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%I #52 Apr 18 2018 23:45:38
%S 1,2,4,6,12,22,40,74,136,250,460,846,1556,2862,5264,9682,17808,32754,
%T 60244,110806,203804,374854,689464,1268122,2332440,4290026,7890588,
%U 14513054,26693668,49097310,90304032,166095010,305496352,561895394
%N 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.
%C Every 0 is next to zero 3's, every 1 is next to one 1, every 2 is next to two 0's, every 3 is next to three 4's, every 4 is next to four 2's.
%C Column 1 of A196707.
%C The perimeter of cuboids with the dimensions of consecutive tribonacci numbers, signature (0,1,0). - _Peter M. Chema_, Feb 03 2017
%H R. H. Hardin, <a href="/A196700/b196700.txt">Table of n, a(n) for n = 1..200</a>
%F Empirical: a(n) = a(n-1) + a(n-2) + a(n-3) for n > 4.
%F 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
%F Empirical: a(n) = 2*(A001590(n) + A001590(n-1) + A001590(n-2)) for n > 1. - _Peter M. Chema_, Feb 03 2017
%F From _Gregory L. Simay_, Jun 23 2017: (Start)
%F 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.
%F 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)
%e All solutions for n=4:
%e 0 0 1 0 0 0
%e 0 0 1 1 0 2
%e 1 0 0 1 2 0
%e 1 0 0 0 0 0
%Y Cf. A000073, A001590, A196707.
%K nonn
%O 1,2
%A _R. H. Hardin_, Oct 05 2011
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