A143454 -id:A143454 - 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).

A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

Original entry on oeis.org

1, 1, 1, 1, 4, 7, 10, 13, 25, 46, 76, 115, 190, 328, 556, 901, 1471, 2455, 4123, 6826, 11239, 18604, 30973, 51451, 85168, 140980, 233899, 388252, 643756, 1066696, 1768393, 2933149, 4864417, 8064505, 13369684, 22169131, 36762382, 60955897, 101064949, 167572342, 277859488, 460727179, 763922026, 1266639052

Offset: 0

Zagros Lalo, Sep 04 2018

The coefficients in the expansion of 1/(1 - x - 3*x^4) are given by the sequence generated by the row sums in triangle A318772.

Coefficients in expansion of 1/(1 - x - 3*x^4) are given by the sum of numbers along "third Layer" skew diagonals pointing top-right in triangle A013610 ((1+3x)^n) and by the sum of numbers along "third Layer" skew diagonals pointing top-left in triangle A027465 ((3+x)^n), see links.

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

G. C. Greubel, Table of n, a(n) for n = 0..1000
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (1 + 3 x)^n
Zagros Lalo, Third layer skew diagonals in center-justified triangle of coefficients in expansion of (3 + x)^n
Index entries for linear recurrences with constant coefficients, signature (1,0,0,3).

Cf. A013610, A027465, A318772, A318773.

Essentially a duplicate of A143454.

[n le 4 select 1 else Self(n-1) +3*Self(n-4): n in [1..51]]; // G. C. Greubel, May 08 2021

CoefficientList[Series[1/(1-x-3x^4), {x, 0, 50}], x]
a[n_]:= a[n]= If[n<4, 1, a[n-1] + 3*a[n-4]]; Table[a[n], {n,0,50}]
LinearRecurrence[{1,0,0,3}, {1,1,1,1}, 51]

my(p=Mod('x,x^4-'x^3-3)); a(n) = vecsum(Vec(lift(p^n))); \\ Kevin Ryde, May 11 2021

def a(n): return 1 if (n<4) else a(n-1) + 3*a(n-4)
[a(n) for n in (0..50)] # G. C. Greubel, May 08 2021

a(n) = a(n-1) + 3*a(n-4) for n >= 0, a(n)=0 for n < 0, with a(0) = a(1) = a(2) = a(3) = 1.

A172369 Triangle read by rows: T(n,k,q) = round(c(n)/(c(k)*c(n-k))) where c are partial products of a sequence defined in comments.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 4, 4, 4, 1, 1, 7, 28, 28, 28, 7, 1, 1, 10, 70, 280, 280, 70, 10, 1, 1, 13, 130, 910, 3640, 910, 130, 13, 1, 1, 25, 325, 3250, 22750, 22750, 3250, 325, 25, 1, 1, 46, 1150, 14950, 149500, 261625, 149500, 14950, 1150, 46, 1

Offset: 0

Roger L. Bagula, Feb 01 2010

Start from A143454 and its partial products c(n) = 1, 1, 1, 1, 1, 4, 28, 280, 3640, 91000, 4186000, ... . Then T(n,k) = round(c(n)/(c(k)*c(n-k))).

			Triangle begins as:
  1;
  1,  1;
  1,  1,    1;
  1,  1,    1,     1;
  1,  1,    1,     1,      1;
  1,  4,    4,     4,      4,      1;
  1,  7,   28,    28,     28,      7,      1;
  1, 10,   70,   280,    280,     70,     10,     1;
  1, 13,  130,   910,   3640,    910,    130,    13,    1;
  1, 25,  325,  3250,  22750,  22750,   3250,   325,   25,  1;
  1, 46, 1150, 14950, 149500, 261625, 149500, 14950, 1150, 46, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A172363 (q=1), A172368 (q=2), this sequence (q=3), A318774.

f[n_, q_]:= f[n, q]= If[n==0, 0, If[n<4, 1, f[n-1, q] +q*f[n-4, q]]];
c[n_, q_]:= Product[f[j, q], {j,n}];
T[n_, k_, q_]:= Round[c[n, q]/(c[k, q]*c[n - k, q])];
Table[T[n, k, 3], {n, 0, 12}, {k, 0, n}]//Flatten (* modified by G. C. Greubel, May 08 2021 *)

@CachedFunction
def f(n,q): return 0 if (n==0) else 1 if (n<4) else f(n-1, q) + q*f(n-4, q)
def c(n,q): return product( f(j,q) for j in (1..n) )
def T(n,k,q): return round(c(n, q)/(c(k, q)*c(n-k, q)))
flatten([[T(n,k,3) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, May 08 2021

T(n, k, q) = round( c(n,q)/(c(k,q)*c(n-k,q)) ), where c(n, q) = Product_{j=1..n} f(j, q), f(n, q) = f(n-1, q) + q*f(n-4, q), f(0, q) = 0, f(1, q) = f(2, q) = f(3, q) = 1, and q = 3. - G. C. Greubel, May 08 2021

Definition corrected to give integral terms, G. C. Greubel, May 08 2021

A193516 T(n,k) = number of ways to place any number of 4X1 tiles of k distinguishable colors into an nX1 grid.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 5, 4, 1, 1, 1, 5, 7, 7, 5, 1, 1, 1, 6, 9, 10, 9, 7, 1, 1, 1, 7, 11, 13, 13, 15, 10, 1, 1, 1, 8, 13, 16, 17, 25, 25, 14, 1, 1, 1, 9, 15, 19, 21, 37, 46, 39, 19, 1, 1, 1, 10, 17, 22, 25, 51, 73, 76, 57, 26, 1, 1, 1, 11, 19, 25, 29, 67, 106, 125

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
..2...3...4...5....6....7....8....9...10...11...12....13....14....15....16
..3...5...7...9...11...13...15...17...19...21...23....25....27....29....31
..4...7..10..13...16...19...22...25...28...31...34....37....40....43....46
..5...9..13..17...21...25...29...33...37...41...45....49....53....57....61
..7..15..25..37...51...67...85..105..127..151..177...205...235...267...301
.10..25..46..73..106..145..190..241..298..361..430...505...586...673...766
.14..39..76.125..186..259..344..441..550..671..804...949..1106..1275..1456
.19..57.115.193..291..409..547..705..883.1081.1299..1537..1795..2073..2371
.26..87.190.341..546..811.1142.1545.2026.2591.3246..3997..4850..5811..6886
.36.137.328.633.1076.1681.2472.3473.4708.6201.7976.10057.12468.15233.18376

			Some solutions for n=9 k=3; colors=1, 2, 3; empty=0
..0....3....0....0....3....3....0....0....0....0....2....2....0....0....1....2
..1....3....0....2....3....3....3....0....0....0....2....2....1....0....1....2
..1....3....0....2....3....3....3....2....0....0....2....2....1....0....1....2
..1....3....3....2....3....3....3....2....1....0....2....2....1....0....1....2
..1....0....3....2....0....3....3....2....1....0....2....0....1....0....0....0
..2....3....3....2....0....3....3....2....1....3....2....2....0....0....0....3
..2....3....3....2....0....3....3....0....1....3....2....2....0....0....0....3
..2....3....0....2....0....3....3....0....0....3....2....2....0....0....0....3
..2....3....0....2....0....0....3....0....0....3....0....2....0....0....0....3

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

Column 1 is A003269(n+1),
Column 2 is A052942,
Column 3 is A143454(n-3),
Row 8 is A082111,
Row 9 is A100536(n+1),
Row 10 is A051866(n+1).

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<4 or k=0, 1, k*T(n-4, 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 < 4 || k == 0, 1, k*T[n-4, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 13}] // 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/4]} (binomial(n-3*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.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A318774 Coefficients in expansion of 1/(1 - x - 3*x^4).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Sage

Formula

A172369 Triangle read by rows: T(n,k,q) = round(c(n)/(c(k)*c(n-k))) where c are partial products of a sequence defined in comments.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Sage

Formula

Extensions

A193516 T(n,k) = number of ways to place any number of 4X1 tiles of k distinguishable colors into an nX1 grid.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula