A133893 Numbers m such that binomial(m+3,m) mod 3 = 0.

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

Offset: 0

Author

Hieronymus Fischer, Oct 20 2007

Keywords

• nonn

Comments

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.

Crossrefs

Cf. A000040, A133620, A133621, A133623, A133630, A133635.

Cf. A133873, A133883, A133890, A133900, A133910.

Programs

Mathematica

Select[Range[200],Mod[Binomial[#+3,#],3]==0&] (* Harvey P. Dale, Aug 27 2023 *)

Formula

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).