| Displaying 1-5 of 5 results found.
page
1
Coefficient triangle for polynomials used for o.g.f.s for unsigned Stirling1 diagonals.
+10
29
1, 2, 1, 6, 8, 1, 24, 58, 22, 1, 120, 444, 328, 52, 1, 720, 3708, 4400, 1452, 114, 1, 5040, 33984, 58140, 32120, 5610, 240, 1, 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 362880, 3733920, 11026296, 12440064, 5765500, 1062500, 67260, 1004, 1
COMMENTS
This is the row reversed second-order Eulerian triangle A008517(k+1,k+1-m). For references see A008517. The o.g.f. for the k-th diagonal, k >= 1, of the unsigned Stirling1 triangle | A008275| is G1(1,x)=1/(1-x) if k=1 and G1(k,x) = g1(k-2,x)/(1-x)^(2*k-1), if k >= 2, with the row polynomials g1(k;x):=Sum_{m=0..k} a(k,m)*x^m. The recurrence eq. for the row polynomials is g1(k,x)=((k+1)+k*x))*g1(k-1,x) + x*(1-x)*(d/dx)g1(k-1,x), k >= 1, with input g1(0,x):=1.
This o.g.f. computation was inspired by Bender et al. article where the Stirling polynomials have been rediscussed.
The A163936 triangle is identical to the triangle given above except for an extra right hand column [1, 0, 0, 0, ... ]. The A163936 triangle is related to the higher order exponential integrals E(x,m,n), see A163931 and A163932. - Johannes W. Meijer, Oct 16 2009
FORMULA
a(k, m) = (k+m+1)*a(k-1, m) + (k-m+1)*a(k-1, m-1), if k >= m >= 0, a(0, 0)=1; a(k, -1):=0, otherwise 0.
a(k,m) = Sum_{n=0..m} (-1)^(k+n+1)*C(2*k+3,n)*Stirling1(m+k-n+2,m+1-n). - Johannes W. Meijer, Oct 16 2009 The compositional inverse (with respect to x) of y = y(t,x) = (x+t*log(1-x)) is x = x(t,y) = 1/(1-t)*y + t/(1-t)^3*y^2/2! + (2*t+t^2)/(1-t)^5*y^3/3! + (6*t+8*t^2+t^3)/(1-t)^7*y^4/4! + .... The numerator polynomials of the rational functions in t are the row polynomials of this triangle. As observed above, the rational functions in t are the generating functions for the diagonals of | A008275|. See the Bala link for a proof. Cf. A008517. - Peter Bala, Dec 02 2011
EXAMPLE
Triangle begins:
1;
2, 1;
6, 8, 1;
24, 58, 22, 1;
120, 444, 328, 52, 1;
...
G.f. for k=3 sequence A000914(n-1), [2,11,35,85,175,322,546,...], is G1(3,x)= g1(1,x)/(1-x)^5= (2+x)/(1-x)^5.
MAPLE
a:= proc(k, m) option remember; if m >= 0 and k >= 0 then (k+m+1)*procname(k-1, m)+(k-m+1)*procname(k-1, m-1) else 0 fi end proc:
a(0, 0):= 1:
MATHEMATICA
a[k_, m_] = Sum[(-1)^(k + n + 1)*Binomial[2k + 3, n]*StirlingS1[m + k - n + 2, m + 1 - n], {n, 0, m}]; Flatten[Table[a[k, m], {k, 0, 8}, {m, 0, k}]][[1 ;; 45]] (* Jean-François Alcover, Jun 01 2011, after Johannes W. Meijer *)
PROG
(PARI) a(k, m)=sum(n=0, m, (-1)^(k + n + 1)*binomial(2*k + 3, n)*stirling(m + k - n + 2, m + 1 - n, 1));
for(k=0, 10, for(m=0, k, print1(a(k, m), ", "))) \\ Indranil Ghosh, Jul 21 2017
CROSSREFS
Row sums give A001147(k+1) = (2*k+1)!!, k>=0.
E2(n, k), the second-order Eulerian numbers with E2(0, k) = δ_{0, k}. Triangle read by rows, E2(n, k) for 0 <= k <= n.
+10
17
1, 0, 1, 0, 1, 2, 0, 1, 8, 6, 0, 1, 22, 58, 24, 0, 1, 52, 328, 444, 120, 0, 1, 114, 1452, 4400, 3708, 720, 0, 1, 240, 5610, 32120, 58140, 33984, 5040, 0, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320, 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880
COMMENTS
The second-order Eulerian number E2(n, k) is the number of Stirling permutations of order n with exactly k descents; here the last index is defined to be a descent. More formally, let Q_n denote the set of permutations of the multiset {1,1,2,2, ..., n,n} in which, for all j, all entries between two occurrences of j are larger than j, then E2(n, k) = card({s in Q_n with des(s) = k}), where des(s) = card({j: s(j) > s(j+1)}) is the number of descents of s.
Also the number of Riordan trapezoidal words of length n with k distinct letters (see Riordan 1976, p. 9).
Also the number of rooted plane trees on n + 1 vertices with k leaves (see Janson 2008, p. 543).
Let b(n) = (1/2)*Sum_{k=0..n-1} (-1)^k*E2(n-1, k+1) / C(2*n-1, k+1). Apparently b(n) = Bernoulli(n, 1) = -n*Zeta(1 - n) = Integral_{x=0..1}F_n(x) for n >= 1. Here F_n(x) are the signed Fubini polynomials ( A278075). (See Rzadkowski and Urlinska, example 4.)
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, 2nd ed. Addison-Wesley, Reading, MA, 1994, p. 270.
FORMULA
E2(n, k) = E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) for n > 0 and 0 <= k <= n, and E2(0, 0) = 1; in all other cases E(n, k) = 0.
E2(n, k) = Sum_{j=0..n-k}(-1)^(n-j)*binomial(2*n+1, j)*Stirling1(2*n-k-j+1, n-k-j+1).
E2(n, k) = Sum_{j=0..k}(-1)^(k-j)*binomial(2*n + 1, k - j)*Stirling2(n + j, j).
Stirling1(x, x - n) = (-1)^n*Sum_{k=0..n} E2(n, k)*binomial(x + k - 1, 2*n).
Stirling2(x, x - n) = Sum_{k=0..n} E2(n, k)*binomial(x + n - k, 2*n).
E2poly(n, x) = Sum_{k=0..n} E2(n, k)*x^k, as row polynomials.
E2poly(n, x) = x*(x-1)^(2*n)*d_{x}((1-x)^(1-2*n)*E2poly(n-1)) for n>=1 and E2poly(0)=1.
E2poly(n, x) = (1 - x)^(2*n + 1)*Sum_{k=0..n}(k^(n + k)/k!*(x*exp(-x))^k.
W(n, k) = [x^k] (1+x)^n*E2poly(n, x/(1 + x)) are the Ward numbers A269939. E2(n, k) = [x^k] (1-x)^n*Wpoly(n, x/(1 - x)); Wpoly(n, x) = Sum_{k=0..n}W(n, k)*x^k.
W(n, k) = Sum_{j=0..k} E2(n, j)*binomial(n - j, n - k).
E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*W(n, j)*binomial(n - j, k - j).
The compositional inverse with respect to x of x - t*(exp(x) - 1) (see B. Drake):
T(n, k) = [t^k](n+1)!*(1-t)^(2*n+1)*[x^(n+1)] InverseSeries(x - t*(exp(x)-1), x).
AS1(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, j+1), where AS1(n, k) are the associated Stirling numbers of the first kind ( A008306, A106828). E2(n, k) = Sum_{j=0..n-k+1} (-1)^(n-k-j+1)*AS1(n+j, j)*binomial(n-j, n-k-j+1), for n >= 1.
AS2(n, k) = Sum_{j=0..n-k} binomial(j, n-2*k)*E2(n-k, n-k-j) for n >=1, where AS2(n, k) are the associated Stirling numbers of the second kind ( A008299, A137375). E2(n, k) = Sum_{j=0..k} (-1)^(k-j)*AS2(n + j, j)*binomial(n - j, k - j).
EXAMPLE
Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1, 2;
[3] 0, 1, 8, 6;
[4] 0, 1, 22, 58, 24;
[5] 0, 1, 52, 328, 444, 120;
[6] 0, 1, 114, 1452, 4400, 3708, 720;
[7] 0, 1, 240, 5610, 32120, 58140, 33984, 5040;
[8] 0, 1, 494, 19950, 195800, 644020, 785304, 341136, 40320;
[9] 0, 1, 1004, 67260, 1062500, 5765500, 12440064, 11026296, 3733920, 362880.
.
To illustrate the generating function for row 3: The expansion of (1 - x)^7*(x*exp(-x) + 16*x^2*exp(-x)^2 + (243*x^3*exp(-x)^3)/2) gives the polynomial x + 8*x^2 + 6*x^3. The coefficients of this polynomial give row 3.
.
Stirling permutations of order 3 with exactly k descents: (When counting the descents one may assume an invisible '0' appended to the permutations.)
T[3, k=0]:
T[3, k=1]: 112233;
T[3, k=2]: 331122; 223311; 221133; 133122; 122331; 122133; 113322; 112332;
T[3, k=3]: 332211; 331221; 233211; 221331; 133221; 123321.
MAPLE
# Using the recurrence:
E2 := proc(n, k) option remember;
if k = 0 and n = 0 then return 1 fi; if n < 0 then return 0 fi;
E2(n-1, k)*k + E2(n-1, k-1)*(2*n - k) end: seq(seq(E2(n, k), k = 0..n), n = 0..9);
# Using the row generating function:
E2egf := n -> (1-x)^(2*n+1)*add(k^(n+k)/k!*(x*exp(-x))^k, k=0..n);
T := (n, k) -> coeftayl(E2egf(n), x=0, k): seq(print(seq(T(n, j), j=0..n)), n=0..7);
# Using the built-in function:
E2 := (n, k) -> `if`(k=0, k^n, combinat:-eulerian2(n, k-1)):
# Using the compositional inverse (series reversion):
E2triangle := proc(N) local r, s, C; Order := N + 2;
s := solve(y = series(x - t*(exp(x) - 1), x), x):
r := n -> -n!*(t - 1)^(2*n - 1)*coeff(s, y, n); C := [seq(expand(r(n)), n = 1..N)];
seq(print(seq(coeff(C[n+1], t, k), k = 0..n)), n = 0..N-1) end: E2triangle(10);
MATHEMATICA
T[n_, k_] := T[n, k] = If[k == 0, Boole[n == 0], If[n < 0, 0, k T[n - 1, k] + (2 n - k) T[n - 1, k - 1]]]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten
(* Via row polynomials: *)
E2poly[n_] := If[n == 0, 1,
Expand[Simplify[x (x - 1)^(2 n) D[((1 - x)^(1 - 2 n) E2poly[n - 1]), x]]]];
Table[CoefficientList[E2poly[n], x], {n, 0, 9}] // Flatten
(* Series reversion *)
Revert[gf_, len_] := Module[{S = InverseSeries[Series[gf, {x, 0, len + 1}], x]},
Table[CoefficientList[(n + 1)! (1 - t)^(2 n + 1) Coefficient[S, x, n + 1], t],
{n, 0, len}] // Flatten]; Revert[x + t - t Exp[x], 6]
PROG
(PARI)
E2poly(n) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1)));
{ for(n = 0, 9, print(Vecrev(E2poly(n)))) }
(PARI) T(n, k) = sum(j=0, n-k, (-1)^(n-j)*binomial(2*n+1, j)*stirling(2*n-k-j+1, n-k-j+1, 1)); \\ Michel Marcus, Feb 11 2021 (SageMath) # See link to notebook.
CROSSREFS
Indexing the second-order Eulerian numbers comes in three flavors: A008517 (following Riordan and Comtet), A201637 (following Graham, Knuth, and Patashnik) and this indexing, extending the definition of Gessel and Stanley. ( A008517 is the main entry of the numbers.) The corresponding triangles of the first-order Eulerian numbers are A008292, A173018, and A123125. Row reversed: A163936 (with offset = 0).
Triangle related to the o.g.f.s. of the right-hand columns of A130534 (E(x,m=1,n)).
+10
16
1, 1, 0, 2, 1, 0, 6, 8, 1, 0, 24, 58, 22, 1, 0, 120, 444, 328, 52, 1, 0, 720, 3708, 4400, 1452, 114, 1, 0, 5040, 33984, 58140, 32120, 5610, 240, 1, 0, 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 0, 362880, 3733920, 11026296, 12440064, 5765500, 1062500
COMMENTS
The asymptotic expansions of the higher-order exponential integral E(x,m=1,n) lead to triangle A130524, see A163931 for information on E(x,m,n). The o.g.f.s. of the right-hand columns of triangle A130534 have a nice structure: gf(p) = W1(z,p)/(1-z)^(2*p-1) with p = 1 for the first right-hand column, p = 2 for the second right-hand column, etc. The coefficients of the W1(z,p) polynomials lead to the triangle given above, n >= 1 and 1 <= m <= n. Our triangle is the same as A112007 with an extra right-hand column, see also the second Eulerian triangle A008517. The row sums of our triangle lead to A001147. We observe that the row sums of the triangles A163936 (m=1), A163937 (m=2), A163938 (m=3) and A163939 (m=4) for z=1 lead to A001147, A001147 (minus a(0)), A001879 and A000457 which are the first four left-hand columns of the triangle of the Bessel coefficients A001497 or, if one wishes, the right-hand columns of A001498. We checked this phenomenon for a few more values of m and found that this pattern persists: m = 5 leads to A001880, m=6 to A001881, m=7 to A038121 and m=8 to A130563 which are the next left- (right-) hand columns of A001497 ( A001498). An interesting phenomenon. If one assumes the triangle not (1,1) based but (0,0) based, one has T(n, k) = E2(n, n-k), where E2(n, k) are the second-order Eulerian numbers A340556. - Peter Luschny, Feb 12 2021
FORMULA
a(n, m) = Sum_{k=0..(m-1)} (-1)^(n+k+1)*binomial(2*n-1,k)*Stirling1(m+n-k-1,m-k), for 1 <= m <= n.
Assuming offset = 0 the T(n, k) are the coefficients of recursively defined polynomials. T(n, k) = [x^k] x^n*E2poly(n, 1/x), where E2poly(n, x) = x*(x - 1)^(2*n)*d_{x}((1 - x)^(1 - 2*n)*E2poly(n - 1, x))) for n >= 1 and E2poly(0, x) = 1. - Peter Luschny, Feb 12 2021
EXAMPLE
Triangle starts:
[ 1] 1;
[ 2] 1, 0;
[ 3] 2, 1, 0;
[ 4] 6, 8, 1, 0;
[ 5] 24, 58, 22, 1, 0;
[ 6] 120, 444, 328, 52, 1, 0;
[ 7] 720, 3708, 4400, 1452, 114, 1, 0;
[ 8] 5040, 33984, 58140, 32120, 5610, 240, 1, 0;
[ 9] 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 0;
The first few W1(z,p) polynomials are
W1(z,p=1) = 1/(1-z);
W1(z,p=2) = (1 + 0*z)/(1-z)^3;
W1(z,p=3) = (2 + 1*z + 0*z^2)/(1-z)^5;
W1(z,p=4) = (6 + 8*z + 1*z^2 + 0*z^3)/(1-z)^7.
MAPLE
with(combinat): a := proc(n, m): add((-1)^(n+k+1)*binomial(2*n-1, k)*stirling1(m+n-k-1, m-k), k=0..m-1) end: seq(seq(a(n, m), m=1..n), n=1..9); # Johannes W. Meijer, revised Nov 27 2012
MATHEMATICA
Table[Sum[(-1)^(n + k + 1)*Binomial[2*n - 1, k]*StirlingS1[m + n - k - 1, m - k], {k, 0, m - 1}], {n, 1, 10}, {m, 1, n}] // Flatten (* G. C. Greubel, Aug 13 2017 *)
PROG
(PARI) for(n=1, 10, for(m=1, n, print1(sum(k=0, m-1, (-1)^(n+k+1)* binomial(2*n-1, k)*stirling(m+n-k-1, m-k, 1)), ", "))) \\ G. C. Greubel, Aug 13 2017 (PARI) \\ assuming offset = 0:
E2poly(n, x) = if(n == 0, 1, x*(x-1)^(2*n)*deriv((1-x)^(1-2*n)*E2poly(n-1, x)));
{ for(n = 0, 9, print(Vec(E2poly(n, x)))) } \\ Peter Luschny, Feb 12 2021
Fifth diagonal of second-order Eulerian triangle A008517. Fifth column (m=4) of triangle A112007.
+10
4
1, 114, 5610, 195800, 5765500, 155357384, 4002695088, 101180433024, 2549865473424, 64728375139872, 1666424486271456, 43708768764064128, 1171582385481357696, 32157753536587053312, 905080567903692754176
FORMULA
a(n)= (n+9)*a(n-1) + (n+1)* A112008(n), n>=1, a(0)=1. a(n) = sum((-1)^(n+k+1)*binomial(2*n+11,k)*stirling1(n-k+10,5-k),k=0..4)
(End)
EXAMPLE
5610= a(2) = 11*114 + 3*1452.
CROSSREFS
Equals fifth left hand column of A163936. (End)
Row reversed version of triangle A201637 (second-order Eulerian triangle).
+10
3
1, 0, 1, 0, 2, 1, 0, 6, 8, 1, 0, 24, 58, 22, 1, 0, 120, 444, 328, 52, 1, 0, 720, 3708, 4400, 1452, 114, 1, 0, 5040, 33984, 58140, 32120, 5610, 240, 1, 0, 40320, 341136, 785304, 644020, 195800, 19950, 494, 1, 0, 362880, 3733920, 11026296, 12440064, 5765500, 1062500, 67260, 1004, 1, 0, 3628800, 44339040, 162186912, 238904904, 155357384, 44765000, 5326160, 218848, 2026, 1
COMMENTS
The row polynomials of this triangle P(n, x) = Sum_{m=0..n} T(n, m)*x^m appear as numerator polynomials in the o.g.f.s for the diagonal sequences of triangle A132393 (|Stirling1| with offset 0 for rows and columns). See the comment and the P. Bala link there.
FORMULA
T(n, m) = A201637(n, n-m), n >= m >= 0. Recurrence: T(0, 0) = 1, T(n, -1) = 0, T(n, m) = 0 if n < m, (n-m+1)*T(n-1, m-1) + (n-1+m)*T(n-1, m), n >= 1, m = 0..n; from the A008517 recurrence. T(0, 0) = 1, T(n, m) = Sum_{p = 0..m-1} (-1)^(n-p)*binomial(2*n+1, p)* A132393(n+m-p, m-p), n >= 1, m = 0..n; from a A008517 program.
EXAMPLE
The triangle T(n, m) begins:
n\m 0 1 2 3 4 5 6 7 8 9 ...
0: 1
1: 0 1
2: 0 2 1
3: 0 6 8 1
4: 0 24 58 22 1
5: 0 120 444 328 52 1
6: 0 720 3708 4400 1452 114 1
7: 0 5040 33984 58140 32120 5610 240 1
8: 0 40320 341136 785304 644020 195800 19950 494 1
9: 0 362880 3733920 11026296 12440064 5765500 1062500 67260 1004 1
...
MAPLE
T:= (n, k)-> combinat[eulerian2](n, n-k):
MATHEMATICA
Table[Boole[n == 0] + Sum[(-1)^(n + k) * Binomial[2 n + 1, k] StirlingS1[2 n - m - k, n - m - k], {k, 0, n - m - 1}], {n, 0, 10}, {m, n, 0, -1}] // Flatten (* Michael De Vlieger, Jul 21 2017, after Jean-François Alcover at A201637 *)
Search completed in 0.010 seconds
| 