A336286 The hafnian of a symmetric Toeplitz matrix of order 2*n, n>=2 with the first row (0,1,2,...,2,0); a(0)=a(1)=1.
1, 1, 5, 57, 859, 16087, 362781, 9593105, 291347603, 9998539791, 382732896853, 16169762600329, 747423640472235, 37523173542935207, 2033249827596197549, 118278700627740322977, 7352204062275501662371, 486343759162888783503775, 34112193002666850227154213
Offset: 0
Examples
A symmetric 4 X 4 Toeplitz matrix A with the first row (0,1,2,0) has the form: 0 1 2 0 1 0 1 2 2 1 0 1 0 2 1 0. Its hafnian equals Hf(A) = a12*a34 + a13*a24 + a14*a23 = 1*1 + 2*2 + 0*1 = 5 = a(2).
Links
- Dmitry Efimov, The hafnian of Toeplitz matrices of a special type, perfect matchings and Bessel polynomials, arXiv:1904.08651 [math.CO], 2020.
Programs
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Maple
[1,1,seq(add((-1)^(n-k-1)*(n+k-1)!*(-3*n+k)/(k!*(n-k)!),k=0..n),n=2..32)] # Georg Fischer, Jun 05 2021
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Mathematica
Join[{1,1},RecurrenceTable[{a[n+1] == (4*n+4)*a[n]-(8*n-13)*a[n-1]-2*a[n-2], a[2]==5, a[3]==57, a[4]==859}, a[n], {n,2,32}]] (* Georg Fischer, Jun 05 2021 *)
Formula
a(n) = Sum_{k=0..n} (-1)^(n-k)*(n+k-1)!*(3*n-k)/(k!*(n-k)!), n>=2.
D-finite with recurrence a(n+1) = (4n+4)*a(n) - (8n-13)*a(n-1) - 2*a(n-2), n>=4.
D-finite with recurrence a(n+1) = ((32*n^2-12*n+2)*a(n) + (8*n+1)*a(n-1))/(8*n-7), n>=3.
a(n) ~ (2*n)!/sqrt(e)*n!.
Comments