A329952 Numbers k such that binomial(k,3) is divisible by 8.

0, 1, 2, 8, 10, 16, 17, 18, 24, 26, 32, 33, 34, 40, 42, 48, 49, 50, 56, 58, 64, 65, 66, 72, 74, 80, 81, 82, 88, 90, 96, 97, 98, 104, 106, 112, 113, 114, 120, 122, 128, 129, 130, 136, 138, 144, 145, 146, 152, 154, 160, 161, 162, 168, 170, 176, 177, 178, 184, 186, 192, 193, 194

Offset

1,3

Comments

These are possible sizes for 3-symmetric graphs.

The possible size of 2-symmetric graphs is sequence A042948.

These numbers are 0, 1, 2, 8, and 10 modulo 16.

Links

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Formula

G.f.: (6*x^4+2*x^3+6*x^2+x+1)*x^2/(x^6-x^5-x+1). - Alois P. Heinz, Nov 29 2019

a(n) = a(n-1) + a(n-5) - a(n-6) for n>6. - Colin Barker, Nov 29 2019

Example

binomial(10, 3) = 120, which is divisible by 8. Thus 10 belongs to this sequence.

Mathematica

Select[Range[200], Mod[Floor[#(#-1)(#-2)/6], 8]==0&] (* Joshua Oliver, Nov 26 2019 *)
LinearRecurrence[{1, 0, 0, 0, 1, -1}, {0, 1, 2, 8, 10, 16}, 80] (* Harvey P. Dale, Jul 03 2022 *)

Prog

(Python)
for n in range(200):
    if (n*(n-1)*(n-2)//6)%8==0:
        print(n, end=' ')
(PARI) for(k=0, 194, my(j=binomial(k, 3)); if(!(j%8), print1(k, ", "))) \\ Hugo Pfoertner, Nov 29 2019
(PARI) concat(0, Vec(x^2*(1 + x + 6*x^2 + 2*x^3 + 6*x^4) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^70))) \\ Colin Barker, Nov 29 2019

Crossrefs

Cf. A042948.

Sequence in context: A161349 A336176 A375709 * A073886 A079930 A047467

Adjacent sequences: A329949 A329950 A329951 * A329953 A329954 A329955

Keyword

nonn,easy

Author

Sebastian Jeon and Tanya Khovanova, Nov 25 2019

Status

approved