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A329952
Numbers k such that binomial(k,3) is divisible by 8.
2
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.
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
KEYWORD
nonn,easy
AUTHOR
Sebastian Jeon and Tanya Khovanova, Nov 25 2019
STATUS
approved