A065238 -id:A065238 - cp's OEIS frontend

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

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

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