A035617 -id:A035617 - cp's OEIS frontend

A323812 a(n) = n*Fibonacci(n-2) + ((-1)^n + 1)/2.

1, 3, 5, 10, 19, 35, 65, 117, 211, 374, 661, 1157, 2017, 3495, 6033, 10370, 17767, 30343, 51681, 87801, 148831, 251758, 425065, 716425, 1205569, 2025675, 3399005, 5696122, 9534331, 15941099, 26625281, 44426877, 74062507, 123360230, 205303933, 341416205, 567353377, 942154863, 1563526761

Offset: 2

Petros Hadjicostas, Sep 01 2019

nonn

For n >= 2, a(n) is one-half the number of length n losing strings with a binary alphabet in the "same game".
In the "same game", winning strings are those that can be reduced to the null string by repeatedly removing an entire run of two or more consecutive symbols.
Sequence A035615 counts the winning strings of length n in a binary alphabet in the "same game", while A309874 counts the losing strings.
Thus, a(n) = A309874(n)/2 for n >= 2. The reason sequence A309874 is divisible by 2 is because the complement of every winning string is also a winning string (where by "complement" we mean 0 is replaced with 1 and vice versa).

			11011001 is a winning string because 110{11}001 -> 11{000}1 -> {111} -> null. Its complement, 00100110 is also a winning string because 001{00}110 -> 00{111}0 -> {000} -> null.

Chris Burns and Benjamin Purcell, A note on Stephan's conjecture 77, preprint, 2005. [Cached copy]
Chris Burns and Benjamin Purcell, Counting the number of winning strings in the 1-dimensional same game, Fibonacci Quarterly, 45(3) (2007), 233-238.
Sascha Kurz, Polynomials for same game, pdf.
Ralf Stephan, Prove or disprove: 100 conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.

Cf. A035615, A035617, A065237, A065238, A065239, A065240, A065241, A065242, A065243, A309874, A323844.

Table[n Fibonacci[n-2]+((-1)^n+1)/2,{n,2,40}] (* Harvey P. Dale, Sep 17 2019 *)

a(n) = A309874(n)/2 for n >= 2.

A324128 a(n) = 2nFibonacci(n) + (-1)^n + 1.

Original entry on oeis.org

2, 2, 6, 12, 26, 50, 98, 182, 338, 612, 1102, 1958, 3458, 6058, 10558, 18300, 31586, 54298, 93026, 158878, 270602, 459732, 779286, 1318222, 2225666, 3751250, 6312438, 10606572, 17797418, 29825282, 49922402, 83468678, 139411778, 232622148, 387796318, 645922550, 1074985346, 1787678458, 2970700846

Offset: 0

N. J. A. Sloane, Feb 20 2019

This sequence is distantly related to the number of losing strings using a binary alphabet in the "same game" described by Burns and Purcell (2007) and Kurz (2001). - Petros Hadjicostas, Sep 01 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Chris Burns and Benjamin Purcell, Counting the number of winning strings in the 1-dimensional same game, Fibonacci Quarterly, 45(3) (2007), 233-238.
Sascha Kurz, Polynomials in "same game", 2001.
Index entries for linear recurrences with constant coefficients, signature (2,2,-4,-2,2,1).

Cf. A000045, A035615, A099920, A309874, A324129.

A324128[n_]:=Fibonacci[n]2n+(-1)^n+1;Array[A324128,50,0] (* Paolo Xausa, Nov 15 2023 *)

Vec(2*(1 - x - x^2 + 2*x^3 + x^4 - x^5) / ((1 - x)*(1 + x)*(1 - x - x^2)^2) + O(x^40)) \\ Colin Barker, Mar 03 2019

From Chai Wah Wu, Feb 20 2019: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3) - 2*a(n-4) + 2*a(n-5) + a(n-6) for n > 5.
G.f.: (2*x^5 - 2*x^4 - 4*x^3 + 2*x^2 + 2*x - 2)/((x - 1)*(x + 1)*(x^2 + x - 1)^2). (End)
From Petros Hadjicostas, Sep 01 2019: (Start)
a(n) = 2*A324129(n) for n >= 0.
a(n) = A309874(n) + 2*A099920(n-1) = 2^n - A035615(n) + 2*A099920(n-1) for n >= 2.[Here A309874 counts the losing strings while A035615 counts the winning strings using a binary alphabet in the "same game". See Burns and Purcell (2007) and Kurz (2001).]
(End)

A324129 a(n) = n*Fibonacci(n) + ((-1)^n + 1)/2.

Original entry on oeis.org

1, 1, 3, 6, 13, 25, 49, 91, 169, 306, 551, 979, 1729, 3029, 5279, 9150, 15793, 27149, 46513, 79439, 135301, 229866, 389643, 659111, 1112833, 1875625, 3156219, 5303286, 8898709, 14912641, 24961201, 41734339, 69705889, 116311074, 193898159, 322961275

Offset: 0

N. J. A. Sloane, Feb 20 2019

Equals A324128(n)/2.

This sequence is distantly related to (one-half) the number of losing strings using a binary alphabet in the "same game" described by Burns and Purcell (2007) and Kurz (2001). - Petros Hadjicostas, Sep 01 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Chris Burns and Benjamin Purcell, Counting the number of winning strings in the 1-dimensional same game, Fibonacci Quarterly, 45(3) (2007), 233-238.
Sascha Kurz, Polynomials in "same game", 2001.
Index entries for linear recurrences with constant coefficients, signature (2,2,-4,-2,2,1).

Cf. A000045, A035615, A099920, A323812, A324128.

[n*Fibonacci(n)+((-1)^n+1)/2:n in [0..35]]; // Marius A. Burtea, Aug 29 2019

A324129[n_]:=Fibonacci[n]n+((-1)^n+1)/2;Array[A324129,50,0] (* Paolo Xausa, Nov 15 2023 *)

Vec((1 - x - x^2 + 2*x^3 + x^4 - x^5) / ((1 - x)*(1 + x)*(1 - x - x^2)^2) + O(x^40)) \\ Colin Barker, Mar 03 2019

From Chai Wah Wu, Feb 20 2019: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - 4*a(n-3) - 2*a(n-4) + 2*a(n-5) + a(n-6) for n > 5.
G.f.: (x^5 - x^4 - 2*x^3 + x^2 + x - 1)/((x - 1)*(x + 1)*(x^2 + x - 1)^2). (End)
a(n) = A309874(n)/2 + A099920(n-1) = 2^(n-1) - A035615(n)/2 + A099920(n-1) = A323812(n) + A099920(n-1) for n >= 2. [Sequence A309874 counts the losing strings while A035615 counts the winning strings using a binary alphabet in the "same game". See Burns and Purcell (2007) and Kurz (2001).] - Petros Hadjicostas, Sep 01 2019

cp's OEIS Frontend

A323812 a(n) = n*Fibonacci(n-2) + ((-1)^n + 1)/2.

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A324128 a(n) = 2nFibonacci(n) + (-1)^n + 1.

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A324129 a(n) = n*Fibonacci(n) + ((-1)^n + 1)/2.

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cp's OEIS Frontend

A323812 a(n) = n*Fibonacci(n-2) + ((-1)^n + 1)/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A324128 a(n) = 2*n*Fibonacci(n) + (-1)^n + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A324129 a(n) = n*Fibonacci(n) + ((-1)^n + 1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A324128 a(n) = 2nFibonacci(n) + (-1)^n + 1.