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A329952
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Numbers k such that binomial(k,3) is divisible by 8.
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2
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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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).
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FORMULA
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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
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EXAMPLE
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binomial(10, 3) = 120, which is divisible by 8. Thus 10 belongs to this sequence.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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Cf. A042948.
Sequence in context: A213535 A161349 A336176 * A073886 A079930 A047467
Adjacent sequences: A329949 A329950 A329951 * A329953 A329954 A329955
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KEYWORD
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nonn,easy
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AUTHOR
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Sebastian Jeon and Tanya Khovanova, Nov 25 2019
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STATUS
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approved
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