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Search: id:A133893
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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
1
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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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COMMENT
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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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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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CROSSREFS
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Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133873, A133883, A133890, A133900, A133910.
Sequence in context: A047275 A047590 A146329 this_sequence A101647 A078745 A087663
Adjacent sequences: A133890 A133891 A133892 this_sequence A133894 A133895 A133896
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KEYWORD
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nonn
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AUTHOR
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Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 20 2007
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