A143455 -id:A143455 - cp's OEIS frontend

A143461 Square array A(n,k) of numbers of length n quaternary words with at least k 0-digits between any other digits (n,k >= 0), read by antidiagonals.

1, 1, 4, 1, 4, 16, 1, 4, 7, 64, 1, 4, 7, 19, 256, 1, 4, 7, 10, 40, 1024, 1, 4, 7, 10, 22, 97, 4096, 1, 4, 7, 10, 13, 43, 217, 16384, 1, 4, 7, 10, 13, 25, 73, 508, 65536, 1, 4, 7, 10, 13, 16, 46, 139, 1159, 262144, 1, 4, 7, 10, 13, 16, 28, 76, 268, 2683, 1048576, 1, 4, 7, 10, 13, 16, 19, 49, 115, 487, 6160, 4194304

Offset: 0

Alois P. Heinz, Aug 16 2008

			A (3,1) = 19, because 19 quaternary words of length 3 have at least 1 0-digit between any other digits: 000, 001, 002, 003, 010, 020, 030, 100, 101, 102, 103, 200, 201, 202, 203, 300, 301, 301, 303.
Square array A(n,k) begins:
       1,   1,   1,  1,  1,  1,  1,  1,  ...
       4,   4,   4,  4,  4,  4,  4,  4,  ...
      16,   7,   7,  7,  7,  7,  7,  7,  ...
      64,  19,  10, 10, 10, 10, 10, 10,  ...
     256,  40,  22, 13, 13, 13, 13, 13,  ...
    1024,  97,  43, 25, 16, 16, 16, 16,  ...
    4096, 217,  73, 46, 28, 19, 19, 19,  ...
   16384, 508, 139, 76, 49, 31, 22, 22,  ...

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

Columns k=0-9 give: A000302, A006130(n+1), A084386(n+2), A143454, A143455, A143456, A143457, A143458, A143459, A143460.

Main diagonal gives A016777.

A:= proc(n, k) option remember; if k=0 then 4^n elif n<=k+1 then 3*n+1 else A(n-1, k) +3*A(n-k-1, k) fi end: seq(seq(A(n, d-n), n=0..d), d=0..13);

a[n_, 0] := 4^n; a[n_, k_] /; n <= k+1 := 3*n+1; a[n_, k_] := a[n, k] = a[n-1, k] + 3*a[n-k-1, k]; Table[a[n-k, k], {n, 0, 13}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jan 15 2014, after Maple *)

G.f. of column k: 1/(x^k*(1-x-3*x^(k+1))).

A(n,k) = 4^n if k=0, else A(n,k) = 3*n+1 if n<=k+1, else A(n,k) = A(n-1,k) + 3*A(n-k-1,k).

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

A143461 Square array A(n,k) of numbers of length n quaternary words with at least k 0-digits between any other digits (n,k >= 0), read by antidiagonals.

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