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A133893
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Numbers m such that binomial(m+3,m) mod 3 = 0.
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0
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6, 7, 8, 15, 16, 17, 24, 25, 26, 33, 34, 35, 42, 43, 44, 51, 52, 53, 60, 61, 62, 69, 70, 71, 78, 79, 80, 87, 88, 89, 96, 97, 98, 105, 106, 107, 114, 115, 116, 123, 124, 125, 132, 133, 134, 141, 142, 143, 150, 151, 152, 159, 160, 161, 168, 169, 170, 177, 178, 179, 186
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OFFSET
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0,1
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COMMENTS
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Also numbers m such that floor(1+(m/3)) mod 3 = 0.
Partial sums of the sequence 6,1,1,7,1,1,7,1,1,7, ... which has period 3.
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LINKS
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FORMULA
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a(n)=3n+6-2*(n mod 3).
G.f.: g(x)=6/(1-x)+x(1+x+7x^2)/((1-x^3)(1-x)) = (6+x+x^2+x^3)/((1-x^3)(1-x)).
G.f.: g(x)=(6-5x-x^4)/((1-x^3)(1-x)^2).
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MATHEMATICA
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Select[Range[200], Mod[Binomial[#+3, #], 3]==0&] (* Harvey P. Dale, Aug 27 2023 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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