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Antidiagonal sums of square array A111910.
3

%I #45 Oct 06 2021 12:56:57

%S 1,2,7,30,146,772,4331,25398,154158,961820,6137734,39909740,263665252,

%T 1765815560,11966535091,81937361702,566185489878,3944202596652,

%U 27676632525362,195481707009220,1388890568962556

%N Antidiagonal sums of square array A111910.

%H Michael De Vlieger, <a href="/A196148/b196148.txt">Table of n, a(n) for n = 0..1000</a>

%H Anthony James Wood, <a href="https://hdl.handle.net/1842/36698">Nonequilibrium steady states from a random-walk perspective</a>, Ph. D. Thesis, The University of Edinburgh (Scotland, UK 2019).

%H Anthony J. Wood, Richard A. Blythe, and Martin R. Evans, <a href="https://arxiv.org/abs/1708.00303">Renyi entropy of the totally asymmetric exclusion process</a>, arXiv:1708.00303 [cond-mat.stat-mech], 2017.

%H Anthony J. Wood, Richard A. Blythe, and Martin R. Evans, <a href="https://arxiv.org/abs/1908.00942">Combinatorial mappings of exclusion processes</a>, arXiv:1908.00942 [cond-mat.stat-mech], 2019.

%F a(n) = Sum_{k = 0..n} S(n-k,k) where S(n,k) = (n+k+1)!*(2*n+2*k+1)!/((n+1)!*(k+1)!*(2*n+1)!*(2*k+1)!).

%F From _Vaclav Kotesovec_, Dec 16 2017: (Start)

%F a(n) ~ 2^(3*n+3) / (sqrt(3*Pi) * n^(5/2)).

%F Recurrence: (n+2)*(2*n+3)*a(n) = 2*(7*n^2 + 7*n + 1)*a(n-1) + 8*(n-1)*(2*n-1)*a(n-2). (End)

%F a(n) = hypergeometric3F2([-n, -n-1/2, -n-1], [3/2, 2], -1). - _G. C. Greubel_, Feb 11 2021

%F Let E(x) = Sum_{n >= 0} x^n/((n+1)!*(2*n+1)!). Then E(x)^2 = 1 + 2*x/(2!*3!) + 7*x^2/(3!*5!) + 30*x^3/(4!*7!) + ... + a(n)*x^n/((n+1)!*(2*n+1)!) + ... is a generating function for the sequence. - _Peter Bala_, Sep 20 2021

%t Table[Sum[(n+1)! * (2*n+1)! / ((n-k+1)! * (k+1)! * (2*n-2*k+1)! * (2*k+1)!), {k,0,n}], {n,0,20}] (* _Vaclav Kotesovec_, Dec 16 2017 *)

%t Table[HypergeometricPFQ[{-n, -n-1/2, -n-1}, {3/2, 2}, -1], {n,0,25}] (* _G. C. Greubel_, Feb 11 2021 *)

%o (PARI) S(n,k) = (n+k+1)!*(2*n+2*k+1)!/((n+1)!*(k+1)!*(2*n+1)!*(2*k+1)!);

%o a(n) = sum(k = 0, n, S(n-k,k)); \\ _Michel Marcus_, Dec 16 2017

%o (Sage) [hypergeometric([-n, -n-1/2, -n-1], [3/2, 2], -1).simplify_hypergeometric() for n in (0..25)] # _G. C. Greubel_, Feb 11 2021

%o (Magma) [(&+[(n-j+1)*Binomial(n+1, j)*Binomial(2*n+4, 2*j+2)/((n+1)*(n+2)*(2*n+3)): j in [0..n]]): n in [0..25]]; // _G. C. Greubel_, Feb 11 2021

%Y Cf. A111910.

%Y Cf. A174119.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Oct 13 2011