A120906 -id:A120906 - cp's OEIS frontend

A120908 Sum of the lengths of the drops in all ternary words of length n on {0,1,2}. The drops of a ternary word on {0,1,2} are the subwords 10,20 and 21, their lengths being the differences 1, 2 and 1, respectively.

0, 4, 24, 108, 432, 1620, 5832, 20412, 69984, 236196, 787320, 2598156, 8503056, 27634932, 89282088, 286978140, 918330048, 2927177028, 9298091736, 29443957164, 92980917360, 292889889684, 920511081864, 2887057484028

Offset: 1

Emeric Deutsch, Jul 15 2006

a(n) = 4*A027471(n).

a(n) = Sum_{k>=0} k*A120907(n,k).

			a(2)=4 because the ternary words 00,01,02,11,12 and 22 have no drops, each of the words 10 and 21 has one drop of length 1 and the word 20 has one drop of length 2.

Vincenzo Librandi, Table of n, a(n) for n = 1..400
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Index entries for linear recurrences with constant coefficients, signature (6,-9).

Cf. A027471, A120906, A120907.

[4*(n-1)*3^(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 09 2011

Maple
```
seq(4*(n-1)*3^(n-2),n=1..27);
```

Table[4*(n-1)*3^(n-2), {n, 30}] (* Wesley Ivan Hurt, Jan 28 2014 *)
LinearRecurrence[{6,-9},{0,4},30] (* Harvey P. Dale, Jul 14 2023 *)

a(n) = 4*(n-1)*3^(n-2); \\ Altug Alkan, May 16 2018

a(n) = 4*(n-1)*3^(n-2).

G.f.: 4*z^2/(1-3*z)^2.

A120907 Triangle read by rows: T(n,k) is the number of ternary words of length n on {0,1,2} having sum of the lengths of the drops equal to k (n>=0, k>=0). The drops of a ternary word on {0,1,2} are the subwords 10,20 and 21, their lengths being the differences 1, 2 and 1, respectively.

Original entry on oeis.org

1, 3, 6, 2, 1, 10, 10, 7, 15, 30, 31, 4, 1, 21, 70, 105, 36, 11, 28, 140, 294, 184, 76, 6, 1, 36, 252, 714, 696, 396, 78, 15, 45, 420, 1554, 2160, 1666, 566, 141, 8, 1, 55, 660, 3102, 5808, 5918, 2990, 995, 136, 19, 66, 990, 5775, 13992, 18348, 12746, 5615, 1280, 226

Offset: 0

Emeric Deutsch, Jul 15 2006

Row n has 2*floor(n/2)+1 terms (i.e. each of the rows 2n and 2n+1 has 2n+1 terms). Row sums are the powers of 3 (A000244). T(n,0)=A000217(n+1) (the triangular numbers). Sum(k*T(n,k),k>=0)=4(n-1)3^(n-2)=A120908(n)=4*A027471(n).

			T(4,3)=4 because we have 1020,2010,2021 and 2120.
Triangle starts:
1;
3;
6,2,1;
10,10,7;
15,30,31,4,1;
21,70,105,36,11;

Cf. A000244, A000217, A120908, A027471, A120906.

G:=1/(1-z+t*z)/(1-2*z+z^2-t*z-t*z^2): Gser:=simplify(series(G,z=0,15)): P[0]:=1: for n from 1 to 12 do P[n]:=sort(coeff(Gser,z^n)) od: for n from 0 to 12 do seq(coeff(P[n],t,j),j=0..2*floor(n/2)) od; # yields sequence in triangular form

G.f.=G(t,z)=1/[(1-z+tz)(1-2z+z^2-tz-tz^2)].

cp's OEIS Frontend

A120908 Sum of the lengths of the drops in all ternary words of length n on {0,1,2}. The drops of a ternary word on {0,1,2} are the subwords 10,20 and 21, their lengths being the differences 1, 2 and 1, respectively.

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