A276027 Number of ways to transform a sequence of n ones to a single number by continually removing two numbers and replacing them with their sum modulo 3.

1, 1, 1, 2, 4, 7, 18, 43, 93, 266, 702, 1687, 5136, 14405, 36898, 117016, 341842, 914064, 2983027, 8972121, 24743851, 82478973, 253555061, 715745648, 2424954125, 7582390623, 21796481477, 74805170349, 237095926682, 691568408221, 2398418942361, 7686495623620

Offset: 1

Author

Caleb Ji, Aug 16 2016

Keywords

• nonn

Comments

Can be considered as the number of maximal chains in a poset whose nodes are the possible states of the sequence. In this sense it counts the same things as A002846 when the elements of that poset are taken modulo 3.

Originally this entry had a reference to a paper on the arXiv by Caleb Ji, Enumerative Properties of Posets Corresponding to a Certain Class of No Strategy Games, arXiv:1608.06025 [math.CO], 2016. However, this article has since been removed from the arXiv. - N. J. A. Sloane, Sep 07 2018

Examples

			For n = 4, the two ways are 1111 -> 211 -> 10 -> 1 and 1111 -> 211 -> 22 -> 1.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1000
• C. Ji, Example for a(7)

Crossrefs

Cf. A002846, A117143.

Similar to A002846 with nodes taken modulo 3.

A117143 is the total number of nodes in this poset.

Programs

Maple

b:= proc(x, y, z) option remember;
      `if`(x+y+z=1, 1, `if`(y>0 and z>0, b(x+1, y-1, z-1), 0)+
      `if`(x>1 or x>0 and y>0 or x>0 and z>0, b(x-1, y, z), 0)+
      `if`(y>1, b(x, y-2, z+1), 0)+`if`(z>1, b(x, y+1, z-2), 0))
    end:
a:= n-> b(0, n, 0):
seq(a(n), n=1..35);  # Alois P. Heinz, Aug 18 2016

Mathematica

b[x_,y_,z_] := b[x, y, z] = If[x+y+z==1, 1,  If[y>0 && z>0, b[x+1, y-1, z-1], 0] + If[x>1 || x>0 && y>0 || x>0 && z>0, b[x-1, y, z], 0] + If[y>1, b[x, y-2, z+1], 0] + If[z>1, b[x, y+1, z-2], 0]]; a[n_]:= b[0, n, 0]; Array[a,35] (* Jean-François Alcover, Aug 07 2017, after Alois P. Heinz *)

Python

from sympy.core.cache import cacheit
@cacheit
def b(x, y, z): return 1 if x + y + z==1 else (b(x + 1, y - 1, z - 1) if y>0 and z>0 else 0) + (b(x - 1, y, z) if x>1 or x>0 and y>0 or x>0 and z>0 else 0) + (b(x, y - 2, z + 1) if y>1 else 0) + (b(x, y + 1, z - 2) if z>1 else 0)
def a(n): return b(0, n, 0)
print([a(n) for n in range(1, 36)]) # Indranil Ghosh, Aug 09 2017, after Maple code

Formula

a(n) = f(0, n, 0) where f(a, b, c) is the number of ways to reach one number beginning with a zeros, b ones, and c twos.

Then f(a, b, c) = f_1 + f_2 + f_3 + f_4 where f_1 = f(a-1, b, c) if a>=2 or a, b >=1 or a,c >=1, f_2 = f(a, b-2, c+1) if b >= 2, f_3 = f(a, b+1, c-2) if c >= 2, and f_4 = f(a+1, b-1, c-1) if b, c >= 1, and each are 0 otherwise. Initial terms: f(a, b, c) = 1 for all 1 <= a+b+c <= 2, where a, b, c >= 0.

Extensions

a(19)-a(32) from Alois P. Heinz, Aug 18 2016