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A036258
Number of inequivalent strings of n digits, when 2 strings are equivalent if turning 1 upside down gives the other.
3
1, 9, 90, 945, 9700, 98475, 992250, 9961125, 99805000, 999024375, 9995118750, 99975590625, 999877937500, 9999389671875, 99996948281250, 999984741328125, 9999923706250000, 99999618530859375, 999998092652343750, 9999990463259765625, 99999952316289062500
OFFSET
0,2
REFERENCES
Nick Baxter, The Burnside di-lemma: combinatorics and puzzle symmetry, in Tribute to a Mathemagician, Peters, 2005, pp. 199-210.
De Bruijn, Polya's theory of counting, in Beckenbach, ed., Applied Combinatorial Math., Wiley, 1964 (p. 182).
FORMULA
a(n+1) = (1/10)*{10^n - 5^n + (4-(-1)^n)*5^[n/2]} (De Bruijn)
From Colin Barker, Jul 03 2017: (Start)
G.f.: (1 - 6*x + 75*x^3) / ((1 - 5*x)*(1 - 10*x)*(1 - 5*x^2)).
a(n) = 5^((n-1)/2+1/2)/2 - 5^n/2 + 10^n for n even.
a(n) = 3*5^((n-1)/2)/2 - 5^n/2 + 10^n for n odd.
a(n) = 15*a(n-1) - 45*a(n-2) - 75*a(n-3) + 250*a(n-4) for n>3.
(End)
MAPLE
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;
MATHEMATICA
LinearRecurrence[{15, -45, -75, 250}, {1, 9, 90, 945}, 30] (* Harvey P. Dale, Jul 05 2023 *)
PROG
(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
CROSSREFS
Sequence in context: A052268 A155199 A147841 * A098399 A264914 A143079
KEYWORD
nonn,easy,base
STATUS
approved