A385252 Number of ternary strings of length 2*n that have at least one 0 but less 0's than the combined number of 1's and 2's.

0, 0, 32, 432, 4608, 45440, 432896, 4051712, 37535744, 345470976, 3165315072, 28905857024, 263303921664, 2393675661312, 21725991600128, 196937443377152, 1783243502256128, 16132632204541952, 145839502212988928, 1317564268289196032, 11896995094093365248, 107375812426273390592

Offset

0,3

Links

Table of n, a(n) for n=0..21.

Formula

a(n) = 9^n - 4^n - Sum_{k=0..n} 2^(n-k)*C(2*n,n-k) for n > 0.

G.f.: (5*x*(sqrt(1-8*x))*(sqrt(1-8*x)+12*x-1)-8*x*(36*x^2-13*x+1))/(sqrt(1-8*x)*(sqrt(1-8*x)+12*x-1)*(36*x^2-13*x+1)) + 1.

a(n) = A001019(n) - A000302(n) - A128418(n), n > 0.

Conjecture D-finite with recurrence n*a(n) +(-37*n+36)*a(n-1) +4*(131*n-245)*a(n-2) +16*(-221*n+605)*a(n-3) +192*(59*n-213)*a(n-4) +6912*(-2*n+9)*a(n-5)=0. - R. J. Mathar, Jul 31 2025

a(n) = 9^n - 4^n - 2^n*binomial(2*n, n)*hypergeom([1, -n], [1+n], -1/2) for n > 0. - Stefano Spezia, Aug 05 2025

Example

a(2)=32 since the strings of length 4 are (number of permutations in parentheses): 1110 (4), 1120 (12), 1220 (12), 2220 (4).
a(3)=432 since the strings of length 6 are (number of permutations in parentheses): 111110 (6), 111120 (30), 111220 (60), 112220 (60), 122220 (30), 222220 (6), 001111 (15), 001112 (60), 001122 (90), 001222 (60), 002222 (15).

Mathematica

a[0]=0; a[n_]:=9^n - 4^n - Sum[2^(n-k)*Binomial[2n, n-k], {k, 0, n}]; Array[a, 22, 0] (* Stefano Spezia, Jul 31 2025 *)

Crossrefs

Cf. A000302, A001019, A128418, A386670.

Sequence in context: A145217 A125444 A022692 * A160142 A159544 A135217

Adjacent sequences: A385249 A385250 A385251 * A385253 A385254 A385255

Keyword

nonn

Author

Enrique Navarrete, Jul 28 2025

Status

approved