A004482 Tersum n + 1 (answer recorded in base 10).
1, 2, 0, 4, 5, 3, 7, 8, 6, 10, 11, 9, 13, 14, 12, 16, 17, 15, 19, 20, 18, 22, 23, 21, 25, 26, 24, 28, 29, 27, 31, 32, 30, 34, 35, 33, 37, 38, 36, 40, 41, 39, 43, 44, 42, 46, 47, 45, 49, 50, 48, 52, 53, 51, 55, 56, 54, 58, 59, 57, 61, 62
Offset: 0
References
- E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 76.
Links
- F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), Article P1.52.
- Andreas Dress, Achim Flammenkamp, and Norbert Pink, Additive periodicity of the Sprague-Grundy function of certain Nim games, Adv. Appl. Math., 22, p. 249-270 (1999).
- Gabriel Nivasch, More on the Sprague-Grundy function for Wythoff's game, pages 377-410 in "Games of No Chance 3", MSRI Publications Volume 56, 2009. See Table 1.
- Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).
Crossrefs
Programs
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Mathematica
LinearRecurrence[{1,0,1,-1},{1,2,0,4},70] (* or *) Table[3*Floor[n/3]+ Mod[ n+1,3],{n,0,70}] (* Harvey P. Dale, Nov 29 2014 *)
Formula
Periodic with period 3 and saltus 3: a(n) = 3*floor(n/3) + ((n+1) mod 3).
a(n) = n - 2*cos(2*(n+1)*Pi/3). - Wesley Ivan Hurt, Sep 29 2017
Sum_{n>=3} (-1)^(n+1)/a(n) = 1/2 - log(2)/3. - Amiram Eldar, Aug 21 2023
Extensions
More terms from Erich Friedman
Comments