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A039755
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Triangle of B-analogs of Stirling numbers of the second kind.
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34
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1, 1, 1, 1, 4, 1, 1, 13, 9, 1, 1, 40, 58, 16, 1, 1, 121, 330, 170, 25, 1, 1, 364, 1771, 1520, 395, 36, 1, 1, 1093, 9219, 12411, 5075, 791, 49, 1, 1, 3280, 47188, 96096, 58086, 13776, 1428, 64, 1, 1, 9841, 239220, 719860, 618870, 209622, 32340, 2388, 81, 1, 1
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OFFSET
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0,5
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COMMENTS
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Let M be an infinite lower triangular bidiagonal matrix with (1,3,5,7,...) in the main diagonal and (1,1,1,...) in the subdiagonal. n-th row = M^n * [1,0,0,0,...]. - Gary W. Adamson, Apr 13 2009
A type B_n set partition is a partition P of the set {1, 2, ..., n, -1, -2, ..., -n} such that for any block B of P, -B is also a block of P, and there is at most one block, called a zero-block, satisfying B = -B. We call (B, -B) a block pair of P if B is not a zero-block. Then T(n,k) is the number of type B_n set partitions with k block pairs. See [Wang].
For example, T(2,1) = 4 since the B_2 set partitions with 1 block pair are {1,2}{-1,-2}, {1,-2}{-1,2}, {1,-1}{2}{-2} and {2,-2}{1}{-1} (the last two partitions contain a zero block).
(End)
Exponential Riordan array [exp(x), (1/2)*(exp(2*x) - 1)]. Triangle of connection constants for expressing the monomial polynomials x^n as a linear combination of the basis polynomials (x-1)*(x-3)*...*(x-(2*k-1)) of A039757. An example is given below. Inverse array is A039757. Equals matrix product A008277 * A122848. - Peter Bala, Jun 23 2014
T(n, k) also gives the (dimensionless) volume of the multichoose(k+1, n-k) = binomial(n, k) polytopes of dimension n-k with side lengths from the set {1, 3, ..., 1+2*k}. See the column g.f.s and the complete homogeneous symmetric function formula for T(n, k) below. - Wolfdieter Lang, May 26 2017
T(n, k) is the number of k-dimensional subspaces (i.e., sets of fixed points like rotation axes and symmetry planes) of the n-cube. See "Sets of fixed points..." in LINKS section. - Tilman Piesk, Oct 26 2019
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LINKS
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T. Piesk, Sets of fixed points of permutations of the n-cube: n = 3 and 4.
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FORMULA
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E.g.f. row polynomials: exp(x + y/2 * (exp(2*x) - 1)).
T(n,k) = T(n-1,k-1) + (2*k+1)*T(n-1,k) with T(0,k) = 1 if k=0 and 0 otherwise. Sum_{k=0..n} T(n,k) = A007405(n). - R. J. Mathar, Oct 30 2009; corrected by Joshua Swanson, Feb 14 2019
T(n,k) = (1/(2^k*k!)) * Sum_{j=0..k} (-1)^(k-j)*C(k,j)*(2*j+1)^n.
The row polynomials R(n,x) satisfy the Dobinski-type identity:
R(n,x) = exp(-x/2)*Sum_{k >= 0} (2*k+1)^n*(x/2)^k/k!, as well as the recurrence equation R(n+1,x) = (1+x)*R(n,x)+2*x*R'(n,x). The polynomial R(n,x) has all real zeros (apply [Liu et al., Theorem 1.1] with f(x) = R(n,x) and g(x) = R'(n,x)). The polynomials R(n,2*x) are the row polynomials of A154537. - Peter Bala, Oct 28 2011
The o.g.f. for the n-th diagonal (with interpolated zeros) is the rational function D^n(x), where D is the operator x/(1-x^2)*d/dx. For example, D^3(x) = x*(1+8*x^2+3*x^4)/(1-x^2)^5 = x + 13*x^3 + 58*x^5 + 170*x^7 + ... . See A214406 for further details.
An alternative formula for the o.g.f. of the n-th diagonal is exp(-x/2)*(Sum_{k >= 0} (2*k+1)^(k+n-1)*(x/2*exp(-x))^k/k!).
(End)
T(n,m) = Sum_{i=0..n-m} 2^(n-m-i)*binomial(n,i)*St2(n-i,m), where St2(n,k) are the Stirling numbers of the second kind, A048993 (also A008277). See p. 755 of Dolgachev and Lunts.
The relation of this entry's e.g.f. above to that of the Bell polynomials, Bell_n(y), of A048993 establishes this formula from a binomial transform of the normalized Bell polynomials, NB_n(y) = 2^n Bell_n(y/2); that is, e^x exp[(y/2)(e^(2x)-1)] = e^x exp[x*2*Bell.(y/2)] = exp[x(1+NB.(y))] = exp(x*P.(y)), so the row polynomials of this entry are given by P_n(y) = [1+NB.(y)]^n = Sum_{k=0..n} C(n,k) NB_k(y) = Sum_{k=0..n} 2^k C(n,k) Bell_k(y/2).
The umbral compositional inverses of the Bell polynomials are the falling factorials Fct_n(y) = y! / (y-n)!; i.e., Bell_n(Fct.(y)) = y^n = Fct_n(Bell.(y)). Since P_n(y) = [1+2Bell.(y/2)]^n, the umbral inverses are determined by [1 + 2 Bell.[ 2 Fct.[(y-1)/2] / 2 ] ]^n = [1 + 2 Bell.[ Fct.[(y-1)/2] ] ]^n = [1+y-1]^n = y^n. Therefore, the umbral inverse sequence of this entry's row polynomials is the sequence IP_n( y) = 2^n Fct_n[(y-1)/2] = (y-1)(y-3) .. (y-2n+1) with IP_0(y) = 1 and, from the binomial theorem, with e.g.f. exp[x IP.(y)]= exp[ x 2Fct.[(y-1)/2] ] = (1+2x)^[(y-1)/2] = exp[ [(y-1)/2] log(1+2x) ].
(End)
Let B(n,k) = T(n,k)*((2*k)!)/(2^k*k!) and P(n,x) = Sum_{k=0..n} B(n,k)*x^(2*k+1). Then (1) P(n+1,x) = (x+x^3)*P'(n,x) for n >= 0, and (2) Sum_{n>=0} B(n,k)/(n!)*t^n = binomial(2*k,k)*exp(t)*(exp(2*t)-1)^k/4^k for k >= 0, and (3) Sum_{n>=0} t^n* P(n,x)/(n!) = x*exp(t)/sqrt(1+x^2-x^2*exp(2*t)). - Werner Schulte, Dec 12 2016
G.f. column k: x^k/Product_{j=0..k} (1 - (1+2*j)*x), k >= 0.
T(n, k) = h^{(k+1)}_{n-k}, the complete homogeneous symmetric function of degree n-k of the k+1 symbols a_j = 1 + 2*j, j = 0, 1, ..., k. (End)
With p(n, x) = Sum_{k=0..n} A001147(k) * T(n, k) * x^k for n >= 0 holds:
(1) Sum_{i=0..n} p(i, x)*p(n-i, x) = 2^n*(Sum_{k=0..n} A028246(n+1, k+1)*x^k);
(2) p(n, -1/2) = (n!) * ([t^n] sqrt(2 / (1 + exp(-2*t)))). - Werner Schulte, Feb 16 2024
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EXAMPLE
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Triangle T(n,k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
0: 1
1: 1 1
2: 1 4 1
3: 1 13 9 1
4: 1 40 58 16 1
5: 1 121 330 170 25 1
6: 1 364 1771 1520 395 36 1
7: 1 1093 9219 12411 5075 791 49 1
8: 1 3280 47188 96096 58086 13776 1428 64 1
9: 1 9841 239220 719860 618870 209622 32340 2388 81 1
10: 1 29524 1205941 5278240 6289690 2924712 630042 68160 3765 100 1
The sequence of row polynomials of A214406 begins [1, 1+x, 1+8*x+3*x^2, ...]. The o.g.f.'s for the diagonals of this triangle thus begin
1/(1-x) = 1 + x + x^2 + x^3 + ...
(1+x)/(1-x)^3 = 1 + 4*x + 9*x^2 + 16*x^3 + ...
(1+8*x+3*x^2)/(1-x)^5 = 1 + 13*x + 58*x^2 + 170*x^3 + ... . - Peter Bala, Jul 20 2012
Connection constants: x^3 = 1 + 13*(x-1) + 9*(x-1)*(x-3) + (x-1)*(x-3)*(x-5). Hence row 3 = [1,13,9,1]. - Peter Bala, Jun 23 2014
Complete homogeneous symmetric functions: T(3, 1) = h^{(2)}_2 = 1^2 + 3^2 + 1^1*3^1 = 13. The three 2D polytopes are two squares and a rectangle. T(3, 2) = h^{(3)}_1 = 1^1 + 3^1 + 5^1 = 9. The 1D polytopes are three lines. - Wolfdieter Lang, May 26 2017
T(4, 3) = 16 is the number of 3-dimensional subspaces (mirror hyperplanes) of the 4-cube. (These are 4 cubes and 12 cuboids.) See "Sets of fixed points..." in LINKS section. - Tilman Piesk, Oct 26 2019
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MAPLE
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A039755 := proc(n, k) if k < 0 or k > n then 0 ; elif n <= 1 then 1; else procname(n-1, k-1)+(2*k+1)*procname(n-1, k) ; end if; end proc:
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MATHEMATICA
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t[n_, k_] = Sum[(-1)^(k-j)*(2j+1)^n*Binomial[k, j], {j, 0, k}]/(2^k*k!); Flatten[Table[t[n, k], {n, 0, 10}, {k, 0, n}]][[1 ;; 56]]
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PROG
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(PARI) T(n, k)=if(k<0 || k>n, 0, n!*polcoeff(polcoeff(exp(x+y/2*(exp(2*x+x*O(x^n))-1)), n), k))
(Magma) [[(&+[(-1)^(k-j)*(2*j+1)^n*Binomial(k, j): j in [0..k]])/( 2^k*Factorial(k)): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 14 2019
(Sage) [[sum((-1)^(k-j)*(2*j+1)^n*binomial(k, j) for j in (0..k))/( 2^k*factorial(k)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 14 2019
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CROSSREFS
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KEYWORD
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AUTHOR
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Ruedi Suter (suter(AT)math.ethz.ch)
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STATUS
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approved
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