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%I #21 Jul 03 2022 16:16:28
%S 0,1,2,8,10,16,17,18,24,26,32,33,34,40,42,48,49,50,56,58,64,65,66,72,
%T 74,80,81,82,88,90,96,97,98,104,106,112,113,114,120,122,128,129,130,
%U 136,138,144,145,146,152,154,160,161,162,168,170,176,177,178,184,186,192,193,194
%N Numbers k such that binomial(k,3) is divisible by 8.
%C These are possible sizes for 3-symmetric graphs.
%C The possible size of 2-symmetric graphs is sequence A042948.
%C These numbers are 0, 1, 2, 8, and 10 modulo 16.
%H Colin Barker, <a href="/A329952/b329952.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,1,-1).
%F 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
%F a(n) = a(n-1) + a(n-5) - a(n-6) for n>6. - _Colin Barker_, Nov 29 2019
%e binomial(10, 3) = 120, which is divisible by 8. Thus 10 belongs to this sequence.
%t Select[Range[200],Mod[Floor[#(#-1)(#-2)/6],8]==0&] (* _Joshua Oliver_, Nov 26 2019 *)
%t LinearRecurrence[{1,0,0,0,1,-1},{0,1,2,8,10,16},80] (* _Harvey P. Dale_, Jul 03 2022 *)
%o (Python)
%o for n in range(200):
%o if (n*(n-1)*(n-2)//6)%8==0:
%o print(n, end=' ')
%o (PARI) for(k=0,194,my(j=binomial(k,3));if(!(j%8),print1(k,", "))) \\ _Hugo Pfoertner_, Nov 29 2019
%o (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
%Y Cf. A042948.
%K nonn,easy
%O 1,3
%A Sebastian Jeon and _Tanya Khovanova_, Nov 25 2019
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