A143447 -id:A143447 - cp's OEIS frontend

A143453 Square array A(n,k) of numbers of length n ternary words with at least k 0-digits between any other digits (n,k >= 0), read by antidiagonals.

1, 1, 3, 1, 3, 9, 1, 3, 5, 27, 1, 3, 5, 11, 81, 1, 3, 5, 7, 21, 243, 1, 3, 5, 7, 13, 43, 729, 1, 3, 5, 7, 9, 23, 85, 2187, 1, 3, 5, 7, 9, 15, 37, 171, 6561, 1, 3, 5, 7, 9, 11, 25, 63, 341, 19683, 1, 3, 5, 7, 9, 11, 17, 39, 109, 683, 59049, 1, 3, 5, 7, 9, 11, 13, 27, 57, 183, 1365, 177147

Offset: 0

Alois P. Heinz, Aug 16 2008

			A(3,1) = 11, because 11 ternary words of length 3 have at least 1 0-digit between any other digits: 000, 001, 002, 010, 020, 100, 101, 102, 200, 201, 202.
Square array A(n,k) begins:
     1,   1,  1,  1,  1,  1,  1,  1, ...
     3,   3,  3,  3,  3,  3,  3,  3, ...
     9,   5,  5,  5,  5,  5,  5,  5, ...
    27,  11,  7,  7,  7,  7,  7,  7, ...
    81,  21, 13,  9,  9,  9,  9,  9, ...
   243,  43, 23, 15, 11, 11, 11, 11, ...
   729,  85, 37, 25, 17, 13, 13, 13, ...
  2187, 171, 63, 39, 27, 19, 15, 15, ...

Alois P. Heinz, Rows n = 0..140, flattened

Column k=0: A000244, k=1: A001045(n+2), k=2: A003229(n+1) and A077949(n+2), k=3: A052942(n+3), k=4: A143447, k=5: A143448, k=6: A143449, k=7: A143450, k=8: A143451, k=9: A143452.

Diagonal: A005408.

A := proc (n::nonnegint, k::nonnegint) option remember; if k=0 then 3^n elif n<=k+1 then 2*n+1 else A(n-1, k) +2*A(n-k-1, k) fi end: seq(seq(A(n,d-n), n=0..d), d=0..14);

a[n_, 0] := 3^n; a[n_, k_] /; n <= k+1 := 2*n+1; a[n_, k_] := a[n, k] = a[n-1, k] + 2*a[n-k-1, k]; Table[a[n-k, k], {n, 0, 14}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Dec 11 2013 *)

G.f. of column k: 1/(x^k*(1-x-2*x^(k+1))).

A(n,k) = 3^n if k=0, else A(n,k) = 2*n+1 if n<=k+1, else A(n,k) = A(n-1,k) + 2*A(n-k-1,k).

A318777 Coefficients in expansion of 1/(1 - x - 2*x^5).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 5, 7, 9, 11, 17, 27, 41, 59, 81, 115, 169, 251, 369, 531, 761, 1099, 1601, 2339, 3401, 4923, 7121, 10323, 15001, 21803, 31649, 45891, 66537, 96539, 140145, 203443, 295225, 428299, 621377, 901667, 1308553, 1899003, 2755601, 3998355, 5801689, 8418795

Offset: 0

Zagros Lalo, Sep 25 2018

The coefficients in the expansion of 1/(1 - x - 2*x^5) are given by the sequence generated by the row sums in triangle A318775.

Coefficients in expansion of 1/(1 - x - 2*x^5) are given by the sum of numbers along "fourth Layer" skew diagonals pointing top-right in triangle A013609 ((1+2x)^n) and by the sum of numbers along "fourth Layer" skew diagonals pointing top-left in triangle A038207 ((2+x)^n), see links.

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3.

Zagros Lalo, Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 2x)^n.
Zagros Lalo, Fourth layer skew diagonals in center-justified triangle of coefficients in expansion of (2 + x)^n.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,2).

Cf. A318775, A318776.
Cf. A013609, A038207.
Essentially a duplicate of A143447.

a:=[1,1,1,1,1,3];; for n in [7..50] do a[n]:=a[n-1]+2*a[n-5]; od; a; # Muniru A Asiru, Sep 26 2018

seq(coeff(series((1-x-2*x^5)^(-1),x,n+1), x, n), n = 0 .. 50); # Muniru A Asiru, Sep 26 2018

a[0] = 1; a[n_] := a[n] = If[n < 0, 0, a[n - 1] + 2 * a[n - 5]];Table[a[n], {n, 0, 45}] // Flatten
LinearRecurrence[{1, 0, 0, 0, 2}, {1, 1, 1, 1, 1}, 46]
CoefficientList[Series[1/(1 - x - 2 x^5), {x, 0, 45}], x]

a(0)=1, a(n) = a(n-1) + 2 * a(n-5) for n = 0,1...; a(n)=0 for n < 0.

G.f.: -1/(2*x^5 + x - 1). - Chai Wah Wu, Aug 03 2020

A193517 T(n,k) = number of ways to place any number of 5X1 tiles of k distinguishable colors into an nX1 grid.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 4, 5, 4, 1, 1, 1, 1, 5, 7, 7, 5, 1, 1, 1, 1, 6, 9, 10, 9, 6, 1, 1, 1, 1, 7, 11, 13, 13, 11, 8, 1, 1, 1, 1, 8, 13, 16, 17, 16, 17, 11, 1, 1, 1, 1, 9, 15, 19, 21, 21, 28, 27, 15, 1, 1, 1, 1, 10, 17, 22, 25, 26, 41, 49, 41, 20, 1, 1, 1

Offset: 1

R. H. Hardin, with proof and formula from Robert Israel in the Sequence Fans Mailing List, Jul 29 2011

Table starts:
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..2..3...4...5...6...7...8....9...10...11...12...13...14...15...16...17...18
..3..5...7...9..11..13..15...17...19...21...23...25...27...29...31...33...35
..4..7..10..13..16..19..22...25...28...31...34...37...40...43...46...49...52
..5..9..13..17..21..25..29...33...37...41...45...49...53...57...61...65...69
..6.11..16..21..26..31..36...41...46...51...56...61...66...71...76...81...86
..8.17..28..41..56..73..92..113..136..161..188..217..248..281..316..353..392
.11.27..49..77.111.151.197..249..307..371..441..517..599..687..781..881..987
.15.41..79.129.191.265.351..449..559..681..815..961.1119.1289.1471.1665.1871
.20.59.118.197.296.415.554..713..892.1091.1310.1549.1808.2087.2386.2705.3044
.26.81.166.281.426.601.806.1041.1306.1601.1926.2281.2666.3081.3526.4001.4506

			Some solutions for n=11 k=3; colors=1, 2, 3; empty=0
..0....2....2....0....0....1....0....3....3....0....0....0....0....3....1....0
..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....1....0....2....1....0....3....3....0....3....2....3....1....0....0....1
..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....0....3....0....1....1....2....0....2....0....0....3....0....3....0....2

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A003520,
Column 2 is A143447(n-4),
Column 3 is A143455(n-4),
Row 10 is A028884.

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<5 or k=0, 1, k*T(n-5, k) +T(n-1, k)))
    end:
seq(seq(T(n, d+1-n), n=1..d), d=1..13); # Alois P. Heinz, Jul 29 2011

T[n_, k_] := T[n, k] = If[n<0, 0, If[n < 5 || k == 0, 1, k*T[n-5, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 14}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. So T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n=0,1,...,z-1. The solution is T(n,k) = sum_r r^(-n-1)/(1 + z k r^(z-1)) where the sum is over the roots of the polynomial k x^z + x - 1.
T(n,k) = sum {s=0..[n/5]} (binomial(n-4*s,s)*k^s).
For z X 1 tiles, T(n,k,z) = sum{s=0..[n/z]} (binomial(n-(z-1)*s,s)*k^s). - R. H. Hardin, Jul 31 2011

cp's OEIS Frontend

A143453 Square array A(n,k) of numbers of length n ternary words with at least k 0-digits between any other digits (n,k >= 0), read by antidiagonals.

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A318777 Coefficients in expansion of 1/(1 - x - 2*x^5).

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A193517 T(n,k) = number of ways to place any number of 5X1 tiles of k distinguishable colors into an nX1 grid.

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