A036258 - OEIS

%I #10 Jul 05 2023 15:26:13

%S 1,9,90,945,9700,98475,992250,9961125,99805000,999024375,9995118750,

%T 99975590625,999877937500,9999389671875,99996948281250,

%U 999984741328125,9999923706250000,99999618530859375,999998092652343750,9999990463259765625,99999952316289062500

%N Number of inequivalent strings of n digits, when 2 strings are equivalent if turning 1 upside down gives the other.

%D Nick Baxter, The Burnside di-lemma: combinatorics and puzzle symmetry, in Tribute to a Mathemagician, Peters, 2005, pp. 199-210.

%D De Bruijn, Polya's theory of counting, in Beckenbach, ed., Applied Combinatorial Math., Wiley, 1964 (p. 182).

%H Colin Barker, <a href="/A036258/b036258.txt">Table of n, a(n) for n = 0..950</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (15,-45,-75,250).

%F a(n+1) = (1/10)*{10^n - 5^n + (4-(-1)^n)*5^[n/2]} (De Bruijn)

%F From _Colin Barker_, Jul 03 2017: (Start)

%F G.f.: (1 - 6*x + 75*x^3) / ((1 - 5*x)*(1 - 10*x)*(1 - 5*x^2)).

%F a(n) = 5^((n-1)/2+1/2)/2 - 5^n/2 + 10^n for n even.

%F a(n) = 3*5^((n-1)/2)/2 - 5^n/2 + 10^n for n odd.

%F a(n) = 15*a(n-1) - 45*a(n-2) - 75*a(n-3) + 250*a(n-4) for n>3.

%F (End)

%p f:=n-> if n mod 2 = 0 then 10^n-(5^n-5^(n/2))/2 else 10^n-(5^n-3*5^((n-1)/2))/2; fi;

%t LinearRecurrence[{15,-45,-75,250},{1,9,90,945},30] (* _Harvey P. Dale_, Jul 05 2023 *)

%o (PARI) Vec((1 - 6*x + 75*x^3) / ((1 - 5*x)*(1 - 10*x)*(1 - 5*x^2)) + O(x^30)) \\ _Colin Barker_, Jul 03 2017

%Y Cf. A036255, A036257.

%K nonn,easy,base

%O 0,2

%A _N. J. A. Sloane_