A291227 -id:A291227 - cp's OEIS frontend

A291219 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^3.

1, 1, 3, 5, 11, 21, 42, 83, 163, 323, 635, 1255, 2473, 4880, 9625, 18985, 37451, 73869, 145715, 287421, 566954, 1118331, 2205947, 4351307, 8583091, 16930447, 33395857, 65874464, 129939569, 256310161, 505580371, 997274197, 1967156763, 3880282533, 7653987242

Offset: 0

Clark Kimberling, Aug 24 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
In the following guide to p-INVERT sequences using s = (1,0,1,0,1,...) = A000035, in some cases t(1,0,1,0,1,...) is a shifted version of the indicated sequence.
p(S)                             t(1,0,1,0,1,...)
1 - S                                A000045 (Fibonacci numbers)
1 - S^2                              A147600
1 - S^3                              A291217
1 - S^5                              A291218
1 - S - S^2                          A289846
1 - S - S^3                          A291219
1 - S - S^4                          A291220
1 - S^3- S^6                         A291221
1 - S^2- S^3                         A291222
1 - S^3- S^4                         A291223
1 - 2S                               A052542
1 - 3S                               A006190
(1 - S)^2                            A239342
(1 - S)^3                            A276129
(1 - S)^4                            A291224
(1 - S)^5                            A291225
(1 - S)^6                            A291226
1 - S - 2 S^2                        A291227
1 - 2 S - 2 S^2                      A291228
1 - 3 S - 2 S^2                      A060801
(1 - S)(1 - 2 S)                     A291229
(1 - S)(1 - 2 S)(1 - 3 S)            A291230
(1 - S)(1 - 2 S)(1 - 3 S)( 1 - 4 S)  A291231
(1 - 2 S)^2                          A291264
(1 - 3 S)^2                          A291232
1 - S - S^2 - S^3                    A291233
1 - S - S^2 - S^3 - S^4              A291234
1 - S - S^2 - S^3 - S^4 - S^5        A291235
(1 - S)(1 - 3 S)                     A291236
(1 - S)(1 - 2S)( 1 - 4S)             A291237
(1 - S)^2 (1 - 2S)                   A291238
(1 - S^2) (1 - 2S)                   A291239
(1 - S^3)^2                          A291240
1 - S - S^2 + S^3                    A291241
1 - 2 S - S^2 + S^3                  A291242
1 - 3 S + S^2                        A291243
1 - 4 S + S^2                        A291244
1 - 5 S + S^2                        A291245
1 - 6 S + S^2                        A291246
1 - S - S^2 - S^3 + S^4              A291247
1 - S - S^2 - S^3 - S^4 + S^5        A291248
1 - S - S^2 - S^3 + S^4 + S^5        A291249
1 - S - 2 S^2 + 2 S^3                A291250
1 - 3 S^2 + 2 S^3                    A291251 (includes negative terms)
(1 - S^3)^3                          A291252
(1 - S - S^2)^2                      A291253
(1 - 2 S - S^2)^2                    A291254
(1 - S - 2 S^2)^2                    A291255

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,3,-1,-3,1,1)

Cf. A000035, A290890, A291000.

I:=[1,1,3,5,11,21]; [n le 6 select I[n] else Self(n-1)+3*Self(n-2)-Self(n-3)-3*Self(n-4)+Self(n-5)+Self(n-6): n in [1..45]]; // Vincenzo Librandi, Aug 25 2017

z = 60; s = x/(1 - x^2); p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291219 *)
LinearRecurrence[{1, 3, -1, -3, 1, 1}, {1, 1, 3, 5, 11, 21}, 50] (* Vincenzo Librandi, Aug 25 2017 *)

G.f.: -(1 - x^2 + x^4)/(-1 + x + 3*x^2 - x^3 - 3*x^4 + x^5 + x^6).

a(n) = a(n-1) + 3*a(n-2) - a(n-3) - 3*a(n-4) + a(n-5) + a(n-6) for n >= 7.

A335747 Number of ways to tile vertically-fault-free 3 X n strip with squares and dominoes.

Original entry on oeis.org

1, 3, 13, 26, 66, 154, 380, 904, 2204, 5286, 12818, 30854, 74636, 179948, 434820, 1049122, 2533818, 6115538, 14766868, 35646080, 86064196, 207766110, 501609946, 1210964110, 2923573588, 7058053972, 17039774268

Offset: 0

Greg Dresden and Oluwatobi Jemima Alabi, Jun 20 2020

nonn

By "vertically-fault-free" we mean that the tilings of the 3 X n strip do not split along any interior vertical line. Here are two of the 66 possible vertically-fault-free tilings of a 3 X 4 strip with squares and dominoes:
.  _       _  
| |_ |   | _|_| |
|| _| |   | | |_|
|| _||   ||_| |

			a(2) = 13 thanks to these thirteen vertically-fault-free tilings of a 3 X 2 strip:
._ _     _ _     _ _     _ _     _ _     _ _     _ _
|_ _|   |_|_|   |_|_|   |_ _|   |_|_|   |_ _|   |_ _|
|_|_|   |_ _|   |_|_|   |_ _|   |_ _|   |_|_|   |_ _|
|_|_|   |_|_|   |_ _|   |_|_|   |_ _|   |_ _|   |_ _|
._ _     _ _     _ _     _ _     _ _     _ _
|_ _|   |_ _|   |_ _|   | |_|   |_| |   | | |
| |_|   |_| |   | | |   |_|_|   |_|_|   |_|_|
|_|_|   |_|_|   |_|_|   |_ _|   |_ _|   |_ _|

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,4,-1,-1).

Cf. A033506 (which gives all tilings of 3 X n strip), A112577, A134438, A291227.

Cf. A000045, A000129.

I:=[26, 66, 154, 380]; [1,3,13] cat [n le 4 select I[n] else Self(n-1) +4*Self(n-2) -Self(n-3) -Self(n-4): n in [1..40]]; // G. C. Greubel, Jan 15 2022

CoefficientList[Series[(1+2x+6x^2+2x^3-8x^4+x^6)/((1+x-x^2)(1-2x-x^2)), {x, 0, 26}], x] (* Michael De Vlieger, Jul 03 2020 *)
LinearRecurrence[{1,4,-1,-1}, {1,3,13,26,66,154,380}, 40] (* G. C. Greubel, Jan 15 2022 *)

def P(n): return lucas_number1(n,2,-1)
def A335747(n): return (1/3)*(-9*bool(n==0) - 3*bool(n==1) + 3*bool(n==2) + 2*(3*P(n+1) + 2*P(n-1)) + 2*(-1)^n*fibonacci(n-1))
[A335747(n) for n in (0..40)] # G. C. Greubel, Jan 15 2022

a(n) = a(n-1) + 4*a(n-2) - a(n-3) - a(n-4) for n >= 7.
a(n) = 2*A291227(n) - 8*A112577(n-2) + 2*A112577(n-4) for n >= 4.
a(n) = (2/3)*(A221174(n+1) + (-1)^n*A000045(n-1)) for n >= 3. - Greg Dresden, Jul 03 2020
G.f.: (1 + 2*x + 6*x^2 + 2*x^3 - 8*x^4 + x^6) / ((1 + x - x^2)*(1 - 2*x - x^2)). - Colin Barker, Jun 21 2020
a(n) = (1/3)*(-9*[n=0] - 3*[n=1] + 3*[n=2] + 2*(3*A000129(n+1) + 2*A000129(n-1)) + 2*(-1)^n*Fibonacci(n-1)). - G. C. Greubel, Jan 15 2022

cp's OEIS Frontend

A291219 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^3.

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