A391470 Number of words of length 3*n that can be formed with a three-letter alphabet when the number of letters of each type is == 2 (mod 3).

0, 0, 90, 2268, 58806, 1592136, 43053282, 1162320516, 31380882462, 847287015120, 22876797237930, 617673439330668, 16677181570526406, 450283904728735896, 12157665462543713202, 328256967425918137236, 8862938119558357917102, 239299329229770240980640

Offset

0,3

Comments

See comments in A391468.

Links

Pablo Serra, Table of n, a(n) for n = 0..100

Pablo Serra, On the number of words of N = 3*M letters with a three-letter alphabet, arXiv:2512.22362 [math.CO], 2025.

Index entries for linear recurrences with constant coefficients, signature (27,-27,729).

Formula

a(n) = Sum_{n1+n2+n3=n-2} trinomial(3*n; 3*n1 + 2, 3*n2 + 2, 3*n3 + 2) for n>=2.

a(n) = 3^(3*n-2) - ((1 + (-1)^n) - (1 - (-1)^n)*i*sqrt(3))*i^n*3^((3*n-2)/2)/2 for n>=1, where i is the imaginary unit.

a(n) = 27*(a(n-1) - a(n-2) + 27*a(n-3)), for n >= 4.

G.f.: 18*x^2*(5 - 9*x)/((1 - 27*x)*(1 + 27*x^2)).

a(n) + A391468(n) + A391469(n) + 2*A013733(n-1) = 3^(3*n) for n>=1.

E.g.f.: (2 + exp(27*x) - 3*cos(3*sqrt(3)*x) - 3*sqrt(3)*sin(3*sqrt(3)*x))/9. - Stefano Spezia, Dec 30 2025

Example

For n=2, the words are permutations of A,A,B,B,C,C and there are a(0) = trinomial(6;2,2,2) = 6!/(2!*2!*2!) = 90 of them.

Mathematica

LinearRecurrence[{27, -27, 729}, {0, 0, 90, 2268}, 20] (* Paolo Xausa, Jan 16 2026 *)

Crossrefs

Cf. A391468, A391469, A013733.

Sequence in context: A257040 A008393 A055603 * A210090 A109124 A013359

Adjacent sequences: A391467 A391468 A391469 * A391471 A391472 A391473

Keyword

nonn,easy

Author

Pablo Serra, Dec 29 2025

Status

approved