4*a(n+3) - a(n+4) - 3*a(n+5) + 3*a(n+6) + a(n+7) - 4*a(n+8) + a(n+9) + 2*a(n+10) - a(n+11) for all n in Z. - _Michael Somos_, Nov 28 2016
%e G.f. = 2*x^8 + 4*x^9 + 10*x^10 + 16*x^11 + 28*x^12 + 42*x^13 + 64*x^14 + ...
%t gf[x_,k_]:=x^k/2 (1/k Plus@@(EulerPhi[#] (1-x^#)^(-(k/#))&/@Divisors[k])-(1+x)/(1-x^2)^Floor[k/2+1]); CoefficientList[Series[gf[x,5],{x,0,50}],x] (* _Herbert Kociemba_, Nov 27 2016 *)
%t Drop[CoefficientList[Series[2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)), {x,0,50}], x], 8] (* _G. C. Greubel_, Feb 25 2019 *)
%o (PARI) {a(n) = if( n<0, n=5-n); polcoeff( 2 * x^8 / ((1-x)^2*(1-x^2)^2*(1-x^5)) + x * O(x^n), n)}; /* _Michael Somos_, Nov 28 2016 */
%o (PARI) Vec(2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)) + O(x^40)) \\ _Colin Barker_, Mar 13 2019
%o (Magma) m:=50; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( 2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5)) )); // _G. C. Greubel_, Feb 25 2019
%o (Sage) a=(2*x^8/((1-x)^2*(1-x^2)^2*(1-x^5))).series(x, 50).coefficients(x, sparse=False); a[8:] # _G. C. Greubel_, Feb 25 2019
%Y Cf. A001399, A008804, A032248. Column k = 5 of A180472.
%K nonn,easy
%O 8,1
%A _Christian G. Bower_