A320919 Positive integers k such that binomial(k, 3) is divisible by 6.

1, 2, 9, 10, 18, 20, 28, 29, 36, 37, 38, 45, 46, 54, 56, 64, 65, 72, 73, 74, 81, 82, 90, 92, 100, 101, 108, 109, 110, 117, 118, 126, 128, 136, 137, 144, 145, 146, 153, 154, 162, 164, 172, 173, 180, 181, 182, 189, 190, 198, 200

Offset: 1

Author

Tanya Khovanova, Oct 24 2018

Keywords

• nonn
• easy

Comments

When taken modulo 36 this sequence is periodic with period is 9.

These are numbers for which a 3-symmetric permutation of size n might exist.

Numbers for which a 2-symmetric permutations might exist are numbers n such that n choose 2 is even. Equivalently, these are numbers that have remainder 0 or 1 modulo 4. This is sequence A042948.

Examples

			For k=8, binomial(8,3) = 56, which is not divisible by 6. Therefore 8 is not in the sequence.
For k=9, binomial(9,3) = 84, which is divisible by 6, so 9 is a term of the sequence.
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Tanya Khovanova, 3-Symmetric Permutations
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,1,-1).

Crossrefs

Cf. A042948, A316775.

Programs

GAP

Filtered([1..200],k->Binomial(k,3) mod 6 = 0); # Muniru A Asiru, Oct 24 2018

Maple

select(k->modp(binomial(k,3),6)=0,[$1..200]); # Muniru A Asiru, Oct 24 2018

Mathematica

Transpose[Select[Table[{n, IntegerQ[Binomial[n, 3]/3!]}, {n, 200}], #[[2]] == True &]][[1]]

PARI

select(n->binomial(n, 3)%6 == 0, vector(100, n, n)) \\ Colin Barker, Oct 24 2018

PARI

Vec(x*(1 + x + 7*x^2 + x^3 + 8*x^4 + 2*x^5 + 8*x^6 + x^7 + 7*x^8) / ((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)) + O(x^40)) \\ Colin Barker, Oct 24 2018

Formula

From Colin Barker, Oct 24 2018: (Start)

G.f.: x*(1 + x + 7*x^2 + x^3 + 8*x^4 + 2*x^5 + 8*x^6 + x^7 + 7*x^8) / ((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)).

a(n) = a(n-1) + a(n-9) - a(n-10) for n>10.

(End)