A065243 -id:A065243 - cp's OEIS frontend

A065242 Number of winning length n strings with a 9-symbol alphabet in "same game".

1, 0, 9, 9, 153, 369, 3393, 12609, 89145, 415161, 2614689, 13684977, 82237185, 457154577, 2704775985, 15524314425, 91659251961

Offset: 0

Sascha Kurz, Oct 23 2001

Strings that can be reduced to null string by repeatedly removing an entire run of two or more consecutive symbols.

For binary strings, the formula for the number of winning strings of length n has been conjectured by Ralf Stephan and proved by Burns and Purcell (2005, 2007). For b-ary strings with b >= 3, the same problem seems to be unsolved. - Petros Hadjicostas, Aug 31 2019

			11011001 is a winning string since 110{11}001 -> 11{000}1 -> {111} -> 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 in "same game", 2001. [ps file]
Sascha Kurz, Polynomials for same game, 2001. [pdf file]
Ralf Stephan, Prove or disprove: 100 conjectures from the OEIS, arXiv:math/0409509 [math.CO], 2004.

Cf. A035615, A035617, A065237, A065238, A065239, A065240, A065241, A065243, A309874, A323812.

Row b=9 of A323844.

a(12)-a(16) from Bert Dobbelaere, Dec 26 2018

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

Original entry on oeis.org

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.

cp's OEIS Frontend

A065242 Number of winning length n strings with a 9-symbol alphabet in "same game".

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