A053602 -id:A053602 - cp's OEIS frontend

A068928 Number of incongruent ways to tile a 3 X 2n room with 1x2 Tatami mats. At most 3 Tatami mats may meet at a point.

2, 2, 2, 4, 5, 9, 12, 21, 30, 51, 76, 127, 195, 322, 504, 826, 1309, 2135, 3410, 5545, 8900, 14445, 23256, 37701, 60813, 98514, 159094, 257608, 416325, 673933, 1089648, 1763581, 2852242, 4615823, 7466468, 12082291, 19546175, 31628466

Offset: 1

Dean Hickerson, Mar 11 2002

R. J. Mathar, Paving rectangular regions with rectangular tiles,...., arXiv:1311.6135 [math.CO], Table 10.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,0,-1,-1).

Cf. A068922 for total number of tilings, A068926 for more info.

Essentially the same as A001224.

For n >= 8, a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-5) - a(n-6).

O.g.f.: x(2-4x^2-x^4+x^6)/((1-x-x^2)(1-x^2-x^4)). a(n) = (A000045(n+1)+A053602(n+1))/2, n>1. [From R. J. Mathar, Aug 30 2008]

A123231 Row sums of A123230.

Original entry on oeis.org

1, 2, 1, 3, 2, 5, 3, 8, 5, 13, 8, 21, 13, 34, 21, 55, 34, 89, 55, 144, 89, 233, 144, 377, 233, 610, 377, 987, 610, 1597, 987, 2584, 1597, 4181, 2584, 6765, 4181, 10946, 6765, 17711, 10946, 28657, 17711, 46368, 28657, 75025, 46368, 121393, 75025, 196418

Offset: 1

Roger L. Bagula, Oct 06 2006

nonn

All terms are Fibonacci numbers A000045: a(2n-1) = Fibonacci(n), a(2n) = Fibonacci(n+2), a(2n-1) = a(2n+2). - Alexander Adamchuk, Oct 08 2006

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Cf. A051792, A000045, A028242, A030451.

a:=[1,2,1,3];; for n in [5..50] do a[n]:=a[n-2]+a[n-4]; od; a; # Muniru A Asiru, Oct 12 2018

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1 + 2*x + x^3)/(1 - x^2 - x^4))); // G. C. Greubel, Oct 12 2018

seq(coeff(series(-x*(1+2*x+x^3)/(x^4+x^2-1),x,n+1), x, n), n = 1 .. 50); # Muniru A Asiru, Oct 12 2018

p[0, x] = 1; p[1, x] = x + 1; p[k_, x_] := p[k, x] = x*p[k - 1, x] + (-1)^(n + 1)p[k - 2, x]; Table[Sum[CoefficientList[p[n, x], x][[m]], {m, 1, n + 1}], {n, 0, 20}]
Rest[Flatten[Reverse/@Partition[Fibonacci[Range[30]],2,1]]] (* Harvey P. Dale, Mar 19 2013 *)

vector(50, n, fibonacci(3/4 -(-1)^(n+1)*3/4 +(n+1)/2)) \\ G. C. Greubel, Oct 12 2018

From Alexander Adamchuk, Oct 08 2006: (Start)
a(n) = Fibonacci(A028242(n+2)).
a(n) = Fibonacci(A030451(n+1)).
a(n) = Fibonacci(3/4 -(-1)^(n+1)*3/4 +(n+1)/2). (End)
a(n) = A053602(n+1) = A097594(n-5). - R. J. Mathar, Mar 08 2011
G.f. -x*(1+2*x+x^3) / ( -1+x^2+x^4 ). - R. J. Mathar, Mar 08 2011
a(n) = a(n-2) + a(n-4). - Muniru A Asiru, Oct 12 2018

More terms from Alexander Adamchuk, Oct 08 2006

A051792 a(n) = (-1)^(n-1)*(a(n-1) - a(n-2)), a(1)=1, a(2)=1.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, -1, 1, 2, -1, -3, 2, 5, -3, -8, 5, 13, -8, -21, 13, 34, -21, -55, 34, 89, -55, -144, 89, 233, -144, -377, 233, 610, -377, -987, 610, 1597, -987, -2584, 1597, 4181, -2584, -6765, 4181, 10946, -6765, -17711, 10946, 28657, -17711, -46368

Offset: 1

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 10 1999

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (0,-1,0,1).

Cf. A000045, A053602.

[Fibonacci(1 -Floor((n-4)/2) -2*((n-4) mod 2)): n in [1..60]]; // G. C. Greubel, Dec 06 2022

LinearRecurrence[{0,-1,0,1},{1,1,0,1},60] (* Harvey P. Dale, May 08 2017 *)

PARI
```
a(n)=fibonacci((3-n)\2+(3-n)%2*2)
```

def A051792():
    x, y, b = 1, 1, true
    while True:
        yield x
        x, y = y, x - y
        y = -y if b else y
        b = not b
a = A051792()
print([next(a) for  in range(51)]) # _Peter Luschny, Mar 19 2020

a(3-n) = A053602(n).
From Michael Somos: (Start)
G.f.: x*(1 + x + x^2 + 2*x^3)/(1 + x^2 - x^4).
a(n) = -a(n-2) + a(n-4). (End)
a(n) = b(n-1) + b(n-2) + b(n-3) + 2*b(n-4), where b(n) = i^n * A079977(n). - G. C. Greubel, Dec 06 2022

A233323 Triangle read by rows: T(n,k) = number of palindromic compositions of n in which the largest part is equal to k, 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, 2, 0, 1, 1, 1, 1, 0, 1, 1, 4, 1, 1, 0, 1, 1, 2, 3, 0, 1, 0, 1, 1, 7, 3, 3, 0, 1, 0, 1, 1, 4, 6, 1, 2, 0, 1, 0, 1, 1, 12, 7, 7, 1, 2, 0, 1, 0, 1, 1, 7, 12, 3, 5, 0, 2, 0, 1, 0, 1, 1, 20, 16, 15, 3, 5, 0, 2, 0, 1, 0, 1

Offset: 1

L. Edson Jeffery, Dec 10 2013

A palindromic composition of a natural number m is an ordered partition of m into N+1 natural numbers (or parts), p_0, p_1, ..., p_N, of the form m = p_0 + p_1 + ... + p_N such that p_j = p_{N-j}, for each j in {0,...,N}. Two palindromic compositions, sum_{j=0..N} p_j and sum_{j=0..N} q_j (say), are identical if and only if p_j = q_j, j = 0,...,N; otherwise they are taken to be distinct.

			There are eight palindromic compositions of n=7, namely, {7}, {3,1,3}, {2,3,2}, {2,1,1,1,2}, {1,5,1}, {1,2,1,2,1}, {1,1,3,1,1}, {1,1,1,1,1,1,1}, and three of them have the largest part equal to 3, so T(7,3) = 3.
Triangle T(n,k) begins:
  1;
  1,  1;
  1,  0, 1,
  1,  2, 0, 1;
  1,  1, 1, 0, 1;
  1,  4, 1, 1, 0, 1;
  1,  2, 3, 0, 1, 0, 1;
  1,  7, 3, 3, 0, 1, 0, 1;
  1,  4, 6, 1, 2, 0, 1, 0, 1;
  1, 12, 7, 7, 1, 2, 0, 1, 0, 1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened (rows n = 1..50 from Charles R Greathouse IV)
V. E. Hoggatt, Jr., and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.

Cf. A016116 (row sums), A233324 (row partial sums).

T(n,2)+1 = A053602(n+1) = A123231(n). T(4n-2,2n) = A011782(n-1). - Alois P. Heinz, Dec 11 2013

b:= proc(n, k) option remember; `if`(n<=k, 1, 0)+
      add(b(n-2*j, k), j=1..min(k, iquo(n, 2)))
    end:
T:= (n, k)-> b(n, k) -b(n, k-1):
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Dec 11 2013

b[n_, k_] := b[n, k] = If[n <= k, 1, 0] + Sum[b[n-2*j, k], { j, 1, Min[k, Quotient[n, 2]]}]; t[n_, k_] := b[n, k] - b[n, k-1]; Table[Table[t[n, k], {k, 1, n}], {n, 1, 14}] // Flatten (* Jean-François Alcover, Dec 13 2013, translated from Alois P. Heinz's Maple code *)

T(n,k,ok=0)={
    if(n<1,return(n==0 && ok));
    if(k>n/2 && !ok,
        n-=k;
        if(n<0||n%2, return(0));
        return(2^max(n/2-1,0))
    );
    sum(i=1,k,
        T(n-2*i,k,ok||i==k)
    )+(n==k || (ok && nCharles R Greathouse IV, Dec 11 2013

A233324 Triangle read by rows: T(n,k) = number of palindromic compositions of n in which no part exceeds k, 1 <= k <= n.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 3, 3, 4, 1, 2, 3, 3, 4, 1, 5, 6, 7, 7, 8, 1, 3, 6, 6, 7, 7, 8, 1, 8, 11, 14, 14, 15, 15, 16, 1, 5, 11, 12, 14, 14, 15, 15, 16, 1, 13, 20, 27, 28, 30, 30, 31, 31, 32, 1, 8, 20, 23, 28, 28, 30, 30, 31, 31, 32, 1, 21, 37, 52, 55, 60, 60, 62, 62, 63, 63, 64

Offset: 1

L. Edson Jeffery, Dec 11 2013

A palindromic composition of a natural number m is an ordered partition of m into N+1 natural numbers (or parts), p_0, p_1, ..., p_N, of the form m = p_0 + p_1 + ... + p_N such that p_j = p_{N-j}, for each j in {0,...,N}. Two palindromic compositions, sum_{j=0..N} p_j and sum_{j=0..N} q_j (say), are identical if and only if p_j = q_j, j = 0,...,N; otherwise they are taken to be distinct.
Partial sums of rows of A233323.
T(n,k) is defined for n,k >= 0.  T(n,k) = T(n,n) = A016116(n) for k>= 0. - Alois P. Heinz, Dec 11 2013

			Triangle T(n,k) begins:
1;
1,  2;
1,  1,  2;
1,  3,  3,  4;
1,  2,  3,  3,  4;
1,  5,  6,  7,  7,  8;
1,  3,  6,  6,  7,  7,  8;
1,  8, 11, 14, 14, 15, 15, 16;
1,  5, 11, 12, 14, 14, 15, 15, 16;
1, 13, 20, 27, 28, 30, 30, 31, 31, 32;

Alois P. Heinz, Rows n = 1..141, flattened
V. E. Hoggatt, Jr., and Marjorie Bicknell, Palindromic compositions, Fibonacci Quart., Vol. 13(4), 1975, pp. 350-356.

Cf. A233323.

T(n,2) = A053602(n+1) = A123231(n). T(2n,3) = A001590(n+3). T(2n,4) = A001631(n+4). - Alois P. Heinz, Dec 11 2013

T:= proc(n, k) option remember; `if`(n<=k, 1, 0)+
      add(T(n-2*j, k), j=1..min(k, iquo(n, 2)))
    end:
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Dec 11 2013

T[n_, k_] := T[n, k] = If[n <= k, 1, 0] + Sum[T[n-2*j, k], {j, 1, Min[k, Quotient[ n, 2]]}]; Table[Table[T[n, k], {k, 1, n}], {n, 1, 14}] // Flatten (* Jean-François Alcover, Mar 09 2015, after Alois P. Heinz *)

T(n,k)=if(n<1,return(n==0));sum(i=1,k,T(n-2*i,k))+(n<=k) \\ Charles R Greathouse IV, Dec 11 2013

A272912 Difference sequence of the sequence A116470 of all distinct Fibonacci numbers and Lucas numbers (A000032).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 2, 5, 3, 8, 5, 13, 8, 21, 13, 34, 21, 55, 34, 89, 55, 144, 89, 233, 144, 377, 233, 610, 377, 987, 610, 1597, 987, 2584, 1597, 4181, 2584, 6765, 4181, 10946, 6765, 17711, 10946, 28657, 17711, 46368, 28657, 75025, 46368, 121393, 75025

Offset: 1

Clark Kimberling, May 10 2016

Every term is a Fibonacci number (A000045).

			A116470 = (1, 2, 3, 4, 5, 7, 8, 11, 13, 18, 21, 29, 34, 47, 55, 76,...), so that (a(n)) = (1,1,1,1,2,1,3,2,5,3,8,5,13,8,12,...).

Clark Kimberling, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Cf. A000045, A000032, A116470, A053602, A123231.

u = Table[Fibonacci[n], {n, 1, 200}]; v = Table[LucasL[n], {n, 1, 200}];
Take[Differences[Union[u, v]], 100]

Vec(x*(1+x-x^5)/(1-x^2-x^4) + O(x^50)) \\ Colin Barker, May 10 2016

From Colin Barker, May 10 2016: (Start)
a(n) = a(n-2)+a(n-4) for n>4.
G.f.: x*(1+x-x^5) / (1-x^2-x^4).
(End)
a(n) = A053602(n-2), n>2. - R. J. Mathar, May 20 2016
a(n) = A123231(n-3), n>3. - Georg Fischer, Oct 23 2018

A032089 "BHK" (reversible, identity, unlabeled) transform of 1,0,1,0...(odds).

Original entry on oeis.org

1, 0, 1, 1, 2, 3, 5, 9, 14, 25, 39, 68, 107, 182, 289, 483, 772, 1275, 2047, 3355, 5402, 8811, 14213, 23112, 37325, 60580, 97905, 158717, 256622, 415715, 672337, 1088661, 1760998, 2850645, 4611643, 7463884, 12075527, 19541994, 31617521, 51163695, 82781216, 133951675

Offset: 1

Christian G. Bower

Andrew Howroyd, Table of n, a(n) for n = 1..1000
Christian G. Bower, Transforms (2)
Index entries for linear recurrences with constant coefficients, signature (1,3,-2,-2,0,-1,1,1).

LinearRecurrence[{1,3,-2,-2,0,-1,1,1},{1,0,1,1,2,3,5,9},42] (* Stefano Spezia, Apr 01 2025 *)

Vec((1-x-2*x^2+2*x^3+x^6)/((1-x)*(1+x)*(1-x-x^2)*(1-x^2-x^4)) + O(x^40)) \\ Andrew Howroyd, Aug 31 2018

G.f.: x*(1-x-2*x^2+2*x^3+x^6)/((1-x)*(1+x)*(1-x-x^2)*(1-x^2-x^4)).
a(n) = a(n-1) + 3*a(n-2) - 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7) + a(n-8) for n > 8. - Andrew Howroyd, Aug 31 2018
2*a(n) = 2*A000035(n) + A000045(n) - A053602(n). - R. J. Mathar, Mar 02 2022

A097594 a(n) = (a(n-1) mod a(n-2)) + a(n-2), a(0) = 3, a(1) = 2.

Original entry on oeis.org

2, 5, 3, 8, 5, 13, 8, 21, 13, 34, 21, 55, 34, 89, 55, 144, 89, 233, 144, 377, 233, 610, 377, 987, 610, 1597, 987, 2584, 1597, 4181, 2584, 6765, 4181, 10946, 6765, 17711, 10946, 28657, 17711, 46368, 28657, 75025, 46368, 121393, 75025, 196418, 121393, 317811, 196418, 514229, 317811, 832040, 514229

Offset: 0

Gerald McGarvey, Aug 29 2004

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1,0,1).

Cf. A000045, A051792, A053602.

[Fibonacci(3 +Floor(n/2) +2*(n mod 2)): n in [0..60]]; // G. C. Greubel, Dec 06 2022

LinearRecurrence[{0,1,0,1}, {2,5,3,8}, 60] (* G. C. Greubel, Dec 06 2022 *)

[fibonacci(3 +(n//2) + 2*(n%2)) for n in range(61)] # G. C. Greubel, Dec 06 2022

a(2n) = Fibonacci(n+4), a(2n+1) = Fibonacci(n+3).
a(n) = A053602(n+6).
a(n) = abs( A051792(n+11) ).
G.f.: (2 + 5*x + x^2 + 3*x^3)/(1 - x^2 - x^4). - G. C. Greubel, Dec 06 2022

A239366 Triangular array read by rows: T(n,k) is the number of palindromic compositions of n having exactly k 1's, n>=0, 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 0, 0, 1, 2, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 3, 0, 3, 0, 1, 0, 1, 2, 1, 1, 2, 1, 0, 0, 1, 5, 0, 5, 0, 4, 0, 1, 0, 1, 3, 2, 3, 2, 1, 3, 1, 0, 0, 1, 8, 0, 10, 0, 7, 0, 5, 0, 1, 0, 1, 5, 3, 5, 5, 4, 3, 1, 4, 1, 0, 0, 1, 13, 0, 18, 0, 16, 0, 9, 0, 6, 0, 1, 0, 1, 8, 5, 10, 8, 7, 9, 5, 4, 1, 5, 1, 0, 0, 1

Offset: 0

Geoffrey Critzer, Mar 20 2014

Row sums = 2^floor(n/2).

T(n,0) = A053602(n-1) for n>0, T(n,1) = A079977(n-5) for n>4, T(2n+1,3) = A006367(n-1) for n>0, both bisections of column k=2 contain A010049. - Alois P. Heinz, Mar 21 2014

			1,
0, 1,
1, 0, 1,
1, 0, 0, 1,
2, 0, 1, 0, 1,
1, 1, 1, 0, 0, 1,
3, 0, 3, 0, 1, 0, 1,
2, 1, 1, 2, 1, 0, 0, 1,
5, 0, 5, 0, 4, 0, 1, 0, 1,
3, 2, 3, 2, 1, 3, 1, 0, 0, 1
There are eight palindromic compositions of 6: T(6,0)=3 because we have: 6, 3+3, 2+2+2.  T(6,2)=3 because we have: 1+4+1, 2+1+1+2, 1+2+2+1.  T(6,4)=1 because we have: 1+1+2+1+1. T(6,6)=1 because we have: 1+1+1+1+1+1.

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

b:= proc(n) option remember;  `if`(n=0, 1, expand(
      add(b(n-j)*`if`(j=1, x^2, 1), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))
    (add(b(i)*`if`(n-2*i=1, x, 1), i=0..n/2)):
seq(T(n), n=0..30);  # Alois P. Heinz, Mar 21 2014

nn=15;Table[Take[CoefficientList[Series[((1+x)*(1-x+x^2+x*y-x^2*y))/(1-x^2-x^4-x^2*y^2+x^4*y^2),{x,0,nn}],{x,y}][[n]],n],{n,1,nn}]//Grid

G.f.: G(x,y) = ((1 + x)*(1 - x + x^2 + x*y - x^2*y))/(1 - x^2 - x^4 - x^2*y^2 + x^4*y^2). Satisfies G(x,y) = 1/(1 - x) - x + y*x + (x^2/(1 - x^2) - x^2 +y^2*x^2)*G(x,y).

A275435 Triangle read by rows: T(n,k) is the number of 00-avoiding binary words of length n having degree of asymmetry equal to k (n>=0; 0<=k<=floor(n/2)).

Original entry on oeis.org

1, 2, 1, 2, 3, 2, 2, 4, 2, 5, 6, 2, 3, 8, 8, 2, 8, 14, 10, 2, 5, 16, 20, 12, 2, 13, 30, 30, 14, 2, 8, 30, 48, 40, 16, 2, 21, 60, 78, 54, 18, 2, 13, 56, 106, 112, 68, 20, 2, 34, 116, 184, 166, 86, 22, 2, 21, 102, 224, 286, 224, 104, 24, 2, 55, 218, 408, 452, 310, 126, 26, 2

Offset: 0

Emeric Deutsch, Aug 15 2016

The degree of asymmetry of a finite sequence of numbers is defined to be the number of pairs of symmetrically positioned distinct entries. Example: the degree of asymmetry of (2,7,6,4,5,7,3) is 2, counting the pairs (2,3) and (6,5).
A sequence is palindromic if and only if its degree of asymmetry is 0.
Number of entries in row n is 1+floor(n/2).
Sum of entries in row n is A000045(n+2) (Fibonacci).
T(n,0) = A053602(n+2) (= number of palindromic 00-avoiding binary words of length n).
Sum(k*T(n,k), k>=0) = A275436(n).

			Row 3 is [3,2] because the 00-avoiding binary words of length 3 are 010, 011, 101, 110, 111, having asymmetry degrees 0, 1, 0, 1, 0, respectively.
Triangle starts:
1;
2;
1,2;
3,2;
2,4,2;
5,6,2.

Cf. A000045, A053602, A275436

G := (1+2*z+t*z^2+z^3-t*z^5)/(1-z^2-t*z^2-z^4-t*z^4+t*z^6): Gser := simplify(series(G, z = 0, 20)): for n from 0 to 18 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 18 do seq(coeff(P[n], t, j), j = 0 .. (1/2)*n) end do; # yields sequence in triangular form

Table[BinCounts[#, {0, Floor[n/2] + 1, 1}] &@ Map[Total@ BitXor[Take[#, Ceiling[Length[#]/2]], Reverse@ Take[#, -Ceiling[Length[#]/2]]] &, Select[PadLeft[IntegerDigits[#, 2], n] & /@ Range[0, 2^n - 1], Length@ SequenceCases[#, {0, 0}] == 0 &]], {n, 0, 15}] // Flatten (* Michael De Vlieger, Aug 15 2016, Version 10.1 *)

G.f.: G(t,z) = (1 + 2z + tz^2 +z^3 -tz^5)/(1 - z^2 - tz^2 - z^4 - tz^4 + tz^6).

cp's OEIS Frontend

A068928 Number of incongruent ways to tile a 3 X 2n room with 1x2 Tatami mats. At most 3 Tatami mats may meet at a point.

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A123231 Row sums of A123230.

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A051792 a(n) = (-1)^(n-1)*(a(n-1) - a(n-2)), a(1)=1, a(2)=1.

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A233323 Triangle read by rows: T(n,k) = number of palindromic compositions of n in which the largest part is equal to k, 1 <= k <= n.

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A233324 Triangle read by rows: T(n,k) = number of palindromic compositions of n in which no part exceeds k, 1 <= k <= n.

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A272912 Difference sequence of the sequence A116470 of all distinct Fibonacci numbers and Lucas numbers (A000032).

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A032089 "BHK" (reversible, identity, unlabeled) transform of 1,0,1,0...(odds).

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