A196700 Number of n X 1 0..4 arrays with each element x equal to the number of its horizontal and vertical neighbors equal to 3,1,0,4,2 for x=0,1,2,3,4.
1, 2, 4, 6, 12, 22, 40, 74, 136, 250, 460, 846, 1556, 2862, 5264, 9682, 17808, 32754, 60244, 110806, 203804, 374854, 689464, 1268122, 2332440, 4290026, 7890588, 14513054, 26693668, 49097310, 90304032, 166095010, 305496352, 561895394
Offset: 1
Keywords
Examples
All solutions for n=4: 0 0 1 0 0 0 0 0 1 1 0 2 1 0 0 1 2 0 1 0 0 0 0 0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
Formula
Empirical: a(n) = a(n-1) + a(n-2) + a(n-3) for n > 4.
G.f.: 1 - 1/x - 1/x^2 + 1/x^2/G(0), where G(k)= 1 - (2*k+1)*x/(1 - x/(x - (2*k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 09 2013
Empirical: a(n) = 2*(A001590(n) + A001590(n-1) + A001590(n-2)) for n > 1. - Peter M. Chema, Feb 03 2017
From Gregory L. Simay, Jun 23 2017: (Start)
a(n) = A000073(n+2) - A000073(n-2), the difference of two tribonacci numbers. The corresponding g.f. is (1 - x^4)/(1 - x - x^2 - x^3). E.g.: a(10) = A000073(12) - A000073(8) = 274 - 24 = 250.
The tribonacci formula arises from considering the number of compositions of n where only the order of parts 1,2,3 matters (part of an upcoming paper), which may be denoted by C(n [4). We are convolving the number of partitions of n with parts >3 with the tribonacci numbers. The number of partitions of n with parts greater than 3 is P(n) - P(n-1) - P(n-2) + P(n-4) + P(n-5) - P(n-6). (Derived from the corresponding gf which is (1-x)(1-x^2)(1-x^3)gfP(x).) The rest is algebra. It looks like C(n, [4) = P(n) + Sum_{j=0..n-3} P(n-3-j)*A196700(j+1). (End)
Comments