A218473 Number of 3n-length 3-ary words, either empty or beginning with the first letter of the alphabet, that can be built by repeatedly inserting triples of identical letters into the initially empty word.
1, 1, 7, 61, 591, 6101, 65719, 729933, 8297247, 96044101, 1128138567, 13411861629, 161066465583, 1950996039669, 23808159962839, 292413627476141, 3611870017079871, 44838216520062117, 559127724970143079, 7000374603097246173, 87964883375131331151
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..300
Programs
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Maple
a:= n-> `if`(n=0, 1, add(binomial(3*n, j)*(n-j)*2^j, j=0..n-1)/n): seq(a(n), n=0..20); # second Maple program a:= proc(n) a(n):= `if`(n<3, [1, 1, 7][n+1], (-81*(3*n-1)*(3*n-5)*a(n-2) +(81*n^2-81*n+15)*a(n-1))/ ((2*n-1)*n)) end: seq(a(n), n=0..20); -
Mathematica
Flatten[{1,Table[1/n*Sum[Binomial[3*n,j]*(n-j)*2^j,{j,0,n-1}],{n,1,20}]}] (* Vaclav Kotesovec, May 22 2013 *) Flatten[{1,Table[FullSimplify[SeriesCoefficient[(1/(81*x-3)+2/((3-81*x)*(1-27*x-3*Sqrt[3*x*(27*x-2)])^(2/3))),{x,0,n}]],{n,1,10}]}] (* Vaclav Kotesovec, Jul 06 2013 *)
Formula
a(n) = (1/n) * Sum_{j=0..n-1} binomial(3*n,j)*(n-j)*2^j for n>0, a(0) = 1.
a(n) ~ 3^(3*n-3/2)/(sqrt(Pi)*2^(n-1)*n^(3/2)). - Vaclav Kotesovec, May 22 2013
G.f. (for n>0): (1/(81*x-3)+2/((3-81*x)*(1-27*x-3*sqrt(3*x*(27*x-2)))^(2/3))). - Vaclav Kotesovec, Jul 06 2013
From Peter Bala, Feb 06 2022: (Start)
The o.g.f. A(x) satisfies the algebraic equation 8*x - 36*x*A(x) + (54*x - 1)*A(x)^2 + (-27*x + 1)*A(x)^3 = 0.
A(x) = (6 - 4*T(2*x))/(2*T(2*x)^2 - 9*T(2*x) + 9), where T(x) = 1 + x*T(x)^3 is the o.g.f. of A001764.
A(x) = 1 + 2*x*B'(2*x)/B(2*x), where B(x) = 2 + x + 2*x^2 + 6*x^3 + 22*x^4 + 91*x^5 + ... is the o.g.f. of A000139.
exp(Sum_{n >= 1} a(n)*x*n/n) = 1 + x + 4*x^2 + 24*x^3 + 176*x^4 + 1456*x^5 + ... is the o.g.f. of A000309, a power series with integral coefficients. It follows that the Gauss congruences a(n*p^k) == a(n*p*(k-1)) (mod p^k) hold for all prime p and positive integers n and k. (End)