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%I #41 Sep 08 2022 08:45:49
%S 1,91,3185,63700,866320,8836464,71954064,488259720,2848181700,
%T 14620666060,67255063876,281248448936,1081724803600,3863302870000,
%U 12914469594000,40680579221100,121443493851225,345280521733875,938920716995625,2451077240157000,6162708489537600
%N a(n) = binomial(m+n-1,n)^2 - binomial(m+n,n+1)*binomial(m+n-2,n-1) with m = 14.
%C 13th column (and diagonal) of the triangle A001263. - _Bruno Berselli_, May 07 2012
%D S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; Prop. 8.4, case n=14.
%H Vincenzo Librandi, <a href="/A169937/b169937.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_25">Index entries for linear recurrences with constant coefficients</a>, signature (25, -300, 2300, -12650, 53130, -177100, 480700, -1081575, 2042975, -3268760, 4457400, -5200300, 5200300, -4457400, 3268760, -2042975, 1081575, -480700, 177100, -53130, 12650, -2300, 300, -25, 1).
%F a(n) = (1/13)*A010965(n+12)^2*(n+13)/(n+1). - _Bruno Berselli_, Nov 09 2011
%F a(n) = Product_{i=1..12} A002378(n+i)/A002378(i). - _Bruno Berselli_, Sep 01 2016
%F From _Amiram Eldar_, Oct 19 2020: (Start)
%F Sum_{n>=0} 1/a(n) = 45997360927193/23100 - 201753552*Pi^2.
%F Sum_{n>=0} (-1)^n/a(n) = 16431564019/23100 - 72072*Pi^2. (End)
%p f:=m->[seq( binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1), n=0..20)]; f(14);
%t Table[Binomial[13+n,n]^2-Binomial[14+n,n+1]Binomial[12+n,n-1],{n,0,20}] (* _Harvey P. Dale_, Nov 09 2011 *)
%o (Magma) [(1/13)*Binomial(n+12,12)^2*(n+13)/(n+1): n in [0..20]]; // _Bruno Berselli_, Nov 09 2011
%o (PARI) a(n)=binomial(n+12,12)^2*(n+13)/(n+1)/13 \\ _Charles R Greathouse IV_, Nov 09 2011
%Y The expression binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1) for the values m = 2 through 14 produces the sequences A000012, A000217, A002415, A006542, A006857, A108679, A134288, A134289, A134290, A134291, A140925, A140935, A169937.
%Y Cf. A002378.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Aug 28 2010
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