A156289 Triangle read by rows: T(n,k) is the number of end rhyme patterns of a poem of an even number of lines (2n) with 1<=k<=n evenly rhymed sounds.
1, 1, 3, 1, 15, 15, 1, 63, 210, 105, 1, 255, 2205, 3150, 945, 1, 1023, 21120, 65835, 51975, 10395, 1, 4095, 195195, 1201200, 1891890, 945945, 135135, 1, 16383, 1777230, 20585565, 58108050, 54864810, 18918900, 2027025, 1, 65535, 16076985
Offset: 1
Examples
The triangle begins n\k|..1.....2......3......4......5......6 ========================================= .1.|..1 .2.|..1.....3 .3.|..1....15.....15 .4.|..1....63....210....105 .5.|..1...255...2205...3150....945 .6.|..1..1023..21120..65835..51975..10395 .. T(3,3) = 15. The 15 partitions of the set [6] into three even blocks are: (12)(34)(56), (12)(35)(46), (12)(36)(45), (13)(24)(56), (13)(25)(46), (13)(26)(45), (14)(23)(56), (14)(25)(36), (14)(26)(35), (15)(23)(46), (15)(24)(36), (15)(26)(34), (16)(23)(45), (16)(24)(35), (16)(25)(34). Examples of recurrence relation T(4,3) = 5*T(3,2) + 9*T(3,3) = 5*15 + 9*15 = 210; T(6,5) = 9*T(5,4) + 25*T(5,5) = 9*3150 + 25*945 = 51975. T(4,2) = 28 + 35 = 63 (M_3 multinomials A036040 for partitions of 8 with 3 even parts, namely (2,6) and (4^2)). - _Wolfdieter Lang_, May 13 2015
References
- L. Comtet, Analyse Combinatoire, Presses Univ. de France, 1970, Vol. II, pages 61-62.
- L. Comtet, Advanced Combinatorics, Reidel, 1974, pages 225-226.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..11325 (rows 1 <= n <= 150, flattened)
- Richard Olu Awonusika, On Jacobi polynomials P_k(alpha, beta) and coefficients c_j^L(alpha, beta) (k >= 0, L = 5,6; 1 <= j <= L; alpha, beta > -1), The Journal of Analysis (2020).
- Thomas Browning, Counting Parabolic Double Cosets in Symmetric Groups, arXiv:2010.13256 [math.CO], 2020.
- J. Riordan, Letter, Jul 06 1978
Crossrefs
Programs
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Maple
T := proc(n,k) option remember; `if`(k = 0 and n = 0, 1, `if`(n < 0, 0, (2*k-1)*T(n-1, k-1) + k^2*T(n-1, k))) end: for n from 1 to 8 do seq(T(n,k), k=1..n) od; # Peter Luschny, Sep 04 2017
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Mathematica
T[n_,k_] := Which[n < k, 0, n == 1, 1, True, 2/Factorial2[2 k] Sum[(-1)^(k + j) Binomial[2 k, k + j] j^(2 n), {j, 1, k}]] (* alternate computation with function triangle[] defined in A257490 *) a[n_]:=Map[Apply[Plus,#]&,triangle[n],{2}] (* Hartmut F. W. Hoft, Apr 26 2015 *)
Formula
Recursion: T(n,1)=1 for 1<=n; T(n,k)=0 for 1<=n
Generating function for the k-th column of the triangle T(i+k,k):
G(k,x) = Sum_{i>=0} T(i+k,k)*x^i = Product_{j=1..k} (2*j-1)/(1-j^2*x).
Closed form expression: T(n,k) = (2/(k!*2^k))*Sum_{j=1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).
From Peter Bala, Feb 21 2011: (Start)
GENERATING FUNCTION
E.g.f. (including a constant 1):
(1)... F(x,z) = exp(x*(cosh(z)-1))
= Sum_{n>=0} R(n,x)*z^(2*n)/(2*n)!
= 1 + x*z^2/2! + (x + 3*x^2)*z^4/4! + (x + 15*x^2 + 15*x^3)*z^6/6! + ....
ROW POLYNOMIALS
The row polynomials R(n,x) begin
... R(1,x) = x
... R(2,x) = x + 3*x^2
... R(3,x) = x + 15*x^2 + 15*x^3.
The egf F(x,z) satisfies the partial differential equation
(2)... d^2/dz^2(F) = x*F + x*(2*x+1)*F' + x^2*F'',
where ' denotes differentiation with respect to x. Hence the row polynomials satisfy the recurrence relation
(3)... R(n+1,x) = x*{R(n,x) + (2*x+1)*R'(n,x) + x*R''(n,x)}
with R(0,x) = 1. The recurrence relation for T(n,k) given above follows from this.
(4)... T(n,k) = (2*k-1)!!*A036969(n,k).
(End)
Comments