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

%I #30 Mar 22 2022 05:53:19

%S 220,4004,43680,371280,2713200,17907120,109830336,637408200,

%T 3543239700,19028509500,99348849600,506679132960,2533395664800,

%U 12454385680800,60338017584000,288616850776800,1365157704174264,6393385628146440,29677938224482240,136674715507484000

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

%C a (k_1,n_1; k_2,n_2)-partition of a chain C is a chain of k_1+k_2 intervals of C, k_1 being of length n_1 and k_2 of length n_2.

%H Vincenzo Librandi, <a href="/A055663/b055663.txt">Table of n, a(n) for n = 9..200</a> [a(67) and a(72) corrected by Georg Fischer, May 29 2021]

%F a(n) = (16/3)*(2*n-7)*(2*n-9)*(2*n-11)*(2*n-13)*(n-8)*(2*n-15)!/(n*(n-1)*(n-2)*(n-8)!^2).

%F D-finite with recurrence: (-9*n+n^2)*a(n) - (42-26*n+4*n^2)*a(n-1) = 0, a(9) = 220 for n >= 10. - _Georg Fischer_, May 29 2021

%F a(n) ~ n^(5/2) * 2^(2*n) / (384*sqrt(Pi)). - _Vaclav Kotesovec_, May 29 2021

%F From _Amiram Eldar_, Mar 22 2022: (Start)

%F Sum_{n>=9} 1/a(n) = sqrt(3)*Pi/20 - 5051/18900.

%F Sum_{n>=9} (-1)^(n+1)/a(n) = 65*sqrt(5)*log(phi)/6 - 15731/1350, where phi is the golden ratio (A001622). (End)

%e a(9)=220 because in the linearly ordered set {1,...,90} we can choose in 220 ways 12 successive blocks, 3 constituted of 3 consecutive elements and 9 of 9 consecutive elements.

%p rec:={(-9*n+n^2)*a(n) - (42-26*n+4*n^2)*a(n-1) = 0, a(9) = 220};

%p f:= gfun:-rectoproc(rec, a(n), remember): map(f, [$9..26]); # _Georg Fischer_, May 29 2021

%o (Magma) [16/3*(2*n-7)*(2*n-9)*(2*n-11)*(2*n-13)*(n-8)*Factorial(2*n-15)/(n*(n-1)*(n-2)*Factorial(n-8)^2): n in [9..30]]; // _Vincenzo Librandi_, Jun 30 2011

%Y Cf. A055660.

%K nonn

%O 9,1

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