K = 1) the technique of recurrence relations is from the second order, Equation (12). The system in Equation (27) makes it possible for one particular to compute the whole permanental polynomial, per C =j0 ,j1 =nPj0 ,j1 c00 c11 ,jj(36)as a function of two valiables (c0 , c1 ) for any significant matrix size n and to seek out its asymptotics analytically. The relevant results are going to be presented elsewhere. Remark 7. Renowned m age numbers [52,53,557,591] Un counting the amount of unique strategies to seat n husbands at a circular table of 2n areas so that males and ladies alternate and no adjacent couples are allowed, that’s, the number of 2-discordant permutations of 1, …, n such that ( j) is just not congruent to any j, j + 1 (mod n), are equal for the permanent on the D-Tyrosine Purity uniform circulant matrix C with a band of two zero diagonals. Which is, they are equal to a certain worth on the permanent when each variables are zero, c0 = c1 = 0, that is the continuous term of your permanental polynomial in Equation (36): Un = Cn |c0 =c1 =0 = P0,0 . (37)A well-known fourth-order recurrence equation for the m age numbers [53,59]: Un = nUn-1 + 2Un-2 – (n – four)Un-3 – Un-4 , (38)right away follows in the recurrence relations for the permanent Cn , Equation (27), as its specific case. Namely, for c0 = c1 = 0 the permanent, and hence, the m age numbers (the sequence A000179 in [56]) are Licoflavone B Epigenetics provided by a very simple formula Un = Cn = An – An-1 (39)via the permanent An with the matrix with the defect A, Equation (four), which coincides with the straight m age number Vn , counting the amount of permutations of 1, …, n such that ( j) is just not congruent to any j, min j + 1, n (mod n) (that corresponds to an analogous trouble of seating n husbands at a straight-line table), and satisfies the thirdorder recurrence (the sequence A000271 in [56]): An = (n – 1)( An-1 + An-2 ) + An-3 . (40)Entropy 2021, 23,ten ofThe latter immediately follows from Equation (27) due to the fact A n = A n + A n -(1)A(n) = A()(n) = An + An-1 + An-for c0 = c1 = 0.(41)As a result, the method of the recursion of permanents with defects offers a basic derivation of your recurrence (40) for the straight m age numbers Vn = An |c0 =c1 =0 , which can be different from a long combinatorial derivation (see Theorem 1 in [53]). six. The Permanent of a Uniform Circulant Matrix having a Band of Three Any-Value Diagonals (k = 3) along with the 3-M age Numbers Ultimately, we think about a really complicated case of a uniform circulant n n matrix using a band of 3 (k = three) diagonals (with all the entry values c0 , c1 , c2 ) inside the matrix J of all 1s. It can be depicted as C in Equation (three) and C in Equation (five) under. c1 c0 1 . . . c2 c1 c0 . . . 1 1 1 1 c2 c1 . . . 1 1C = Circ(c1 , c2 , 1, …, 1, c0 ) = 1 1 c .. .1 1 1 . . .1 1 1 . . . c2 c1 cc1 c0 . 1 c2 cc0 1 1 . . .(42)Theorem 1. The permanent of this circulant with a band of 3 diagonals isCn per Circ(c0 , c1 , c2 , 1, …, 1) = Bn + (c0 – 1) An-1 + (c2 – 1) A()n-1 + (c0 – 1)(c2 – 1) Bn-2 (43)and is determined by a resolution of your following program of recurrence relations: An = A(n) – (1 – c2 ) An-1 + (c0 + c1 – 1) An-1 – (1 – c0 )(1 – c1 ) An-2 , Bn = B(n)(1) (two) (two)(44) (45)+c1 Bn-1 – (1 – c0 ) An-1 – (1 – c2 ) A()n-1 -(1 – c0 )2 An-2 – (1 – c2 )two A()n-2 + (1 – c0 )(1 – c2 ) Bn-2 ,A(n) =(n – 3)Cn-1 + An-1 + An-1 + 2(1 – c2 ) A()(n-1) + (1 – c1 ) B(n-1) +(1 – c0 ) An-2 + (c1 + c2 – two) A()n-2 – c2 (1 – c2 ) A()(n-2) – two(1 – c0 )(1 – c2 ) B(n-2) -(2 + c0 – c1 – c2 )(1 – c2 ) A()n-3 + (1 – c0 )(1 – c2 )two A()n-4 ,B(n) =(1)(46.