A035615 Number of winning length n strings with a 2-symbol alphabet in "same game".

1, 0, 2, 2, 6, 12, 26, 58, 126, 278, 602, 1300, 2774, 5878, 12350, 25778, 53470, 110332, 226610, 463602, 945214, 1921550, 3896642, 7885092, 15927086, 32121582, 64697726, 130166378, 261637446, 525478668, 1054673162, 2115601450, 4241716734, 8501080838, 17031744170

Offset: 0

Author

Erich Friedman

Keywords

• nonn
• nice
• easy

Comments

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

Examples

			11011001 is a winning string since 110{11}001 -> 11{000}1 -> {111} -> null.
		

Links

• Robert Price, Table of n, a(n) for n = 0..1000
• 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.
• Index entries for linear recurrences with constant coefficients, signature (4, -2, -8, 6, 6, -3, -2).

Crossrefs

Cf. A035617, A065237, A065238, A065239, A065240, A065241, A065242, A065243.

See A309874 for the losing strings.

For some similar questions in base 10, see A323830, A323831, A320487. - N. J. A. Sloane, Feb 04 2019

Row b=2 of A323844.

Programs

Mathematica

Join[{1}, Rest[CoefficientList[Series[x (2x^6 - 6x^5 + 8x^4 + 2x^3 - 6x^2 + 2x)/((1 - x^2)(1 - 2x)(1 - x - x^2)^2), {x, 0, 40}], x]]] (* or *) Join[{1}, LinearRecurrence[{4, -2, -8, 6, 6, -3, -2}, {0, 2, 2, 6, 12, 26, 58}, 40]] (* Harvey P. Dale, Sep 26 2012 *)

PARI

a(n)=if(n, ([0,1,0,0,0,0,0; 0,0,1,0,0,0,0; 0,0,0,1,0,0,0; 0,0,0,0,1,0,0; 0,0,0,0,0,1,0; 0,0,0,0,0,0,1; -2,-3,6,6,-8,-2,4]^(n-1)*[0;2;2;6;12;26;58])[1,1], 1) \\ Charles R Greathouse IV, Jun 15 2015

Formula

G.f.: x(2x^6 - 6x^5 + 8x^4 + 2x^3 - 6x^2 + 2x)/[(1 - x^2)(1 - 2x)(1 - x - x^2)^2] (conjectured). - Ralf Stephan, May 11 2004. Established by Burns and Purcell - see link.

a(0) = 1, a(1) = 0, a(2) = 2, a(3) = 2, a(4) = 6, a(5) = 12, a(6) = 26, a(7) = 58, a(n) = 4*a(n-1) - 2*a(n-2) - 8*a(n-3) + 6*a(n-4) + 6*a(n-5) - 3*a(n-6) - 2*a(n-7). - Harvey P. Dale, Sep 26 2012

a(n) = 2^n - 2 * n * Fibonacci(n-2) - (-1)^n - 1 for n >= 2 (proved by Burns and Purcell (2005, 2007)). - Petros Hadjicostas, Jul 04 2018

Extensions

More terms from Naohiro Nomoto, Jul 09 2001

Further terms from Sascha Kurz, Oct 19 2001

a(27)-a(36) from Robert Price, Apr 08 2019