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 A060524 Triangle read by rows: T(n,k) = number of degree-n permutations with k odd cycles, k=0..n, n >= 0. 14
 1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 9, 0, 14, 0, 1, 0, 89, 0, 30, 0, 1, 225, 0, 439, 0, 55, 0, 1, 0, 3429, 0, 1519, 0, 91, 0, 1, 11025, 0, 24940, 0, 4214, 0, 140, 0, 1, 0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1, 893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1, 0 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,8 COMMENTS The row polynomials t(n,x):=Sum_{k=0..n} T(n,k)*x^k satisfy the recurrence relation t(n,x) = x*t(n-1,x) + ((n-1)^2)*t(n-2,x); t(-1,x)=0, t(0,x)=1. - Wolfdieter Lang, see above. This is an example of a Sheffer triangle (coefficient triangle for Sheffer polynomials). In the umbral calculus (see the Roman reference given under A048854) s(n,x) := Sum_{k=0..n} T(n,k)*x^k would be called Sheffer polynomials for (1/cosh(t),tanh(t)), which translates to the e.g.f. for column number k>=0 given by (1/sqrt(1-x^2))*((arctanh(x))^k)/k!. The e.g.f. given below is rewritten in this Sheffer context as (1/sqrt(1-x^2))*exp(y*log(sqrt((1+x)/(1-x))))= (1/sqrt(1-x^2))*exp(y*arctanh(x)). The rows of the Jabotinsky type triangle |A049218| provide the coefficients of the associated polynomials. - Wolfdieter Lang, Feb 24 2005 The solution of the differential-difference relation f(n+1,x)= (d/dx)f(n,x) + (n^2)*f(n-1,x), n >= 1, with inputs f(0,x) and f(1,x) = (d/dx)f(0,x) is f(n,x) = t(n,d_x)*f(0,x), with the differential operator d_x:=d/dx and the row polynomials t(n,x) defined above. This problem appears in a computation of thermo field dynamics where f(0,x)=1/cosh(x). See the triangle A060081. - Wolfdieter Lang, Feb 24 2005 The inverse of the Sheffer matrix T with elements T(n,k) is the Sheffer matrix A060081. - Wolfdieter Lang, Jul 22 2005 T(n,k)=0 if n-k= 1(mod 2), else T(n,k) = sum of M2(n,p), p from {1,...,A000041(n)} restricted to partitions with exactly k odd parts and any nonnegative number of even parts. For the M2-multinomial numbers in A-St order see A036039(n,p). - Wolfdieter Lang, Aug 07 2007 LINKS Alois P. Heinz, Rows n = 0..140, flattened Peter Bala, Meixner polynomials and Brouncker's continued fraction for Pi Paul L. Butzer and Tom H. Koornwinder, Josef Meixner: His life and his orthogonal polynomials, Indagationes Mathematicae, Volume 30, Issue 1, January 2019, Pages 250-264. A. Hamdi and J. Zeng, Orthogonal polynomials and operator orderings, J. Math. Phys., 51:043506, (2010); arXiv:1006.0808 [math.CO], 2010. Eric Weisstein's World of Mathematics, Meixner polynomial of the first kind. FORMULA E.g.f.: (1+x)^((y-1)/2)/(1-x)^((y+1)/2). T(n, k) = T(n-1, k-1) + ((n-1)^2)*T(n-2, k); T(-1, k):=0, T(n, -1):=0, T(0, 0)=1, T(n, k)=0 if n= 0 there holds ((2*n + 1)!!/(2^n*n!))^2 * Pi = (4*n + 3) + 4*((2*n + 1)!!^4) * Sum_{k >= 1} (-1)^(k+1)/((2*k + 1)*R(2*n+1, 2*k)*R(2*n+1, 2*k+2)). Cf. A142979 and A142983. R(2*n, 0) = A001147(n)^2 = A001818(n); R(2*n+1, 0) = 0. R(n, 1) = n! = A000142(n). R(2*n, 2) = (4*n + 1)*A001147(n)^2 = (4*n + 1)*((2*n)!/(2^n*n!))^2; R(2*n+1, 2) = 2*A001447(n+1)^2 = 2*(2*n + 1)!^2/(n!^2*4^n). R(n, 3) = (2*n + 1)*n! = A007680(n). (End) EXAMPLE Triangle begins: [1], [0, 1], [1, 0, 1], [0, 5, 0, 1], [9, 0, 14, 0, 1], [0, 89, 0, 30, 0, 1], [225, 0, 439, 0, 55, 0, 1], [0, 3429, 0, 1519, 0, 91, 0, 1], [11025, 0, 24940, 0, 4214, 0, 140, 0, 1], [0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1], [893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1], [0, 23941125, 0, 14466221, 0, 1467290, 0, 41778, 0, 385, 0, 1], ... Signed version begins: [1], [0, 1], [-1, 0, 1], [0, -5, 0, 1], [9, 0, -14, 0, 1], [0, 89, 0, -30, 0, 1], [-225, 0, 439, 0, -55, 0, 1], [0, -3429, 0, 1519, 0, -91, 0, 1], ... From Peter Bala, Feb 23 2024: (Start) Maple can verify the following series for Pi: Row 1 polynomial R(1, x) = x: Pi = 3 + 4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(1, 2*n)*R(1, 2*n+2)). Row 3 polynomial R(3, x) = 5*x + x^3: (3/2)^2 * Pi = 7 + 4*(3^4)*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(3, 2*n)*R(3, 2*n+2)). Row 5 polynomial R(5, x) = 89*x + 30*x^3 + x^5: ((3*5)/(2*4))^2 * Pi = 11 + 4*(3*5)^4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(5, 2*n)*R(5, 2*n+2)). (End) MAPLE with(combinat): b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0, add(multinomial(n, n-i*j, i\$j)*(i-1)!^j/j!*b(n-i*j, i-1)* `if`(irem(i, 2)=1, x^j, 1), j=0..n/i)))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n\$2)): seq(T(n), n=0..12); # Alois P. Heinz, Mar 09 2015 # alternative A060524 := proc(n, k) option remember; if n

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