cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 12 results. Next

A112000 One half of third column (k=2) of triangle A111999.

Original entry on oeis.org

-3, 65, -1190, 22050, -433125, 9144135, -208107900, 5099994900, -134219460375, 3781060408125, -113633468798850, 3631422078033750, -123022987568105625, 4405418319999571875, -166312279434175875000, 6602853358582065585000, -275059081486584416896875
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Crossrefs

Cf. Second (k=1) column: A000906(n+2)*(-1)^n = 2*A000457(n+2)*(-1)^n, n>=0.

Formula

a(n)=A111999(n+3, 2)/2, n>=0.
Conjecture: +n*(4*n+5)*a(n) +(2*n+3)*(n+2)*(4*n+9)*a(n-1)=0. - R. J. Mathar, Jul 09 2017

A112001 One fourth of fourth column (k=3) of triangle A111999.

Original entry on oeis.org

6, -231, 6608, -176330, 4714710, -130014885, 3745101360, -113311778580, 3608430175350, -120977128347075, 4267052705916000, -158151857368254750, 6150573832264611750, -250598489767931930625, 10680148143520635780000, -475373543975596731465000
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Formula

a(n)=A112492(n+4, 3)/4, n>=0.
Conjecture: +n*(20*n^2+75*n+67)*a(n) +(2*n+5)*(n+2)*(20*n^2+115*n+162)*a(n-1)=0. - R. J. Mathar, Jul 09 2017

A112003 One fourth of fifth column (k=4) of triangle A111999.

Original entry on oeis.org

-30, 1827, -75915, 2774695, -97362265, 3410657250, -121647896370, 4464233824050, -169525943987400, 6681570663692925, -273715584240223125, 11660420923428902625, -516518847915111916875, 23780637468174359272500, -1137205925185934836732500
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Formula

a(n)=A111999(n+5, 4)/4, n>=0.

A112004 One eighth of sixth column (k=5) of triangle A111999.

Original entry on oeis.org

90, -8028, 459855, -22166430, 991865875, -43155112000, 1871623083330, -82074799907100, 3670430981817600, -168269191527937500, 7933301574849275625, -385378337023784354250, 19309132829440209380625, -998382450032876448540000, 53277504029278691474812500
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Formula

a(n)=A111999(n+6, 5)/8, n>=0.

A145271 Coefficients for expansion of (g(x)d/dx)^n g(x); refined Eulerian numbers for calculating compositional inverse of h(x) = (d/dx)^(-1) 1/g(x); iterated derivatives as infinitesimal generators of flows.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 1, 11, 4, 7, 1, 1, 26, 34, 32, 15, 11, 1, 1, 57, 180, 122, 34, 192, 76, 15, 26, 16, 1, 1, 120, 768, 423, 496, 1494, 426, 294, 267, 474, 156, 56, 42, 22, 1, 1, 247, 2904, 1389, 4288, 9204, 2127, 496, 5946, 2829, 5142, 1206, 855, 768, 1344, 1038, 288, 56, 98, 64, 29, 1
Offset: 0

Views

Author

Tom Copeland, Oct 06 2008

Keywords

Comments

For more detail, including connections to Legendre transformations, rooted trees, A139605, A139002 and A074060, see Mathemagical Forests p. 9.
For connections to the h-polynomials associated to the refined f-polynomials of permutohedra see my comments in A008292 and A049019.
From Tom Copeland, Oct 14 2011: (Start)
Given analytic functions F(x) and FI(x) such that F(FI(x))=FI(F(x))=x about 0, i.e., they are compositional inverses of each other, then, with g(x) = 1/dFI(x)/dx, a flow function W(s,x) can be defined with the following relations:
W(s,x) = exp(s g(x)d/dx)x = F(s+FI(x)) ,
W(s,0) = F(s) ,
W(0,x) = x ,
dW(0,x)/ds = g(x) = F'[FI(x)] , implying
dW(0,F(x))/ds = g(F(x)) = F'(x) , and
W(s,W(r,x)) = F(s+FI(F(r+FI(x)))) = F(s+r+FI(x)) = W(s+r,x) . (See MF link below.) (End)
dW(s,x)/ds - g(x)dW(s,x)/dx = 0, so (1,-g(x)) are the components of a vector orthogonal to the gradient of W and, therefore, tangent to the contour of W, at (s,x) . - Tom Copeland, Oct 26 2011
Though A139605 contains A145271, the op. of A145271 contains that of A139605 in the sense that exp(s g(x)d/dx) w(x) = w(F(s+FI(x))) = exp((exp(s g(x)d/dx)x)d/du)w(u) evaluated at u=0. This is reflected in the fact that the forest of rooted trees assoc. to (g(x)d/dx)^n, FOR_n, can be generated by removing the single trunk of the planted rooted trees of FOR_(n+1). - Tom Copeland, Nov 29 2011
Related to formal group laws for elliptic curves (see Hoffman). - Tom Copeland, Feb 24 2012
The functional equation W(s,x) = F(s+FI(x)), or a restriction of it, is sometimes called the Abel equation or Abel's functional equation (see Houzel and Wikipedia) and is related to Schröder's functional equation and Koenigs functions for compositional iterates (Alexander, Goryainov and Kudryavtseva). - Tom Copeland, Apr 04 2012
g(W(s,x)) = F'(s + FI(x)) = dW(s,x)/ds = g(x) dW(s,x)/dx, connecting the operators here to presentations of the Koenigs / Königs function and Loewner / Löwner evolution equations of the Contreras et al. papers. - Tom Copeland, Jun 03 2018
The autonomous differential equation above also appears with a change in variable of the form x = log(u) in the renormalization group equation, or Beta function. See Wikipedia, Zinn-Justin equations 2.10 and 3.11, and Krajewski and Martinetti equation 21. - Tom Copeland, Jul 23 2020
A variant of these partition polynomials appears on p. 83 of Petreolle et al. with the indeterminates e_n there related to those given in the examples below by e_n = n!*(n'). The coefficients are interpreted as enumerating certain types of trees. See also A190015. - Tom Copeland, Oct 03 2022

Examples

			From _Tom Copeland_, Sep 19 2014: (Start)
Let h(x) = log((1+a*x)/(1+b*x))/(a-b); then, g(x) = 1/(dh(x)/dx) = (1+ax)(1+bx), so (0')=1, (1')=a+b, (2')=2ab, evaluated at x=0, and higher order derivatives of g(x) vanish. Therefore, evaluated at x=0,
R^0 g(x) =  1
R^1 g(x) =  a+b
R^2 g(x) = (a+b)^2 + 2ab = a^2 + 4 ab + b^2
R^3 g(x) = (a+b)^3 + 4*(a+b)*2ab = a^3 + 11 a^2*b + 11 ab^2 + b^3
R^4 g(x) = (a+b)^4 + 11*(a+b)^2*2ab + 4*(2ab)^2
         =  a^4 + 26 a^3*b + 66 a^2*b^2 + 26 ab^3 + b^4,
etc., and these bivariate Eulerian polynomials (A008292) are the first few coefficients of h^(-1)(x) = (e^(ax) - e^(bx))/(a*e^(bx) - b*e^(ax)), the inverse of h(x). (End)
Triangle starts:
  1;
  1;
  1,   1;
  1,   4,    1;
  1,  11,    4,    7,    1;
  1,  26,   34,   32,   15,   11,    1;
  1,  57,  180,  122,   34,  192,   76,  15,   26,   16,    1;
  1, 120,  768,  423,  496, 1494,  426, 294,  267,  474,  156,   56,  42,  22,    1;
  1, 247, 2904, 1389, 4288, 9204, 2127, 496, 5946, 2829, 5142, 1206, 855, 768, 1344, 1038, 288, 56, 98, 64, 29, 1;
		

References

  • D. S. Alexander, A History of Complex Dynamics: From Schröder to Fatou to Julia, Friedrich Vieweg & Sohn, 1994.
  • T. Mansour and M. Schork, Commutation Relations, Normal Ordering, and Stirling Numbers, Chapman and Hall/CRC, 2015.

Crossrefs

Cf. (A133437, A086810, A181289) = (LIF, reduced LIF, associated g(x)), where LIF is a Lagrange inversion formula. Similarly for (A134264, A001263, A119900), (A134685, A134991, A019538), (A133932, A111999, A007318).
Second column is A000295, subdiagonal is A000124, row sums are A000142, row lengths are A000041. - Peter Luschny, Jul 21 2016

Programs

  • Maple
    with(LinearAlgebra): with(ListTools):
    A145271_row := proc(n) local b, M, V, U, G, R, T;
    if n < 2 then return 1 fi;
    b := (n,k) -> `if`(k=1 or k>n+1,0,binomial(n-1,k-2)*g[n-k+1]);
    M := n -> Matrix(n, b):
    V := n -> Vector[row]([1, seq(0,i=2..n)]):
    U := n -> VectorMatrixMultiply(V(n), M(n)^(n-1)):
    G := n -> Vector([seq(g[i], i=0..n-1)]);
    R := n -> VectorMatrixMultiply(U(n), G(n)):
    T := Reverse([op(sort(expand(R(n+1))))]);
    seq(subs({seq(g[i]=1, i=0..n)},T[j]),j=1..nops(T)) end:
    for n from 0 to 9 do A145271_row(n) od; # Peter Luschny, Jul 20 2016

Formula

Let R = g(x)d/dx; then
R^0 g(x) = 1 (0')^1
R^1 g(x) = 1 (0')^1 (1')^1
R^2 g(x) = 1 (0')^1 (1')^2 + 1 (0')^2 (2')^1
R^3 g(x) = 1 (0')^1 (1')^3 + 4 (0')^2 (1')^1 (2')^1 + 1 (0')^3 (3')^1
R^4 g(x) = 1 (0')^1 (1')^4 + 11 (0')^2 (1')^2 (2')^1 + 4 (0')^3 (2')^2 + 7 (0')^3 (1')^1 (3')^1 + 1 (0')^4 (4')^1
R^5 g(x) = 1 (0') (1')^5 + 26 (0')^2 (1')^3 (2') + (0')^3 [34 (1') (2')^2 + 32 (1')^2 (3')] + (0')^4 [ 15 (2') (3') + 11 (1') (4')] + (0')^5 (5')
R^6 g(x) = 1 (0') (1')^6 + 57 (0')^2 (1')^4 (2') + (0')^3 [180 (1')^2 (2')^2 + 122 (1')^3 (3')] + (0')^4 [ 34 (2')^3 + 192 (1') (2') (3') + 76 (1')^2 (4')] + (0')^5 [15 (3')^2 + 26 (2') (4') + 16 (1') (5')] + (0')^6 (6')
where (j')^k = ((d/dx)^j g(x))^k. And R^(n-1) g(x) evaluated at x=0 is the n-th Taylor series coefficient of the compositional inverse of h(x) = (d/dx)^(-1) 1/g(x), with the integral from 0 to x.
The partitions are in reverse order to those in Abramowitz and Stegun p. 831. Summing over coefficients with like powers of (0') gives A008292.
Confer A190015 for another way to compute numbers for the array for each partition. - Tom Copeland, Oct 17 2014
Equivalent matrix computation: Multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n g(x) to obtain the matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then R^n g(x) = (1, 0, 0, 0, ...) [VP * S]^n (g_0, g_1, g_2, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. - Tom Copeland, Feb 10 2016 (An evaluation removed by author on Jul 19 2016. Cf. A139605 and A134685.)
Also, R^n g(x) = (1, 0, 0, 0, ...) [VP * S]^(n+1) (0, 1, 0, ...)^T in agreement with A139605. - Tom Copeland, Jul 21 2016
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the cycle index polynomials of A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences". - Tom Copeland, Feb 06 2018
A formula for computing the polynomials of each row of this matrix is presented as T_{n,1} on p. 196 of the Ihara reference in A139605. - Tom Copeland, Mar 25 2020
Indeterminate substitutions as illustrated in A356145 lead to [E] = [L][P] = [P][E]^(-1)[P] = [P][RT] and [E]^(-1) = [P][L] = [P][E][P] = [RT][P], where [E] contains the refined Eulerian partition polynomials of this entry; [E]^(-1), A356145, the inverse set to [E]; [P], the permutahedra polynomials of A133314; [L], the classic Lagrange inversion polynomials of A134685; and [RT], the reciprocal tangent polynomials of A356144. Since [L]^2 = [P]^2 = [RT]^2 = [I], the substitutional identity, [L] = [E][P] = [P][E]^(-1) = [RT][P], [RT] = [E]^(-1)[P] = [P][L][P] = [P][E], and [P] = [L][E] = [E][RT] = [E]^(-1)[L] = [RT][E]^(-1). - Tom Copeland, Oct 05 2022

Extensions

Title amplified by Tom Copeland, Mar 17 2014
R^5 and R^6 formulas and terms a(19)-a(29) added by Tom Copeland, Jul 11 2016
More terms from Peter Luschny, Jul 20 2016

A008306 Triangle T(n,k) read by rows: associated Stirling numbers of first kind (n >= 2, 1 <= k <= floor(n/2)).

Original entry on oeis.org

1, 2, 6, 3, 24, 20, 120, 130, 15, 720, 924, 210, 5040, 7308, 2380, 105, 40320, 64224, 26432, 2520, 362880, 623376, 303660, 44100, 945, 3628800, 6636960, 3678840, 705320, 34650, 39916800, 76998240, 47324376, 11098780, 866250, 10395
Offset: 2

Views

Author

Keywords

Comments

Also, T(n,k) is the number of derangements (permutations with no fixed points) of {1..n} with k cycles.
The sum of the n-th row is the n-th subfactorial: A000166(n). - Gary Detlefs, Jul 14 2010

Examples

			Rows 2 through 7 are:
    1;
    2;
    6,   3;
   24,  20;
  120, 130,  15;
  720, 924, 210;
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 256.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 75.

Crossrefs

Cf. A000166, A106828 (another version), A079510 (rearranged triangle), A235706 (specializations).
Diagonals: A000142, A000276, A000483.
Diagonals give reversed rows of A111999.

Programs

  • Haskell
    a008306 n k = a008306_tabf !! (n-2) !! (k-1)
    a008306_row n = a008306_tabf !! (n-2)
    a008306_tabf = map (fst . fst) $ iterate f (([1], [2]), 3) where
       f ((us, vs), x) =
         ((vs, map (* x) $ zipWith (+) ([0] ++ us) (vs ++ [0])), x + 1)
    -- Reinhard Zumkeller, Aug 05 2013
  • Maple
    A008306 := proc(n,k) local j;
    add(binomial(j,n-2*k)*A008517(n-k,j),j=0..n-k) end;
    seq(print(seq(A008306(n,k),k=1..iquo(n,2))),n=2..12):
    # Peter Luschny, Apr 20 2011
  • Mathematica
    t[0, 0] = 1; t[n_, 0] = 0; t[n_, k_] /; k > n/2 = 0; t[n_, k_] := t[n, k] = (n - 1)*(t[n - 1, k] + t[n - 2, k - 1]); A008306 = Flatten[ Table[ t[n, k], {n, 2, 12}, {k, 1, Quotient[n, 2]}]] (* Jean-François Alcover, Jan 25 2012, after David Callan *)
  • PARI
    { A008306(n,k) = (-1)^(n+k) * sum(i=0,k, (-1)^i * binomial(n,i) * stirling(n-i,k-i,1) ); } \\ Max Alekseyev, Sep 08 2018
    

Formula

T(n,k) = Sum_{i=0..k} (-1)^i * binomial(n,i) * |stirling1(n-i,k-i)| = (-1)^(n+k) * Sum_{i=0..k} (-1)^i * binomial(n,i) * A008275(n-i,k-i). - Max Alekseyev, Sep 08 2018
E.g.f.: 1 + Sum_{1 <= 2*k <= n} T(n, k)*t^n*u^k/n! = exp(-t*u)*(1-t)^(-u).
Recurrence: T(n, k) = (n-1)*(T(n-1, k) + T(n-2, k-1)) for 1 <= k <= n/2 with boundary conditions T(0,0) = 1, T(n,0) = 0 for n >= 1, and T(n,k) = 0 for k > n/2. - David Callan, May 16 2005
E.g.f. for column k: B(A(x)) where A(x) = log(1/1-x)-x and B(x) = x^k/k!.
From Tom Copeland, Jan 05 2016: (Start)
The row polynomials of this signed array are the orthogonal NL(n,x;x-n) = n! Sum_{k=0..n} binomial(x,n-k)*(-x)^k/k!, the normalized Laguerre polynomials of order (x-n) as discussed in Gautschi (the Temme, Carlitz, and Karlin and McGregor references come from this paper) in regard to asymptotic expansions of the upper incomplete gamma function--Tricomi's Cinderella of special functions.
e^(x*t)*(1-t)^x = Sum_{n>=0} NL(n,x;x-n)*x^n/n!.
The first few are
NL(0,x) = 1
NL(1,x) = 0
NL(2,x) = -x
NL(3,x) = 2*x
NL(4,x) = -6*x + 3*x^2.
With D=d/dx, :xD:^n = x^n D^n, :Dx:^n = D^n x^n, and K(a,b,c), the Kummer confluent hypergeometric function, NL(n,x;y-n) = n!*e^x binomial(xD+y,n)*e^(-x) = n!*e^x Sum_{k=0..n} binomial(k+y,n) (-x)^k/k! = e^x x^(-y+n) D^n (x^y e^(-x)) = e^x x^(-y+n) :Dx:^n x^(y-n)*e^(-x) = e^x*x^(-y+n)*n!*L(n,:xD:,0)*x^(y-n)*e^(-x) = n! binomial(y,n)*K(-n,y-n+1,x) = n!*e^x*(-1)^n*binomial(-xD-y+n-1,n)*e^(-x). Evaluate these expressions at y=x after the derivative operations to obtain NL(n,x;x-n). (End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Feb 16 2001

A032188 Number of labeled series-reduced mobiles (circular rooted trees) with n leaves (root has degree 0 or >= 2).

Original entry on oeis.org

1, 1, 5, 41, 469, 6889, 123605, 2620169, 64074901, 1775623081, 54989743445, 1882140936521, 70552399533589, 2874543652787689, 126484802362553045, 5977683917752887689, 301983995802099667861, 16239818347465293071401, 926248570498763547197525, 55847464116157184894240201
Offset: 1

Views

Author

Keywords

Comments

With offset 0, a(n) = number of partitions of the multiset {1,1,2,2,...,n,n} into lists of strictly decreasing lists, called blocks, such that the concatenation of all blocks in a list has the Stirling property: all entries between the two occurrences of i exceed i, 1<=i<=n. For example, with slashes separating blocks, a(2) = 5 counts 1/1/2/2; 1/2/2/1; 2/2/1/1; 1/2/2 1; 2/2 1/1, but not, for instance, 2 1/2/1 because it fails the Stirling test for i=2. - David Callan, Nov 21 2011

Examples

			D^3(1) = (24*x^2-64*x+41)/(2*x-1)^6. Evaluated at x = 0 this gives a(4) = 41.
a(3) = 5: Denote the colors of the vertices by the letters a,b,c, .... The 5 possible increasing plane trees on 3 vertices with vertices of outdegree k coming in 2^(k-1) colors are
.
   1a       1a        1b        1a        1b
   |       /  \      /  \      /  \      /  \
   2a     2    3    2    3    3    2    3    2
   |
   3
		

Crossrefs

Programs

  • Maple
    Order := 20; t1 := solve(series((ln(1-A)+2*A),A)=x,A); A000311 := n->n!*coeff(t1,x,n);
    # With offset 0:
    a := n -> add(combinat:-eulerian2(n,k)*2^k,k=0..n):
    seq(a(n),n=0..19); # Peter Luschny, Jul 09 2015
  • Mathematica
    For[y=x+O[x]^21; oldy=0, y=!=oldy, oldy=y; y=((1-y)Log[1-y]+x*y+y-x)/(2y-1), Null]; Table[n!Coefficient[y, x, n], {n, 1, 20}]
    Rest[CoefficientList[InverseSeries[Series[2*x + Log[1-x],{x,0,20}],x],x] * Range[0,20]!] (* Vaclav Kotesovec, Jan 08 2014 *)
  • Maxima
    a(n):=sum((n+k-1)!*sum(1/(k-j)!*sum((2^l*(-1)^(n+l+1)*stirling1(n-l+j-1,j-l))/(l!*(n-l+j-1)!),l,0,j),j,0,k),k,0,n-1); /* Vladimir Kruchinin, Feb 06 2012 */
    
  • PARI
    N = 66; x = 'x + O('x^N);
    Q(k) = if(k>N, 1,  1 + (k+1)*x - 2*x*(k+1)/Q(k+1) );
    gf = 1/Q(0); Vec(gf) \\ Joerg Arndt, May 01 2013
    
  • PARI
    {my(n=20); Vec(serlaplace(serreverse(2*x+log(1-x + O(x*x^n)))))} \\ Andrew Howroyd, Jan 16 2018

Formula

Doubles (index 2+) under "CIJ" (necklace, indistinct, labeled) transform.
E.g.f. A(x) satisfies log(1-A(x))+2*A(x)-x = 0. - Vladeta Jovovic, Dec 06 2002
With offset 0, second Eulerian transform of the powers of 2 [A000079]. See A001147 for definition of SET. - Ross La Haye, Feb 14 2005
From Peter Bala, Sep 05 2011: (Start)
The generating function A(x) satisfies the autonomous differential equation A'(x) = (1-A)/(1-2*A) with A(0) = 0. Hence the inverse function A^-1(x) = int {t = 0..x} (1-2*t)/(1-t) = 2*x+log(1-x). The expansion of A(x) can be found by inverting the above integral using the method of [Dominici, Theorem 4.1] to arrive at the result a(n) = D^(n-1)(1) evaluated at x = 0, where D denotes the operator g(x) -> d/dx((1-x)/(1-2*x)*g(x)). Compare with A006351.
Applying [Bergeron et al., Theorem 1] to the result x = int {t = 0..A(x)} 1/phi(t), where phi(t) = (1-t)/(1-2*t) = 1+t+2*t^2+4*t^3+8*t^4+... leads to the following combinatorial interpretation for this sequence: a(n) gives the number of plane increasing trees on n vertices where each vertex of outdegree k >= 1 can be colored in 2^(k-1) ways. An example is given below. (End)
The integral from 0 to infinity w.r.t. w of exp(-2w)(1-z*w)^(-1/z) gives an o.g.f. for the series with offset 0. Consequently, a(n)= sum(j=1 to infinity): St1d(n,j)/(2^(n+j-1)) where St1d(n,j) is the j-th element of the n-th diagonal of A132393 with offset=1; e.g., a(3)= 5 = 0/2^3 + 2/2^4 + 11/2^5 + 35/2^6 + 85/2^7 + ... . - Tom Copeland, Sep 15 2011
A signed o.g.f., with Γ(v,x) the incomplete gamma function (see A111999 with u=1), is g(z) = (2/z)^(-(1/z)-1) exp(2/z) Γ((1/z)+1,2/z)/z. - Tom Copeland, Sep 16 2011
With offset 0, a(n) = Sum[T(n+k,k), k=1..n] where T(n,k) are the associated Stirling numbers of the first kind (A008306). For example, a(3) = 41 = 6 + 20 + 15. - David Callan, Nov 21 2011
a(n) = sum(k=0..n-1, (n+k-1)!*sum(j=0..k, 1/(k-j)!*sum(l=0..j, (2^l*(-1)^(n+l+1)*stirling1(n-l+j-1,j-l))/(l!*(n-l+j-1)!)))), n>0. - Vladimir Kruchinin, Feb 06 2012
G.f.: 1/Q(0), where Q(k)= 1 + (k+1)*x - 2*x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
a(n) ~ n^(n-1) / (2 * exp(n) * (1-log(2))^(n-1/2)). - Vaclav Kotesovec, Jan 08 2014
a(n) = A032034(n)/2. - Alois P. Heinz, Jul 04 2018
E.g.f: series reversion of 2*x + log(1-x). - Andrew Howroyd, Sep 19 2018

A133932 Coefficients of a partition transform for Lagrange inversion of -log(1 - u(.)*t), complementary to A134685 for an e.g.f.

Original entry on oeis.org

1, -1, 3, -2, -15, 20, -6, 105, -210, 40, 90, -24, -945, 2520, -1120, -1260, 420, 504, -120, 10395, -34650, 25200, 18900, -2240, -15120, -9072, 1260, 2688, 3360, -720, -135135, 540540, -554400, -311850, 123200, 415800, 166320, -50400, -56700, -120960, -75600, 18144, 20160, 25920, -5040
Offset: 1

Views

Author

Tom Copeland, Jan 27 2008

Keywords

Comments

Let f(t) = -log(1 - u(.)*t) = Sum_{n>=1} (u_n / n) * t^n.
If u_1 is not equal to 0, then the compositional inverse for f(t) is given by g(t) = Sum_{j>=1} P(n,t) where, with u_n denoted by (n'),
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -1 (2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - 2 (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 20 (1')(2')(3') - 6 (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 210 (1') (2')^2 (3') + 40 (1')^2 (3')^2 + 90 (1')^2 (2') (4') - 24 (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 2520 (1') (2')^3 (3') - 1120 (1')^2 (2') (3')^2 - 1260 (1')^2 (2')^2 (4') + 420 (1')^3 (3')(4') + 504 (1')^3 (2')(5') - 120 (1')^4 (6') ] * t^6 / 6!
See A134685 for more information.
From Tom Copeland, Sep 28 2016: (Start)
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 34650 (1')(2')^4(3') + (1')^2 [25200 (2')^2(3')^2 + 18900 (2')^3(4')] - (1')^3 [2240 (3')^3 + 15120 (2')(3')(4') + 9072 (2')^2(5')] + (1')^4 [1260 (4')^2 + 2688 (3')(5') + 3360 (2')(6')] - 720 (1')^5(7')] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 540540 (1')(2')^5(3') - (1')^2 [554400 (2')^3(3')^2 + 311850 (2')^4(4')] + (1')^3 [123200 (2')(3')^3 + 415800 (2')^2(3')(4') + 166320 (2')^3(5')] - (1')^4 [50400 (3')^2(4') + 56700 (2')(4')^2 + 120960 (2')(3')(5') + 75600 (2')^2(6')] + (1')^5 [18144 (4')(5') + 20160 (3')(6') + 25920 (2')(7')] - 5040 (1')^6(8')] * t^8 / 8! (End)

Examples

			From _Tom Copeland_, Sep 18 2014: (Start)
Let f(x) = log((1-ax)/(1-bx))/(b-a) = -log(1-u.*x) = x + (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 + (a^3+a^2b+ab^2+a^3)x^4/4 + ... . with (u.)^n = u_n = h_(n-1)(a,b) the complete homogeneous polynomials in two indeterminates.
Then the inverse g(x) = (e^(ax)-e^(bx))/(a*e^(ax)-b*e^(bx)) = x - (a+b)x^2/2! + (a^2+4ab+b^2)x^3/3! - (a^3+11a^2b+11ab^2+b^3)x^4/4! + ... , where the bivariate polynomials are the Eulerian polynomials of A008292.
The inversion formula gives, e.g., P(3,x) = 3(u_2)^2 - 2u_3 = 3(h_1)^2 - 2h_2 = 3(a+b)^2 - 2(a^2 + ab + b^2) = a^2 + 4ab + b^2. (End)
		

Crossrefs

Cf. A145271 (A111999, A007318) = (reduced array, associated g(x)).

Programs

  • Mathematica
    rows[nn_] := {{1}}~Join~With[{s = InverseSeries[t (1 + Sum[u[k] t^k/(k+1), {k, nn}] + O[t]^(nn+1))]}, Table[(n+1)! Coefficient[s, t^(n+1) Product[u[w], {w, p}]], {n, nn}, {p, Reverse[Sort[Sort /@ IntegerPartitions[n]]]}]];
    rows[7] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

The bracketed partitions of P(n,t) are of the form (u_1)^e(1) (u_2)^e(2) ... (u_n)^e(n) with coefficients given by (-1)^(n-1+e(1)) * [2*(n-1)-e(1)]! / [ 2^e(2) (e(2))! * 3^e(3) (e(3))! * ... n^e(n) * (e(n))! ].
From Tom Copeland, Sep 06 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u./(1-u.t)]
= 1/((u_1) + (u_2)*t + (u_3)*t^2 + (u_4)*t^3 + ...),
where Ev denotes umbral evaluation.
Then for the partition polynomials of A133932,
n!*P(n,t) = ((t*h(y)*d/dy)^n) y evaluated at y=0,
and the compositional inverse of f(t) is
g(t) = exp(t*h(y)*d/dy) y evaluated at y=0.
Also, dg(t)/dt = h(g(t)). (End)
From Tom Copeland, Oct 20 2011: (Start)
With exp[x* PS(.,t)] = exp[t*g(x)] = exp[x*h(y)d/dy] exp(t*y) eval. at y=0, the raising/creation and lowering/annihilation operators defined by R PS(n,t)=PS(n+1,t) and L PS(n,t)= n*PS(n-1,t) are
R = t*h(d/dt) = t* 1/[(u_1) + (u_2)*d/dt + (u_3)*(d/dt)^2 + ...], and
L = f(d/dt) = (u_1)*d/dt + (u_2)*(d/dt)^2/2 + (u_3)*(d/dt)^3/3 + ....
Then P(n,t) = (t^n/n!) dPS(n,z)/dz eval. at z=0. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.) (End)
The bracketed partition polynomials of P(n,t) are also given by (d/dx)^(n-1) 1/[u_1 + u_2 * x/2 + u_3 * x^2/3 + ... + u_n * x^(n-1)/n]^n evaluated at x=0. - Tom Copeland, Jul 07 2015
From Tom Copeland, Sep 19 2016: (Start)
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix A007318 by f_m = (m-1)! u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) f_{n+1-k}, or equivalently, multiply the diagonals of A094587 by u_m. Then P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^(n-1) FC * t^n/n!, where S is the shift matrix A129185, representing differentiation in the basis x^n//n!, and FC is the first column of UP^(-1), the inverse matrix of UP. These results follow from A145271 and A133314.
With u_1 = 1, the first column of UP^(-1) with u_1 = 1 (with initial indices [0,0]) is composed of the row polynomials n! * OP_n(-u_2,...,-u_(n+1)), where OP_n(x[1],...,x[n]) are the row polynomials of A263633 for n > 0 and OP_0 = 1, which are related to those of A133314 as reciprocal o.g.f.s are related to reciprocal e.g.f.s; e.g., UP^(-1)[0,0] = 1, Up^(-1)[1,0] = -u_2, UP^(-1)[2,0] = 2! * (-u_3 + u_2^2) = 2! * OP_2(-u_2,-u_3).
Also, P(n,t) = (1, 0, 0, 0,..) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. (End)
From Tom Copeland, Sep 20 2016: (Start)
Let PS(n,u1,u2,...,un) = P(n,t) / (t^n/n!), i.e., the square-bracketed part of the partition polynomials in the expansion for the inverse in the comment section, with u_k = uk.
Also let PS(n,u1=1,u2,...,un) = PB(n,b1,b2,...,bK,...) where each bK represents the partitions of PS, with u1 = 1, that have K components or blocks, e.g., PS(5,1,u2,...,u5) = PB(5,b1,b2,b3,b4) = b1 + b2 + b3 + b4 with b1 = -24 u5, b2 = 90 u2 u4 + 40 u3^2, b3 = -210 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers's equation and compositional inverse pairs (cf. link and A086810) implies, for n > 2, PB(n, 0 * b1, 1 * b2, ..., (K-1) * bK, ...) = (1/2) * Sum_{k = 2..n-1} binomial(n+1,k) * PS(n-k+1, u_1=1, u_2, ..., u_(n-k+1)) * PS(k,u_1=1,u_2,...,u_k).
For example, PB(5,0 * b1, 1 * b2, 2 * b3, 3 * b4) = 3 * 105 u2^4 - 2 * 210 u2^2 u3 + 1 * 90 u2 u4 + 1 * 40 u3^2 - 0 * -24 u5 = 315 u2^4 - 420 u2^2 u3 + 90 u2 u4 + 40 u3^2 = (1/2) [2 * 6!/(4!*2!) * PS(2,1,u2) * PS(4,1,u2,...,u4) + 6!/(3!*3!) * PS(3,1,u2,u3)^2] = (1/2) * [ 2 * 6!/(4!*2!) * (-u2) (-15 u2^3 + 20 u2 u3 - 6 u4) + 6!/(3!*3!) * (3 u2^2 - 2 u3)^2].
Also, PB(n,0*b1,1*b2,...,(K-1)*bK,...) = d/dt t^(n-2)*PS(n,u1=1/t,u2,...,un)|{t=1} = d/dt (1/t)*PS(n,u1=1,t*u2,...,t*un)|{t=1}.
(End)
A recursion relation for computing each partition polynomial of this entry from the lower order polynomials and the coefficients of the refined Stirling polynomials of the first kind A036039 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

Terms ordered according to the reversed Abramowitz-Stegun ordering of partitions (with every k' replaced by (k-1)') by Andrey Zabolotskiy, Mar 08 2024

A269940 Triangle read by rows, T(n, k) = Sum_{m=0..k} (-1)^(m + k)*binomial(n + k, n + m) * |Stirling1(n + m, m)|, for n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 2, 3, 0, 6, 20, 15, 0, 24, 130, 210, 105, 0, 120, 924, 2380, 2520, 945, 0, 720, 7308, 26432, 44100, 34650, 10395, 0, 5040, 64224, 303660, 705320, 866250, 540540, 135135, 0, 40320, 623376, 3678840, 11098780, 18858840, 18288270, 9459450, 2027025
Offset: 0

Views

Author

Peter Luschny, Mar 27 2016

Keywords

Comments

We propose to call this sequence the 'Ward cycle numbers' and sequence A269939 the 'Ward set numbers'. - Peter Luschny, Nov 25 2022

Examples

			Triangle T(n,k) starts:
  [1]
  [0,   1]
  [0,   2,      3]
  [0,   6,     20,     15]
  [0,  24,    130,    210,    105]
  [0,  120,   924,   2380,   2520,    945]
  [0,  720,  7308,  26432,  44100,  34650,  10395]
  [0, 5040, 64224, 303660, 705320, 866250, 540540, 135135]
		

Crossrefs

Variants: A111999, A259456.
Cf. A269939 (Stirling2 counterpart), A268438, A032188 (row sums).

Programs

  • Maple
    T := (n, k) -> add((-1)^(m+k)*binomial(n+k,n+m)*abs(Stirling1(n+m, m)), m=0..k):
    seq(print(seq(T(n, k), k=0..n)), n=0..6);
    # Alternatively:
    T := proc(n, k) option remember;
        `if`(k=0, k^n,
        `if`(k<=0 or k>n, 0,
        (n+k-1)*(T(n-1, k)+T(n-1, k-1)))) end:
    for n from 0 to 6 do seq(T(n, k), k=0..n) od;
  • Mathematica
    T[n_, k_] := Sum[(-1)^(m+k)*Binomial[n+k, n+m]*Abs[StirlingS1[n+m, m]], {m, 0, k}];
    Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 12 2022 *)
  • Sage
    T = lambda n, k: sum((-1)^(m+k)*binomial(n+k, n+m)*stirling_number1(n+m, m) for m in (0..k))
    for n in (0..7): print([T(n, k) for k in (0..n)])
    
  • Sage
    # uses[PtransMatrix from A269941]
    PtransMatrix(8, lambda n: n/(n+1), lambda n, k: (-1)^k*falling_factorial(n+k,n))

Formula

T(n,k) = (-1)^k*FF(n+k,n)*P[n,k](n/(n+1)) where P is the P-transform and FF the falling factorial function. For the definition of the P-transform see the link.
T(n,k) = A268438(n,k)*FF(n+k,n)/(2*n)!.

Extensions

Name corrected after notice from Ed Veling by Peter Luschny, Jun 14 2022

A259456 Triangle read by rows, giving coefficients in an expansion of absolute values of Stirling numbers of the first kind in terms of binomial coefficients.

Original entry on oeis.org

1, 2, 3, 6, 20, 15, 24, 130, 210, 105, 120, 924, 2380, 2520, 945, 720, 7308, 26432, 44100, 34650, 10395, 5040, 64224, 303660, 705320, 866250, 540540, 135135, 40320, 623376, 3678840, 11098780, 18858840, 18288270, 9459450, 2027025, 362880, 6636960, 47324376, 177331440, 389449060, 520059540, 416215800
Offset: 0

Views

Author

N. J. A. Sloane, Jun 30 2015

Keywords

Examples

			Triangle begins:
1,
2,3,
6,20,15,
24,130,210,105,
120,924,2380,2520,945,
...
For k=4 and j=2 in Knuth's equation, |S1(4,4-2)| = |S1(4,2)| = |A008275(4,2)| = 11 = p_{2,1}*C(4,3) +p_{2,2}*C(4,4) = 2*4+3*1. - _R. J. Mathar_, Jul 16 2015
		

References

  • L. Comtet, Advanced Combinatorics (1974), Chapter VI, page 256.
  • DJ Jeffrey, GA Kalugin, N Murdoch, Lagrange inversion and Lambert W, Preprint 2015; http://www.apmaths.uwo.ca/~djeffrey/Offprints/JeffreySYNASC2015paper17.pdf
  • Charles Jordan, Calculus of Finite Differences, Chelsea 1965, p. 152. Table C_{m, nu}.

Crossrefs

Cf. This is a row reversed and unsigned version of A111999.
Cf. A008275, A000276 (2nd column), A000483 (3rd column), A000142 (1st column).
Cf. A133932.

Programs

  • Maple
    A259456 := proc(n,k)
        option remember;
        if k < 1 or k > n  then
            0 ;
        elif n = 1 then
            1;
        else
            procname(n-1,k-1)+procname(n-1,k);
            %*(n+k-1) ;
        end if;
    end proc:
    seq(seq(A259456(n,k),k=1..n),n=1..10) ; # R. J. Mathar, Jul 18 2015
  • Mathematica
    T[n_, k_] := T[n, k] = If[k < 1 || k > n, 0, If[n == 1, 1, (T[n-1, k-1] + T[n-1, k])(n+k-1)]];
    Table[T[n, k], {n, 1, 10}, { k, 1, n}] // Flatten (* Jean-François Alcover, Sep 26 2019, from Maple *)

Formula

T(n,k) = (n-k-1)*( T(n-1,k-1)+T(n-1,k) ), n>=1, 1<=k<=n. [Berg, Eq. 6]
The general results on the convolution of the refined partition polynomials of A133932, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these unsigned polynomials. - Tom Copeland, Sep 20 2016
Showing 1-10 of 12 results. Next