A282252 Exponential Riordan array [Bessel_I(0,2*x)^2, x].
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, 2800, 0, 1260, 0, 84, 0, 1, 4900, 0, 11200, 0, 2520, 0, 112, 0, 1, 0, 44100, 0, 33600, 0, 4536, 0, 144, 0, 1, 63504, 0, 220500, 0, 84000, 0, 7560, 0, 180, 0, 1
Offset: 0
Examples
The triangle begins 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; ... 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 loop L R, loop R L, loop U D, loop D U, L loop R, R loop L, U loop D, D loop U, L R loop, R L loop, U D loop, D U loop. The infinitesimal generator of this array has integer entries and begins 0; 0, 0; 4, 0, 0; 0, 12, 0, 0; -12, 0, 24, 0, 0; 0, -60, 0, 40, 0, 0; 160, 0, -180, 0, 60, 0, 0; 0, 1120, 0, -420, 0, 84, 0, 0; -4620, 0, 4480, 0, -840, 0, 112, 0, 0; ... It is the generalized exponential Riordan array [ 2*log(Bessel_I(0,2*x)), x ].
Links
- G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Programs
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Maple
T := (n, k) -> (1/2)*binomial(n, k)*binomial(n-k, floor((1/2)*n-(1/2)*k))^2*(1+(-1)^(n-k)): seq(seq(T(n, k), k = 0..n), n = 0..9);
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Mathematica
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 *)
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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
Formula
T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))^2*(1 + (-1)^(n-k))/2.
T(n,n-2*k) = n/(n - 2*k)*T(n-1,n-2*k-1).
T(n,k) = the coefficient of t^k in the expansion of (t + X + 1/X + Y + 1/Y)^n.
T(n,k) = 1/Pi^2 * Integral_{y = 0..Pi} Integral_{x = 0..Pi} ( t + 2*cos(x) + 2*cos(y) )^n dx dy.
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! + ....
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).
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.
d/dt(R(n,t)) = n*R(n-1,t).
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.
Comments