A178205 -id:A178205 - cp's OEIS frontend

A193515 T(n,k) = number of ways to place any number of 3X1 tiles of k distinguishable colors into an nX1 grid.

1, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 7, 6, 1, 1, 6, 9, 10, 13, 9, 1, 1, 7, 11, 13, 22, 23, 13, 1, 1, 8, 13, 16, 33, 43, 37, 19, 1, 1, 9, 15, 19, 46, 69, 73, 63, 28, 1, 1, 10, 17, 22, 61, 101, 121, 139, 109, 41, 1, 1, 11, 19, 25, 78, 139, 181, 253, 268, 183, 60, 1, 1, 12

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
..2...3...4...5...6....7....8....9...10...11...12...13....14....15....16....17
..3...5...7...9..11...13...15...17...19...21...23...25....27....29....31....33
..4...7..10..13..16...19...22...25...28...31...34...37....40....43....46....49
..6..13..22..33..46...61...78...97..118..141..166..193...222...253...286...321
..9..23..43..69.101..139..183..233..289..351..419..493...573...659...751...849
.13..37..73.121.181..253..337..433..541..661..793..937..1093..1261..1441..1633
.19..63.139.253.411..619..883.1209.1603.2071.2619.3253..3979..4803..5731..6769
.28.109.268.529.916.1453.2164.3073.4204.5581.7228.9169.11428.14029.16996.20353

			Some solutions for n=7 k=3; colors=1,2,3 and empty=0
..3....0....0....2....0....1....3....0....0....0....1....0....3....1....0....0
..3....0....0....2....2....1....3....2....1....0....1....3....3....1....0....0
..3....1....0....2....2....1....3....2....1....2....1....3....3....1....0....3
..1....1....3....0....2....0....0....2....1....2....3....3....0....2....0....3
..1....1....3....0....0....2....2....2....1....2....3....2....1....2....1....3
..1....0....3....0....0....2....2....2....1....0....3....2....1....2....1....0
..0....0....0....0....0....2....2....2....1....0....0....2....1....0....1....0

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

Column 1 is A000930,
Column 2 is A003229(n-1),
Column 3 is A084386,
Column 4 is A089977,
Column 10 is A178205,
Row 6 is A028872(n+2),
Row 7 is A144390(n+1),
Row 8 is A003154(n+1).

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<3 or k=0, 1, k*T(n-3, 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

nmax = 13; t[?Negative, ] = 0; t[n_, k_] /; (n < 3 || k == 0) = 1; t[n_, k_] := t[n, k] = k*t[n-3, k] + t[n-1, k]; Flatten[ Table[ t[n-k+1, k], {n , 1, nmax}, {k, n, 1, -1}]](* Jean-François Alcover, Nov 28 2011, after Maple *)

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/3]} (binomial(n-2*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

A242763 a(n) = 1 for n <= 7; a(n) = a(n-5) + a(n-7) for n>7.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 7, 7, 8, 9, 9, 12, 12, 15, 16, 17, 21, 21, 27, 28, 32, 37, 38, 48, 49, 59, 65, 70, 85, 87, 107, 114, 129, 150, 157, 192, 201, 236, 264, 286, 342, 358, 428, 465, 522, 606, 644, 770, 823, 950, 1071, 1166, 1376

Offset: 1

Vicente Jesús Maniega Granado, Oct 03 2016

Generalized Fibonacci growth sequence using i = 2 as maturity period, j = 5 as conception period, and k = 2 as growth factor.

Maturity period is the number of periods that a Fibonacci tree node needs for being able to start developing branches. Conception period is the number of periods in a Fibonacci tree node needed to develop new branches since its maturity. Growth factor is the number of additional branches developed by a Fibonacci tree node, plus 1, and equals the base of the exponential series related to the given tree if maturity factor would be zero. Standard Fibonacci would use 1 as maturity period, 1 as conception period, and 2 as growth factor as the series becomes equal to 2^n with a maturity period of 0. Related to Lucas sequences.

			For n = 13 the a(13) = a(8) + a(6) = 2 + 1 = 3.

Colin Barker, Table of n, a(n) for n = 1..1000
Julia Collins, Fibonacci Tree
Fractal Foundation, Fibonacci Fractals
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1,0,1).

Cf. A000079 (i = 0, j = 1, k = 2), A000244 (i = 0, j = 1, k = 3), A000302 (i = 0, j = 1, k = 4), A000351 (i = 0, j = 1, k = 5), A000400 (i = 0, j = 1, k = 6), A000420 (i = 0, j = 1, k = 7), A001018 (i = 0, j = 1, k = 8), A001019 (i = 0, j = 1, k = 9), A011557 (i = 0, j = 1, k = 10), A001020 (i = 0, j = 1, k = 11), A001021 (i = 0, j = 1, k = 12), A016116 (i = 0, j = 2, k = 2), A108411 (i = 0, j = 2, k = 3), A213173 (i = 0, j = 2, k = 4), A074872 (i = 0, j = 2, k = 5), A173862 (i = 0, j = 3, k = 2), A127975 (i = 0, j = 3, k = 3), A200675 (i = 0, j = 4, k = 2), A111575 (i = 0, j = 4, k = 3), A000045 (i = 1, j = 1, k = 2), A001045 (i = 1, j = 1, k = 3), A006130 (i = 1, j = 1, k = 4), A006131 (i = 1, j = 1, k = 5), A015440 (i = 1, j = 1, k = 6), A015441 (i = 1, j = 1, k = 7), A015442 (i = 1, j = 1, k = 8), A015443 (i = 1, j = 1, k = 9), A015445 (i = 1, j = 1, k = 10), A015446 (i = 1, j = 1, k = 11), A015447 (i = 1, j = 1, k = 12), A000931 (i = 1, j = 2, k = 2), A159284 (i = 1, j = 2, k = 3), A238389 (i = 1, j = 2, k = 4), A097041 (i = 1, j = 2, k = 10), A079398 (i = 1, j = 3, k = 2), A103372 (i = 1, j = 4, k = 2), A103373 (i = 1, j = 5, k = 2), A103374 (i = 1, j = 6, k = 2), A000930 (i = 2, j = 1, k = 2), A077949 (i = 2, j = 1, k = 3), A084386 (i = 2, j = 1, k = 4), A089977 (i = 2, j = 1, k = 5), A178205 (i = 2, j = 1, k = 11), A103609 (i = 2, j = 2, k = 2), A077953 (i = 2, j = 2, k = 3), A226503 (i = 2, j = 3, k = 2), A122521 (i = 2, j = 6, k = 2), A003269 (i = 3, j = 1, k = 2), A052942 (i = 3, j = 1, k = 3), A005686 (i = 3, j = 2, k = 2), A237714 (i = 3, j = 2, k = 3), A238391 (i = 3, j = 2, k = 4), A247049 (i = 3, j = 3, k = 2), A077886 (i = 3, j = 3, k = 3), A003520 (i = 4, j = 1, k = 2), A108104 (i = 4, j = 2, k = 2), A005708 (i = 5, j = 1, k = 2), A237716 (i = 5, j = 2, k = 3), A005709 (i = 6, j = 1, k = 2), A122522 (i = 6, j = 2, k = 2), A005710 (i = 7, j = 1, k = 2), A237718 (i = 7, j = 2, k = 3), A017903 (i = 8, j = 1, k = 2).

[n le 7 select 1 else Self(n-5)+Self(n-7): n in [1..70]]; // Vincenzo Librandi, Nov 30 2016

LinearRecurrence[{0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 1, 1, 1}, 70] (*  or *)
CoefficientList[ Series[(1+x+x^2+x^3+x^4)/(1-x^5-x^7), {x, 0, 70}], x] (* Robert G. Wilson v, Nov 25 2016 *)
nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,a+c}; NestList[nxt,{1,1,1,1,1,1,1},70][[;;,1]] (* Harvey P. Dale, Oct 22 2024 *)

Vec(x*(1+x+x^2+x^3+x^4)/((1-x+x^2)*(1+x-x^3-x^4-x^5)) + O(x^100)) \\ Colin Barker, Oct 27 2016

@CachedFunction # a = A242763
def a(n): return 1 if n<8 else a(n-5) +a(n-7)
[a(n) for n in range(1,76)] # G. C. Greubel, Oct 23 2024

Generic a(n) = 1 for n <= i+j; a(n) = a(n-j) + (k-1)*a(n-(i+j)) for n>i+j where i = maturity period, j = conception period, k = growth factor.
G.f.: x*(1+x+x^2+x^3+x^4) / ((1-x+x^2)*(1+x-x^3-x^4-x^5)). - Colin Barker, Oct 09 2016
Generic g.f.: x*(Sum_{l=0..j-1} x^l) / (1-x^j-(k-1)*x^(i+j)), with i > 0, j > 0 and k > 1.

cp's OEIS Frontend

A193515 T(n,k) = number of ways to place any number of 3X1 tiles of k distinguishable colors into an nX1 grid.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula