%I #39 Nov 23 2023 08:02:17
%S 0,0,6,48,204,624,1554,3360,6552,11808,19998,32208,49764,74256,107562,
%T 151872,209712,283968,377910,495216,639996,816816,1030722,1287264,
%U 1592520,1953120,2376270,2869776,3442068,4102224,4859994,5725824,6710880
%N Number of 5-bead necklaces of n colors not allowing reversal, with no adjacent beads having the same color.
%C This sequence would be better defined as a(n) = (n^5-n)/5 with offset 0, which is an integer by Fermat's little theorem. - _N. J. A. Sloane_, Nov 13 2023
%H R. H. Hardin, <a href="/A208536/b208536.txt">Table of n, a(n) for n = 1..210</a>
%H Jack Jeffries, <a href="https://doi.org/10.1090/noti2833">Differentiating by prime numbers</a>, Notices Amer. Math. Soc., 70:11 (2023), 1772-1779.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/P-derivation">p-derivation</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F Empirical: a(n) = (1/5)*n^5 - 1*n^4 + 2*n^3 - 2*n^2 + (4/5)*n.
%F Equivalently: a(n) = ((n-1)^5 - (n-1))/5. - _M. F. Hasler_, Mar 05 2016
%F Empirical formula confirmed by _Petros Hadjicostas_, Nov 05 2017 (see A208535).
%F a(n+2) = delta(-n) = -delta(n) for n >= 0, where delta is the p-derivation over the integers with respect to prime p = 5. - _Danny Rorabaugh_, Nov 10 2017
%F From _Colin Barker_, Nov 11 2017: (Start)
%F G.f.: 6*x^3*(1 + x)^2 / (1 - x)^6.
%F a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>6.
%F (End)
%e All solutions for n=3:
%e ..1....1....1....1....1....1
%e ..3....3....2....2....2....2
%e ..1....2....1....3....3....1
%e ..3....3....3....2....1....2
%e ..2....2....2....3....3....3
%t A208536[n_]:=((n-1)^5-(n-1))/5;Array[A208536,50] (* _Paolo Xausa_, Nov 14 2023 *)
%o (PARI) Vec(6*x^3*(1 + x)^2 / (1 - x)^6 + O(x^40)) \\ _Colin Barker_, Nov 11 2017
%Y Row 5 of A208535.
%Y Also, row 5 (with different offset) of A074650. - _Eric M. Schmidt_, Dec 08 2017
%Y Cf. A208537.
%K nonn,easy
%O 1,3
%A _R. H. Hardin_, Feb 27 2012