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A320919 Positive integers k such that binomial(k, 3) is divisible by 6. 2
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
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.
LINKS
Tanya Khovanova, 3-Symmetric Permutations
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)
EXAMPLE
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.
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]]
PROG
(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
(GAP) Filtered([1..200], k->Binomial(k, 3) mod 6 = 0); # Muniru A Asiru, Oct 24 2018
CROSSREFS
Sequence in context: A047468 A032929 A226832 * A046975 A304441 A005695
KEYWORD
nonn,easy
AUTHOR
Tanya Khovanova, Oct 24 2018
STATUS
approved

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Last modified April 23 16:28 EDT 2024. Contains 371916 sequences. (Running on oeis4.)