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Number of (3,n)-partitions of a chain of length n^2.
2

%I #28 Sep 08 2022 08:45:01

%S 0,0,1,35,286,1330,4495,12341,29260,62196,121485,221815,383306,632710,

%T 1004731,1543465,2303960,3353896,4775385,6666891,9145270,12347930,

%U 16435111,21592285,28032676,35999900,45770725,57657951,72013410

%N Number of (3,n)-partitions of a chain of length n^2.

%C a (k,n)-partition of a chain C is a chain of k intervals of C of length n.

%H Vincenzo Librandi, <a href="/A055658/b055658.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F a(n) = (1/6)*(n-1)*(n-2)*(n^2-3*n+3)*(n^2-3*n+1).

%F G.f.: -x^3*(1+28*x+62*x^2+28*x^3+x^4) / (x-1)^7. - _R. J. Mathar_, Mar 14 2016

%e a(3)=1 because in the linearly ordered set {1,..,9} we can choose in just one way 3 successive blocks of 3 consecutive elements.

%o (Magma) [1/6*(n-1)*(n-2)*(n^2-3*n+3)*(n^2-3*n+1): n in [1..35]]; // _Vincenzo Librandi_, Jun 30 2011

%o (PARI) a(n) = (n-1)*(n-2)*(n^2-3*n+3)*(n^2-3*n+1)/6; \\ _Altug Alkan_, Oct 04 2018

%Y Cf. A055659.

%K nonn,easy

%O 1,4

%A Paolo Dominici (pl.dm(AT)libero.it), Jun 07 2000