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%I #20 Aug 21 2017 05:49:19
%S 1,0,1,4,0,1,0,12,0,1,36,0,24,0,1,0,180,0,40,0,1,400,0,540,0,60,0,1,0,
%T 2800,0,1260,0,84,0,1,4900,0,11200,0,2520,0,112,0,1,0,44100,0,33600,0,
%U 4536,0,144,0,1,63504,0,220500,0,84000,0,7560,0,180,0,1
%N Exponential Riordan array [Bessel_I(0,2*x)^2, x].
%C Bessel_I(0,2*x) = Sum_{n >= 0} binomial(2*n,n)*x^(2*n)/(2*n)! is a modified Bessel function of the first kind.
%C Consider the infinite 2-dimensional square lattice Z x Z with an oriented self-loop at each vertex. Then the triangle entry T(n,k) equals the number of walks of length n from the origin to itself having k loops. An example is given below.
%C See A069466 for walks an infinite 2-dimensional square lattice without self-loops.
%C This is the square of triangle A109187, whose entries give the number of walks of length n from a vertex to itself having k loops on a 1-dimensional integer lattice with an oriented self-loop at each vertex.
%C A109187 is the exponential Riordan array [Bessel_I(0,2*x), x]. Note that Bessel_I(0,2*x)^2 = (Sum_{n >= 0} binomial(2*n,n)* x^(2*n)/(2*n)!)^2 = Sum_{n >= 0} binomial(2*n,n)^2*x^(2*n) /(2*n)!.
%H G. C. Greubel, <a href="/A282252/b282252.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%F T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))^2*(1 + (-1)^(n-k))/2.
%F T(n,n-2*k) = n/(n - 2*k)*T(n-1,n-2*k-1).
%F T(n,k) = the coefficient of t^k in the expansion of (t + X + 1/X + Y + 1/Y)^n.
%F T(n,k) = 1/Pi^2 * Integral_{y = 0..Pi} Integral_{x = 0..Pi} ( t + 2*cos(x) + 2*cos(y) )^n dx dy.
%F E.g.f.: exp(x*t)*Bessel_I(0,2*x)^2 = 1 + t*x + (4 + t^2)*x^2/2! + (12*t + t^3)*x^3/3! + (36 + 24*t^2 + t^4)*x^4/4! + ....
%F The n-th row polynomial R(n,t) = Sum_{k = 0..floor(n/2)} binomial(n,2*k)*binomial(2*k,k)^2 * t^(n-2*k).
%F Recurrence: n^2*R(n,t) = t*(3*n^2 - 3*n + 1)*R(n-1,t) + (16 - 3*t^2)*(n - 1)^2*R(n-2,t) + t*(t^2 - 16)*(n - 1)*(n - 2)*R(n-3,t) with R(0,t) = 1, R(1,t) = t and R(2,t) = 4 + t^2.
%F d/dt(R(n,t)) = n*R(n-1,t).
%F The zeros of the row polynomials appear to lie on the imaginary axis in the complex plane. Also, the zeros of R(n,t) and R(n+1,t) appear to interlace on the imaginary axis.
%e The triangle begins
%e 1;
%e 0, 1;
%e 4, 0, 1;
%e 0, 12, 0, 1;
%e 36, 0, 24, 0, 1;
%e 0, 180, 0, 40, 0, 1;
%e 400, 0, 540, 0, 60, 0, 1;
%e ...
%e T(3,1) = 12: on the square lattice, let L, R, U, D denote a left step, right step, up step and down step respectively. The 12 walks of length 3 containing a single loop are
%e loop L R, loop R L, loop U D, loop D U,
%e L loop R, R loop L, U loop D, D loop U,
%e L R loop, R L loop, U D loop, D U loop.
%e The infinitesimal generator of this array has integer entries and begins
%e 0;
%e 0, 0;
%e 4, 0, 0;
%e 0, 12, 0, 0;
%e -12, 0, 24, 0, 0;
%e 0, -60, 0, 40, 0, 0;
%e 160, 0, -180, 0, 60, 0, 0;
%e 0, 1120, 0, -420, 0, 84, 0, 0;
%e -4620, 0, 4480, 0, -840, 0, 112, 0, 0;
%e ...
%e It is the generalized exponential Riordan array [ 2*log(Bessel_I(0,2*x)), x ].
%p T := (n, k) -> (1/2)*binomial(n, k)*binomial(n-k, floor((1/2)*n-(1/2)*k))^2*(1+(-1)^(n-k)):
%p seq(seq(T(n, k), k = 0..n), n = 0..9);
%t Table[Binomial[n, k] Binomial[n - k, Floor[(n - k)/2]]^2*(1 + (-1)^(n - k))/2, {n, 0, 10}, {k, 0, n}] // Flatten (* _Michael De Vlieger_, Feb 12 2017 *)
%o (PARI) for(n=0,10, for(k=0,n, print1(binomial(n,k)*binomial(n-k,floor((n-k)/2))^2*(1 + (-1)^(n-k))/2, ", "))) \\ _G. C. Greubel_, Aug 16 2017
%Y A201805 gives row sums. Cf. A069466, A109187.
%K nonn,tabl,easy
%O 0,4
%A _Peter Bala_, Feb 12 2017
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