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A300760
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Number of ways to select 4 numbers from the set of the first n natural numbers avoiding 3-term arithmetic progressions.
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3
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0, 1, 4, 10, 25, 51, 98, 165, 267, 407, 601, 849, 1175, 1580, 2089, 2703, 3452, 4338, 5395, 6622, 8058, 9706, 11606, 13758, 16210, 18963, 22066, 25520, 29379, 33645, 38376, 43571, 49293, 55545, 62391, 69831, 77937, 86710, 96223, 106477, 117550, 129444, 142241
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OFFSET
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4,3
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (2,0,-1,0,-2,2,0,1,0,-2,1).
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FORMULA
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a(n) = (n^4 - 12*n^3 + 51*n^2 - 78*n + 32)/24 + b(n) + c(n), where
b(n) = 0 for n even
b(n) = (-n + 2)/4 for n odd
c(n) = 0 for n == 1,2,5,7,10,11 (mod 12)
c(n) = -1/3 for n == 3,6,9 (mod 12)
c(n) = -4/3 for n == 0 (mod 12)
c(n) = -1 for n == 4,8 (mod 12).
a(n) = (n^4 - 12*n^3 + 51*n^2 - 78*n + 32)/24 + (n == 1 (mod 2))*(-n + 2)/4 - (n == 0 (mod 3))/3 - (n == 0 (mod 4)).
G.f.: x^5*(1 + 2*x + 2*x^2 + 6*x^3 + 5*x^4 + 8*x^5) / ((1 - x)^5*(1 + x)^2*(1 + x^2)*(1 + x + x^2)).
a(n) = 2*a(n-1) - a(n-3) - 2*a(n-5) + 2*a(n-6) + a(n-8) - 2*a(n-10) + a(n-11) for n>14.
(End)
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EXAMPLE
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There are 4 selections of 4 natural numbers from the set {1,2,3,4,5,6} free of 3-term arithmetic progressions: {1,2,4,5}, {1,2,5,6}, {1,3,4,6}, {2,3,5,6}.
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MATHEMATICA
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Array[(#^4 - 12 #^3 + 51 #^2 - 78 # + 32)/24 + Boole[OddQ@ #] (-# + 2)/4 - Boole[Mod[#, 3] == 0]/3 - Boole[Mod[#, 4] == 0] &, 43, 4] (* Michael De Vlieger, Mar 14 2018 *)
LinearRecurrence[{2, 0, -1, 0, -2, 2, 0, 1, 0, -2, 1}, {0, 1, 4, 10, 25, 51, 98, 165, 267, 407, 601}, 50] (* Harvey P. Dale, Feb 18 2024 *)
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PROG
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(PARI) concat(0, Vec(x^5*(1 + 2*x + 2*x^2 + 6*x^3 + 5*x^4 + 8*x^5) / ((1 - x)^5*(1 + x)^2*(1 + x^2)*(1 + x + x^2)) + O(x^60))) \\ Colin Barker, Aug 06 2018
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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