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.

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A223168 Triangle S(n, k) by rows: coefficients of 2^((n-1)/2)*(x^(1/2)*d/dx)^n when n is odd, and of 2^(n/2)*(x^(1/2)*d/dx)^n when n is even.

Original entry on oeis.org

1, 1, 2, 3, 2, 3, 12, 4, 15, 20, 4, 15, 90, 60, 8, 105, 210, 84, 8, 105, 840, 840, 224, 16, 945, 2520, 1512, 288, 16, 945, 9450, 12600, 5040, 720, 32, 10395, 34650, 27720, 7920, 880, 32, 10395, 124740, 207900, 110880, 23760, 2112, 64, 135135, 540540, 540540, 205920, 34320, 2496, 64
Offset: 0

Views

Author

Udita Katugampola, Mar 17 2013

Keywords

Comments

Also coefficients in the expansion of k-th derivative of exp(n*x^2), see Mathematica program. - Vaclav Kotesovec, Jul 16 2013

Examples

			Triangle begins:
       1;
       1,      2;
       3,      2;
       3,     12,      4;
      15,     20,      4;
      15,     90,     60,      8;
     105,    210,     84,      8;
     105,    840,    840,    224,    16;
     945,   2520,   1512,    288,    16;
     945,   9450,  12600,   5040,   720,   32;
   10395,  34650,  27720,   7920,   880,   32;
   10395, 124740, 207900, 110880, 23760, 2112, 64;
  135135, 540540, 540540, 205920, 34320, 2496, 64;
  .
Expansion takes the form:
2^0 (x^(1/2)*d/dx)^1 = 1*x^(1/2)*d/dx.
2^1 (x^(1/2)*d/dx)^2 = 1*d/dx + 2*x*d^2/dx^2.
2^1 (x^(1/2)*d/dx)^3 = 3*x^(1/2)*d^2/dx^2 + 2*x^(3/2)*d^3/dx^3.
2^2 (x^(1/2)*d/dx)^4 = 3*d^2/dx^2 + 12*x*d^3/dx^3 + 4*x^2*d^4/dx^4.
2^2 (x^(1/2)*d/dx)^5 = 15*x^(1/2)*d^3/dx^3 + 20*x^(3/2)*d^4/dx^4 + 4*x^(5/2)*d^5/dx^5.
`
`
		

Crossrefs

Odd rows includes absolute values of A098503 from right to left.

Programs

  • Maple
    a[0]:= f(x);
    for i from 1 to 13 do
    a[i]:= simplify(2^((i+1)mod 2)*x^(1/2)*(diff(a[i-1],x$1)));
    end do;
  • Mathematica
    Flatten[CoefficientList[Expand[FullSimplify[Table[D[E^(n*x^2),{x,k}]/(E^(n*x^2)*(2*n)^Floor[(k+1)/2]),{k,1,13}]]]/.x->1,n]] (* Vaclav Kotesovec, Jul 16 2013 *)

A223172 Triangle S(n,k) by rows: coefficients of 6^((n-1)/2)*(x^(1/6)*d/dx)^n when n is odd, and of 6^(n/2)*(x^(5/6)*d/dx)^n when n is even.

Original entry on oeis.org

1, 1, 6, 7, 6, 7, 84, 36, 91, 156, 36, 91, 1638, 1404, 216, 1729, 4446, 2052, 216, 1729, 41496, 53352, 16416, 1296, 43225, 148200, 102600, 21600, 1296, 43225, 1296750, 2223000, 1026000, 162000, 7776, 1339975, 5742750, 5301000, 1674000, 200880, 7776
Offset: 0

Views

Author

Udita Katugampola, Mar 20 2013

Keywords

Examples

			Triangle begins:
        1;
        1,        6;
        7,        6;
        7,       84,        36;
       91,      156,        36;
       91,     1638,      1404,      216;
     1729,     4446,      2052,      216;
     1729,    41496,     53352,    16416,     1296;
    43225,   148200,    102600,    21600,     1296;
    43225,  1296750,   2223000,  1026000,   162000,    7776;
  1339975,  5742750,   5301000,  1674000,   200880,    7776;
  1339975, 48239100, 103369500, 63612000, 15066000, 1446336, 46656;
		

Crossrefs

Programs

  • Maple
    a[0]:= f(x):
    for i from 1 to 13 do
    a[i] := simplify(6^((i+1)mod 2)*x^((4((i+1)mod 2)+1)/6)*(diff(a[i-1],x$1 )));
    end do;

A241171 Triangle read by rows: Joffe's central differences of zero, T(n,k), 1 <= k <= n.

Original entry on oeis.org

1, 1, 6, 1, 30, 90, 1, 126, 1260, 2520, 1, 510, 13230, 75600, 113400, 1, 2046, 126720, 1580040, 6237000, 7484400, 1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400, 1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000, 1, 131070, 96461910, 8203431600, 196556560200, 1882311631200, 8266953895200, 16672848192000, 12504636144000
Offset: 1

Views

Author

N. J. A. Sloane, Apr 22 2014

Keywords

Comments

T(n,k) gives the number of ordered set partitions of the set {1,2,...,2*n} into k even sized blocks. An example is given below. Cf. A019538 and A156289. - Peter Bala, Aug 20 2014

Examples

			Triangle begins:
1,
1, 6,
1, 30, 90,
1, 126, 1260, 2520,
1, 510, 13230, 75600, 113400,
1, 2046, 126720, 1580040, 6237000, 7484400,
1, 8190, 1171170, 28828800, 227026800, 681080400, 681080400,
1, 32766, 10663380, 494053560, 6972966000, 39502663200, 95351256000, 81729648000,
...
From _Peter Bala_, Aug 20 2014: (Start)
Row 2: [1,6]
k  Ordered set partitions of {1,2,3,4} into k blocks    Number
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
1   {1,2,3,4}                                             1
2   {1,2}{3,4}, {3,4}{1,2}, {1,3}{2,4}, {2,4}{1,3},       6
    {1,4}{2,3}, {2,3}{1,4}
(End)
		

References

  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 283.
  • S. A. Joffe, Calculation of the first thirty-two Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 47 (1914), 103-126.
  • S. A. Joffe, Calculation of eighteen more, fifty in all, Eulerian numbers from central differences of zero, Quart. J. Pure Appl. Math. 48 (1917-1920), 193-271.

Crossrefs

Case m=2 of the polynomials defined in A278073.
Cf. A000680 (diagonal), A094088 (row sums), A000364 (alternating row sums), A281478 (central terms), A327022 (refinement).
Diagonals give A002446, A213455, A241172, A002456.

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->1/(2^(k-1))*Sum([1..k],j->(-1)^(k-j)*Binomial(2*k,k-j)*j^(2*n))))); # Muniru A Asiru, Feb 27 2019
  • Maple
    T := proc(n,k) option remember;
    if k > n then 0
    elif k=0 then k^n
    elif k=1 then 1
    else k*(2*k-1)*T(n-1,k-1)+k^2*T(n-1,k); fi;
    end: # Minor edit to make it also work in the (0,0)-offset case. Peter Luschny, Sep 03 2022
    for n from 1 to 12 do lprint([seq(T(n,k), k=1..n)]); od:
  • Mathematica
    T[n_, k_] /; 1 <= k <= n := T[n, k] = k(2k-1) T[n-1, k-1] + k^2 T[n-1, k]; T[, 1] = 1; T[, ] = 0; Table[T[n, k], {n, 1, 9}, {k, 1, n}] (* _Jean-François Alcover, Jul 03 2019 *)
  • Sage
    @cached_function
    def A241171(n, k):
        if n == 0 and k == 0: return 1
        if k < 0 or k > n: return 0
        return (2*k^2 - k)*A241171(n - 1, k - 1) + k^2*A241171(n - 1, k)
    for n in (1..6): print([A241171(n, k) for k in (1..n)]) # Peter Luschny, Sep 06 2017
    

Formula

T(n,k) = 0 if k <= 0 or k > n, = 1 if k=1, otherwise T(n,k) = k*(2*k-1)*T(n-1,k-1) + k^2*T(n-1,k).
Related to Euler numbers A000364 by A000364(n) = (-1)^n*Sum_{k=1..n} (-1)^k*T(n,k). For example, A000364(3) = 61 = 90 - 30 + 1.
From Peter Bala, Aug 20 2014: (Start)
T(n,k) = 1/(2^(k-1))*Sum_{j = 1..k} (-1)^(k-j)*binomial(2*k,k-j)*j^(2*n).
T(n,k) = k!*A156289(n,k) = k!*(2*k-1)!!*A036969.
E.g.f.: A(t,z) := 1/( 1 - t*(cosh(z) - 1) ) = 1 + t*z^2/2! + (t + 6*t^2)*z^4/4! + (t + 30*t^2 + 90*t^3)*z^6/6! + ... satisfies the partial differential equation d^2/dz^2(A) = D(A), where D = t^2*(2*t + 1)*d^2/dt^2 + t*(5*t + 1)*d/dt + t.
Hence the row polynomials R(n,t) satisfy the differential equation R(n+1,t) = t^2*(2*t + 1)*R''(n,t) + t*(5*t + 1)*R'(n,t) + t*R(n,t) with R(0,t) = 1, where ' indicates differentiation w.r.t. t. This is equivalent to the above recurrence equation.
Recurrence for row polynomials: R(n,t) = t*( Sum_{k = 1..n} binomial(2*n,2*k)*R(n-k,t) ) with R(0,t) := 1.
Row sums equal A094088(n) for n >= 1.
A100872(n) = (1/2)*R(n,2). (End)

A001117 a(n) = 3^n - 3*2^n + 3.

Original entry on oeis.org

1, 0, 0, 6, 36, 150, 540, 1806, 5796, 18150, 55980, 171006, 519156, 1569750, 4733820, 14250606, 42850116, 128746950, 386634060, 1160688606, 3483638676, 10454061750, 31368476700, 94118013006, 282379204836, 847187946150, 2541664501740, 7625194831806
Offset: 0

Views

Author

Keywords

Comments

Differences of 0. Labeled ordered partitions into 3 parts.
Number of surjections from an n-element set onto a three-element set, with n >= 3. - Mohamed Bouhamida, Dec 15 2007
Let P(A) be the power set of an n-element set A and R be a relation on P(A) such that for all x, y of P(A), xRy if either 0) x is not a subset of y and y is not a subset of x and x and y are disjoint, or 1) x is a proper subset of y or y is a proper subset of x and x and y are intersecting. Then a(n+1) = |R|. - Ross La Haye, Mar 19 2009
For n>0, the number of rows of n colors using exactly three colors. For n=3, the six rows are ABC, ACB, BAC, BCA, CAB, and CBA. - Robert A. Russell, Sep 25 2018

References

  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • J. F. Steffensen, Interpolation, 2nd ed., Chelsea, NY, 1950, see p. 54.
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.

Crossrefs

Column 3 of A019538 (n>0).

Programs

  • Maple
    with(combstruct):ZL:=[S,{S=Sequence(U,card=r),U=Set(Z,card>=1)}, labeled]: 1, seq(count(subs(r=3,ZL),size=m),m=1..25); # Zerinvary Lajos, Mar 09 2007
    A001117:=-6/(z-1)/(3*z-1)/(2*z-1); # Conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence except for three leading terms.
  • Mathematica
    k=3; Prepend[Table[k!StirlingS2[n,k],{n,1,30}],1] (* Robert A. Russell, Sep 25 2018 *)
  • PARI
    a(n)=3^n-3*2^n+3 \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = [n=0] + 3!*S(n, 3).
E.g.f.: 1 + (exp(x)-1)^3.
For n>=3: a(n+1) = 3*a(n) + 3*(2^n - 2) = 3*a(n) + 3*A000918(n). - Geoffrey Critzer, Feb 27 2009
G.f.: (-1-11*x^2+6*x)/((x-1)*(3*x-1)*(2*x-1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009

Extensions

Extended with formula and alternate description by Christian G. Bower, Aug 15 1998
Simpler description from Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001

A048594 Triangle T(n,k) = k! * Stirling1(n,k), 1<=k<=n.

Original entry on oeis.org

1, -1, 2, 2, -6, 6, -6, 22, -36, 24, 24, -100, 210, -240, 120, -120, 548, -1350, 2040, -1800, 720, 720, -3528, 9744, -17640, 21000, -15120, 5040, -5040, 26136, -78792, 162456, -235200, 231840, -141120, 40320, 40320, -219168, 708744, -1614816, 2693880, -3265920, 2751840, -1451520, 362880
Offset: 1

Views

Author

Oleg Marichev (oleg(AT)wolfram.com)

Keywords

Comments

Row sums (unsigned) give A007840(n), n>=1; (signed): A006252(n), n>=1.
Apart from signs, coefficients in expansion of n-th derivative of 1/log(x).

Examples

			Triangle begins
   1;
  -1,    2;
   2,   -6,   6;
  -6,   22, -36,   24;
  24, -100, 210, -240, 120; ...
The 2nd derivative of 1/log(x) is -2/x^3*log(x)^2 - 6/x^3*log(x)^3 - 6/x^3*log(x)^4.
		

Crossrefs

Cf. A133942 (left edge), A000142 (right edge), A006252 (row sums), A238685 (central terms).
Row sums: A007840 (unsigned), A006252 (signed).

Programs

  • Haskell
    a048594 n k = a048594_tabl !! (n-1) !! (k-1)
    a048594_row n = a048594_tabl !! (n-1)
    a048594_tabl = map snd $ iterate f (1, [1]) where
       f (i, xs) = (i + 1, zipWith (-) (zipWith (*) [1..] ([0] ++ xs))
                                       (map (* i) (xs ++ [0])))
    -- Reinhard Zumkeller, Mar 02 2014
    
  • Magma
    /* As triangle: */ [[Factorial(k)*StirlingFirst(n,k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Dec 15 2015
    
  • Maple
    with(combinat): A048594 := (n,k)->k!*stirling1(n,k);
  • Mathematica
    Flatten[Table[k!*StirlingS1[n,k], {n,10}, {k,n}]] (* Harvey P. Dale, Aug 28 2011 *)
    Join @@ CoefficientRules[ -Table[ D[ 1/Log[z], {z, n}], {n, 9}] /. Log[z] -> -Log[z], {1/z, 1/Log[z]}, "NegativeLexicographic"][[All, All, 2]] (* Oleg Marichev (oleg(AT)wolfram.com) and Maxim Rytin (m.r(AT)inbox.ru); submitted by Robert G. Wilson v, Aug 29 2011 *)
  • PARI
    {T(n, k)= if(k<1 || k>n, 0, stirling(n, k)* k!)} /* Michael Somos Apr 11 2007 */
    
  • SageMath
    def A048594(n,k): return (-1)^(n-k)*factorial(k)*stirling_number1(n,k)
    flatten([[A048594(n,k) for k in range(1,n+1)] for n in range(1,13)]) # G. C. Greubel, Oct 24 2023

Formula

T(n, k) = k*T(n-1, k-1) - (n-1)*T(n-1, k) if n>=k>=1, T(n, 0) = 0 and T(1, 1)=1, else 0.
E.g.f. k-th column: log(1+x)^k, k>=1.
From Peter Bala, Nov 25 2011: (Start):
E.g.f.: 1/(1-t*log(1+x)) = 1 + t*x + (-t+2*t^2)*x^2/2! + ....
The row polynomials are given by D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator exp(-x)*d/dx.
(End)

A223532 Triangle S(n,k) by rows: coefficients of 6^(n/2)*(x^(5/6)*d/dx)^n when n=0,2,4,6,...

Original entry on oeis.org

1, 1, 6, 7, 84, 36, 91, 1638, 1404, 216, 1729, 41496, 53352, 16416, 1296, 43225, 1296750, 2223000, 1026000, 162000, 7776, 1339975, 48239100, 103369500, 63612000, 15066000, 1446336, 46656, 49579075, 2082321150, 5354540100, 4118877000, 1300698000, 187300512
Offset: 1

Views

Author

Udita Katugampola, Mar 23 2013

Keywords

Examples

			Triangle begins:
1;
1, 6;
7, 84, 36;
91, 1638, 1404, 216;
1729, 41496, 53352, 16416, 1296;
43225, 1296750, 2223000, 1026000, 162000, 7776;
1339975, 48239100, 103369500, 63612000, 15066000, 1446336, 46656;
49579075, 2082321150, 5354540100, 4118877000, 1300698000, 187300512, 12083904, 279936;
		

Crossrefs

Programs

  • Maple
    a[0]:= f(x):
    for i from 1 to 20 do
    a[i] := simplify(6^((i+1)mod 2)*x^((4((i+1)mod 2)+1)/6)*(diff(a[i-1],x$1 )));
    end do:
    for j from 1 to 10 do
    b[j]:=a[2j];
    end do;

A046802 T(n, k) = Sum_{j=k..n} binomial(n, j)*E1(j, j-k), where E1 are the Eulerian numbers A173018. Triangle read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 33, 15, 1, 1, 31, 131, 131, 31, 1, 1, 63, 473, 883, 473, 63, 1, 1, 127, 1611, 5111, 5111, 1611, 127, 1, 1, 255, 5281, 26799, 44929, 26799, 5281, 255, 1, 1, 511, 16867, 131275, 344551, 344551, 131275, 16867, 511, 1, 1, 1023, 52905
Offset: 0

Views

Author

Keywords

Comments

T(n,k) is the number of positroid cells of the totally nonnegative Grassmannian G+(k,n) (cf. Postnikov/Williams). It is the triangle of the h-vectors of the stellahedra. - Tom Copeland, Oct 10 2014
See A248727 for a simple transformation of the row polynomials of this entry that produces the umbral compositional inverses of the polynomials of A074909, related to the face polynomials of the simplices. - Tom Copeland, Jan 21 2015
From Tom Copeland, Jan 24 2015: (Start)
The reciprocal of this entry's e.g.f. is [t e^(-xt) - e^(-x)] / (t-1) = 1 - (1+t) x + (1+t+t^2) x^2/2! - (1+t+t^2+t^3) x^3/3! + ... = e^(q.(0;t)x), giving the base sequence (q.(0;t))^n = q_n(0;t) = (-1)^n [1-t^(n+1)] / (1-t) for the umbral compositional inverses (q.(0;t)+z)^n = q_n(z;t) of the Appell polynomials associated with this entry, p_n(z;t) below, i.e., q_n(p.(z;t)) = z^n = p_n(q.(z;t)), in umbral notation. The relations in A133314 then apply between the two sets of base polynomials. (Inserted missing index in a formula - Mar 03 2016.)
The associated o.g.f. for the umbral inverses is Og(x) = x / (1-x q.(0:t)) = x / [(1+x)(1+tx)] = x / [1+(1+t)x+tx^2]. Applying A134264 to h(x) = x / Og(x) = 1 + (1+t) x + t x^2 leads to an o.g.f. for the Narayana polynomials A001263 as the comp. inverse Oginv(x) = [1-(1+t)x-sqrt[1-2(1+t)x+((t-1)x)^2]] / (2xt). Note that Og(x) gives the signed h-polynomials of the simplices and that Oginv(x) gives the h-polynomials of the simplicial duals of the Stasheff polynomials, or type A associahedra. Contrast this with A248727 = A046802 * A007318, which has o.g.f.s related to the corresponding f-polynomials. (End)
The Appell polynomials p_n(x;t) in the formulas below specialize to the Swiss-knife polynomials of A119879 for t = -1, so the Springer numbers A001586 are given by 2^n p_n(1/2;-1). - Tom Copeland, Oct 14 2015
The row polynomials are the h-polynomials associated to the stellahedra, whose f-polynomials are the row polynomials of A248727. Cf. page 60 of the Buchstaber and Panov link. - Tom Copeland, Nov 08 2016
The row polynomials are the h-polynomials of the stellohedra, which enumerate partial permutations according to descents. Cf. Section 10.4 of the Postnikov-Reiner-Williams reference. - Lauren Williams, Jul 05 2022
From p. 60 of the Buchstaber and Panov link, S = P * C / T where S, P, C, and T are the bivariate e.g.f.s of the h vectors of the stellahedra, permutahedra, hypercubes, and (n-1)-simplices, respectively. - Tom Copeland, Jan 09 2017
The number of Le-diagrams of type (k, n) this means the diagram uses the bounding box size k x (n-k), equivalently the number of Grassmann necklaces of type (k, n) and also the number of decorated permutations with k anti-exceedances. - Thomas Scheuerle, Dec 29 2024

Examples

			The triangle T(n, k) begins:
n\k 0   1     2      3      4      5      6     7
0:  1
1:  1   1
2:  1   3     1
3:  1   7     7      1
4:  1  15    33     15      1
5:  1  31   131    131     31      1
6:  1  63   473    883    473     63      1
7:  1 127  1611   5111   5111   1611    127     1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, Holland, 1974, page 245 [From Roger L. Bagula, Nov 21 2009]
  • D. Singh, The numbers L(m,n) and their relations with prepared Bernoulli and Eulerian numbers, Math. Student, 20 (1952), 66-70.

Crossrefs

Programs

  • Maple
    T := (n, k) -> add(binomial(n, r)*combinat:-eulerian1(r, r-k), r = k .. n):
    for n from 0 to 8 do seq(T(n, k), k=0..n) od; # Peter Luschny, Jun 27 2018
  • Mathematica
    t[, 1] = 1; t[n, n_] = 1; t[n_, 2] = 2^(n-1)-1;
    t[n_, k_] = Sum[((i-k+1)^i*(k-i)^(n-i-1) - (i-k+2)^i*(k-i-1)^(n-i-1))*Binomial[n-1, i], {i, 0, k-1}];
    T[n_, k_] := t[n+1, k+1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten
    (* Jean-François Alcover, Jan 22 2015, after Tom Copeland *)
    T[ n_, k_] := Coefficient[n! SeriesCoefficient[(1-x) Exp[t] / (1 - x Exp[(1-x) t]), {t, 0, n}] // Simplify, x, k];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] (* Michael Somos, Jan 22 2015 *)

Formula

E.g.f.: (y-1)*exp(x*y)/(y-exp((y-1)*x)). - Vladeta Jovovic, Sep 20 2003
p(t,x) = (1 - x)*exp(t)/(1 - x*exp(t*(1 - x))). - Roger L. Bagula, Nov 21 2009
With offset=0, T(n,0)=1 otherwise T(n,k) = sum_{i=0..k-1} C(n,i)((i-k)^i*(k-i+1)^(n-i) - (i-k+1)^i*(k-i)^(n-i)) (cf. Williams). - Tom Copeland, Oct 10 2014
With offset 0, T = A007318 * A123125. Second column is A000225; 3rd, appears to be A066810. - Tom Copeland, Jan 23 2015
A raising operator (with D = d/dx) associated with this entry's row polynomials is R = x + t + (1-t) / [1-t e^{(1-t)D}] = x + t + 1 + t D + (t+t^2) D^2/2! + (t+4t^2+t^3) D^3/3! + ... , containing the e.g.f. for the Eulerian polynomials of A123125. Then R^n 1 = (p.(0;t)+x)^n = p_n(x;t) are the Appell polynomials with this entry's row polynomials p_n(0;t) as the base sequence. Examples of this formalism are given in A028246 and A248727. - Tom Copeland, Jan 24 2015
With offset 0, T = A007318*(padded A090582)*(inverse of A097805) = A007318*(padded A090582)*(padded A130595) = A007318*A123125 = A007318*(padded A090582)*A007318*A097808*A130595, where padded matrices are of the form of padded A007318, which is A097805. Inverses of padded matrices are just the padded versions of inverses of the unpadded matrices. This relates the face vectors, or f-vectors, and h-vectors of the permutahedra / permutohedra to those of the stellahedra / stellohedra. - Tom Copeland, Nov 13 2016
Umbrally, the row polynomials (offset 0) are r_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A123125. - Tom Copeland, Nov 16 2016
From the previous umbral statement, OP(x,d/dy) y^n = (y + q.(x))^n, where OP(x,y) = exp[y * q.(x)] = (1-x)/(1-x*exp((1-x)y)), the e.g.f. of A123125, so OP(x,d/dy) y^n evaluated at y = 1 is r_n(x), the n-th row polynomial of this entry, with offset 0. - Tom Copeland, Jun 25 2018
Consolidating some formulas in this entry and A248727, in umbral notation for concision, with all offsets 0: Let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry (A046802, the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
From Peter Luschny, Apr 30 2021: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A122045(n).
Sum_{k=0..n} 2^(n-k)*T(n,k) = A007047(n).
Sum_{k=0..n} T(n, n-k) = A000522(n).
Sum_{k=0..n} T(n-k, k) = Sum_{k=0..n} (n - k)^k = A026898(n-1) for n >= 1.
Sum_{k=0..n} k*T(n, k) = A036919(n) = floor(n*n!*e/2).
(End)

Extensions

More terms from Vladeta Jovovic, Sep 20 2003
First formula corrected by Wolfdieter Lang, Feb 14 2015
Offset set to 0 and edited by Peter Luschny, Apr 30 2021

A223511 Triangle T(n,k) represents the coefficients of (x^9*d/dx)^n, where n=1,2,3,...;generalization of Stirling numbers of second kind A008277, Lah-numbers A008297.

Original entry on oeis.org

1, 9, 1, 153, 27, 1, 3825, 855, 54, 1, 126225, 32895, 2745, 90, 1, 5175225, 1507815, 150930, 6705, 135, 1, 253586025, 80565975, 9205245, 499590, 13860, 189, 1, 14454403425, 4926412575, 623675430, 39180645, 1345050, 25578, 252, 1
Offset: 1

Views

Author

Udita Katugampola, Mar 23 2013

Keywords

Comments

Also the Bell transform of A045755(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 29 2016

Examples

			1;
9,1;
153,27,1;
3825,855,54,1;
126225,32895,2745,90,1;
5175225,1507815,150930,6705,135,1;
253586025,80565975,9205245,499590,13860,189,1;
14454403425,4926412575,623675430,39180645,1345050,25578,252,1;
		

Crossrefs

Programs

  • Maple
    b[0]:=g(x):
    for j from 1 to 10 do
    b[j]:=simplify(x^9*diff(b[j-1],x$1);
    end do;
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0, ..) as column 0.
    BellMatrix(n -> mul(8*k+1, k=0..n), 10); # Peter Luschny, Jan 29 2016
  • Mathematica
    rows = 8;
    t = Table[Product[8k+1, {k, 0, n}], {n, 0, rows}];
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)

A099594 Array read by antidiagonals: poly-Bernoulli numbers B(-k,n).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 8, 14, 8, 1, 1, 16, 46, 46, 16, 1, 1, 32, 146, 230, 146, 32, 1, 1, 64, 454, 1066, 1066, 454, 64, 1, 1, 128, 1394, 4718, 6902, 4718, 1394, 128, 1, 1, 256, 4246, 20266, 41506, 41506, 20266, 4246, 256, 1, 1, 512, 12866, 85310, 237686, 329462, 237686, 85310, 12866, 512, 1
Offset: 0

Views

Author

Ralf Stephan, Oct 27 2004

Keywords

Comments

B_n^{(-k)} is the number of distinct n by k "lonesum matrices" where a matrix of entries 0 or 1 is called lonesum when it is uniquely reconstructible from its row and column sums. [Brewbaker]
B_n^{(-k)} is the cardinality of the set { sigma in S_{n+k}: -k <= i-sigma(i) <= n for all i=1,2,...,n+k }. [Launois]
T(n,k) is also the number of permutations on [n+k] in which each substring whose support belongs to {1, 2, ..., n} or {n+1, n+2, ..., n+k} is increasing. For example, with n = 2 and k = 3, the permutation 41532 does not qualify because the substring 53 has support in {n+1, n+2, ..., n+k} = {3,4,5} but is not increasing. T(2,1) = 4 counts 123, 132, 231, 312 while the permutations satisfying Launois' condition above are 123, 132, 213, 231. A bijection between these sets of permutations would be interesting. - David Callan, Jul 22 2008. (Corrected by Norman Do, Sep 01 2008)
T(n,k) is also the number of acyclic orientations of the complete bipartite graph K_{n,k}. - Vincent Pilaud, Sep 15 2020
When indexed as a triangular array, T(n,k) is the number of permutations of [n] in which 1 is in position k and the excedance entries are precisely the entries to the left of 1. See link. - David Callan, Dec 12 2021
T(n,k) is also the number of max-closed relations between an ordered n-element set and an ordered k-element set (see the paper by Jeavons and Cooper 1995). - Don Knuth, Feb 12 2024

Examples

			Square array begins:
  1,  1,   1,    1,     1,      1, ...
  1,  2,   4,    8,    16,     32, ...
  1,  4,  14,   46,   146,    454, ...
  1,  8,  46,  230,  1066,   4718, ...
  1, 16, 146, 1066,  6902,  41506, ...
  1, 32, 454, 4718, 41506, 329462, ...
  ...
		

Crossrefs

Main diagonal is A048163. Another diagonal is A188634.
Antidiagonal sums are in A098830.

Programs

  • Maple
    A:= (n, k)-> add(Stirling2(n+1, i+1)*Stirling2(k+1, i+1)*
                 i!^2, i=0..min(n, k)):
    seq(seq(A(n, d-n), n=0..d), d=0..10);  # Alois P. Heinz, Jan 02 2016
  • Mathematica
    T[n_, k_] := Sum[(-1)^(j+n)*(1+j)^k*j!*StirlingS2[n, j], {j, 0, n}]; Table[ T[n-k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 30 2016 *)
  • PARI
    T(n,k)=sum(j=0,n,(j+1)^k*sum(i=0,j,(-1)^(n+j-i)*binomial(j,i)*(j-i)^n))
    
  • PARI
    T(n,k)=sum(j=0,min(n,k), j!^2*stirling(n+1, j+1, 2)*stirling(k+1, j+1, 2)); \\ Michel Marcus, Mar 05 2017

Formula

pB(k, n) = (-1)^n * Sum[i=0..n, (-1)^i * i! * Stirling2(n, i) / (i+1)^k ].
E.g.f.: e^(x+y) / [e^x + e^y - e^(x+y)].
T(n, k) = Sum_{j=0..n} (j+1)^k*Sum_{i=0..j} (-1)^(n+j-i)*C(j, i)*(j-i)^n. - Paul D. Hanna, Nov 04 2004
n-th row of the array = row sums of n-th power of triangle A210381. - Gary W. Adamson, Mar 21 2012

A134685 Irregular triangle read by rows: coefficients C(j,k) of a partition transform for direct Lagrange inversion.

Original entry on oeis.org

1, -1, 3, -1, -15, 10, -1, 105, -105, 10, 15, -1, -945, 1260, -280, -210, 35, 21, -1, 10395, -17325, 6300, 3150, -280, -1260, -378, 35, 56, 28, -1, -135135, 270270, -138600, -51975, 15400, 34650, 6930, -2100, -1575, -2520, -630, 126, 84, 36, -1
Offset: 1

Views

Author

Tom Copeland, Jan 26 2008, Sep 13 2008

Keywords

Comments

Let f(t) = u(t) - u(0) = Ev[exp(u.* t) - u(0)] = log{Ev[(exp(z.* t))/z_0]} = Ev[-log(1- a.* t)], where the operator Ev denotes umbral evaluation of the umbral variables u., z. or a., e.g., Ev[a.^n + a.^m] = a_n + a_m . The relation between z_n and u_n is given in reference in A127671 and u_n = (n-1)! * a_n .
If u_1 is not equal to 0, then the compositional inverse for these expressions is given by g(t) = Sum_{j>=1} P(j,t) where, with u_n denoted by (n') for brevity,
P(1,t) = (1')^(-1) * [ 1 ] * t
P(2,t) = (1')^(-3) * [ -(2') ] * t^2 / 2!
P(3,t) = (1')^(-5) * [ 3 (2')^2 - (1')(3') ] * t^3 / 3!
P(4,t) = (1')^(-7) * [ -15 (2')^3 + 10 (1')(2')(3') - (1')^2 (4') ] * t^4 / 4!
P(5,t) = (1')^(-9) * [ 105 (2')^4 - 105 (1') (2')^2 (3') + 10 (1')^2 (3')^2 + 15 (1')^2 (2') (4') - (1')^3 (5') ] * t^5 / 5!
P(6,t) = (1')^(-11) * [ -945 (2')^5 + 1260 (1') (2')^3 (3') - 280 (1')^2 (2') (3')^2 - 210 (1')^2 (2')^2 (4') + 35 (1')^3 (3')(4') + 21 (1')^3 (2')(5') - (1')^4 (6') ] * t^6 / 6!
P(7,t) = (1')^(-13) * [ 10395 (2')^6 - 17325 (1') (2')^4 (3') + (1')^2 [ 6300 (2')^2 (3')^2 + 3150 (2')^3 (4')] - (1')^3 [280 (3')^3 + 1260 (2')(3')(4') + 378 (2')^2(5')] + (1')^4 [35 (4')^2 + 56 (3')(5') + 28 (2')(6')] - (1')^5 (7') ] * t^7 / 7!
P(8,t) = (1')^(-15) * [ -135135 (2')^7 + 270270 (1') (2')^5 (3') - (1')^2 [ 138600 (2')^3 (3')^2 + 51975 (2')^4 (4')] + (1')^3 [15400 (2')(3')^3 + 34650 (2')^2(3')(4') + 6930 (2')^3(5')] - (1')^4 [2100 (3')^2(4') + 1575 (2')(4')^2 + 2520 (2')(3')(5') + 630 (2')^2(6') ] + (1')^5 [126 (4')(5') + 84 (3')(6') + 36 (2')(7')] - (1')^6 (8') ] * t^8 / 8!
...
Substituting ((m-1)') for (m') in each partition and ignoring the (0') factors, the partitions in the brackets of P(n,t) become those of n-1 listed in Abramowitz and Stegun on page 831 (in the reversed order) and the number of partitions in P(n,t) is given by A000041(n-1).
Combinatorial interpretations are given in the link.
From Tom Copeland, Jul 10 2018: (Start)
Coefficients occurring in prolongation for the special Euclidean group SE(2) and special affine group SA(2) in the Olver presentation on moving frames (MFP) in slides 33 and 42. These are a result of applying an iterated derivative of the form h(x)d/dx = d/dy as in this entry (more generally as g(x) d/dx as discussed in A145271). See also p. 6 of Olver's paper on contact forms, but note that the 12 should be a 15 in the formula for the compositional inverse of S(t).
Change variables in the MFP to obtain connections to the partition polynomials Prt_n = n! * P(n,1) above. Let delta and beta in the formulas for the equi-affine curves in MFP be L and B, respectively, and D_y = (1/(L-B*u_x)) d/dx = (1/w_x) d/dx. Then v_(yy) = (1/B) [-w_(xx)/(w_x)^3] in MFP (there is an overall sign error in MFP for v_(yy) and higher derivatives w.r.t. y), and (d/dy)^n v = v_n = (1/B)* [(1/w_1)*(d/dx)]^(n-2) [-w_2/(w_1)^3] for n > 1, with w_n = (d/dx)^n w. Consequently, in the partition polynomials Prt_n for n > 1 here substitute (n') = -B*u_n = w_n for n > 1 and (1') = L-B*u_1 = w_1, where u_n = (d/dx)^n u, and then divide by B. For example, v_4 = (1/B)*Prt_4 = (1/B)*4!*P(4,1) = (1/B) (L-B*u_n)^(-7) [-15*(-B*u_2)^3 + 10 (L-B*u_1)(-B*u_2)(-B*u_3) - (L-B*u_1)^2 (-B*u_4)], agreeing with v_4 in MFP except for the overall sign.
For the SE(2) transformation formulas in MFP, let w_x = cos(phi) + sin(phi)*u_x, and then the same transformations apply as above with cos(phi) and sin(phi) substituted for L and -B, respectively. (End)

Examples

			Examples and checks:
1) Let u_1 = -1 and u_n = 1 for n>1,
then f(t) = exp(u.*t) - u(0) = exp(t)-2t-1
and g(t) = [e.g.f. of signed A000311];
therefore, the row sums of unsigned [C(j,k)] are A000311 =
(0,1,1,4,26,236,2752,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),4!*P(4,1),...).
2) Let u_1 = -1 and u_n = (n-1)! for n>1,
then f(t) = -log(1-t)-2t
and g(t) = [e.g.f. of signed (0,A032188)]
with (0,A032188) = (0,1,1,5,41,469,6889,...) = (0,-P(1,1),2!*P(2,1),-3!P(3,1),...).
3) Let u_1 = -1 and u_n = (-1)^n (n-2)! for n>1, then
f(t) = (1+t) log(1+t) - 2t
and g(t) = [e.g.f. of signed (0,A074059)]
with (0,A074059) = (0,1,1,2,7,34,213,...) = (0,-P(1,1),2!*P(2,1),-3!*P(3,1),...).
4) Let u_1 = 1, u_2 = -1 and u_n = 0 for n>2,
then f(t) = t(1-t/2)
and g(t) = [e.g.f. of (0,A001147)] = 1 - (1-2t)^(1/2)
with (0,A001147) = (0,1,1,3,15,105,945...) =(0,P(1,1),2!*P(2,1),3!*P(3,1),...).
5) Let u_1 = 1, u_2 = -2 and u_n = 0 for n>2,
then f(t)= t(1-t)
and g(t) = t * [o.g.f. of A000108] = [1 - (1-4t)^(1/2)] / 2
with (0,A000108) = (0,1,1,2,5,14,42,...) = (0,P(1,1),P(2,1),P(3,1),...).
.
From _Peter Luschny_, Feb 19 2021: (Start)
Triangle starts:
 [1]  1;
 [2] -1;
 [3]  3,     -1;
 [4] -15,     10,    -1;
 [5]  105,   -105,   [10, 15],  -1;
 [6] -945,    1260,  [-280, -210], [35, 21],  -1;
 [7]  10395, -17325, [6300, 3150], [-280, -1260, -378], [35, 56, 28], -1;
 [8] -135135, 270270, [-138600, -51975], [15400, 34650, 6930], [-2100, -1575, -2520, -630], [126, 84, 36], -1
The coefficients can be seen as a refinement of the Ward numbers: Let R(n, k) = Sum T(n, k), where the sum collects adjacent terms with equal sign, as indicated by the square brackets in the table, then R(n+1, k+1) = (-1)^(n-k)*W(n, k), where W(n, k) are the Ward numbers A181996, for n >= 0 and 0 <= k <= n-1.  (End)
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 831.
  • D. S. Alexander, A History of Complex Dynamics: From Schröder to Fatou to Julia, Friedrich Vieweg & Sohn, 1994, p. 10.
  • J. Riordan, Combinatorial Identities, Robert E. Krieger Pub. Co., 1979, (unsigned partition polynomials in Table 5.2 on p. 181, but may have errors).

Crossrefs

Cf. A145271, (A134991, A019538) = (reduced array, associated g(x)).
Cf. A181996 (Ward numbers).

Programs

  • Mathematica
    rows[n_] := {{1}}~Join~Module[{h = 1/(1 + Sum[u[k] y^k/k!, {k, n-1}] + O[y]^n), g = y, r}, r = Reap[Do[g = h D[g, y]; Sow[Expand[Normal@g /. {y -> 0}]], {k, n}]][[2, 1, ;;]]; Table[Coefficient[r[[k]], Product[u[t], {t, p}]], {k, 2, n}, {p, Reverse@Sort[Sort /@ IntegerPartitions[k-1]]}]];
    rows[8] // Flatten (* Andrei Zabolotskii, Feb 19 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 05 2011: (Start)
Let h(t) = 1/(df(t)/dt)
= 1/Ev[u.*exp(u.*t)]
= 1/(u_1+(u_2)*t+(u_3)*t^2/2!+(u_4)*t^3/3!+...),
an e.g.f. for the partition polynomials of A133314
(signed A049019) with an index shift.
Then for the partition polynomials of A134685,
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)). (Cf. A000311 and A134991)(End)
From Tom Copeland, Oct 30 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)^2/2!+...], 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
Equivalent matrix computation: Multiply the m-th diagonal (with m=1 the index of the main diagonal) of the lower triangular Pascal matrix by u_m = (d/dx)^m f(x) evaluated at x=0 to obtain the matrix UP with UP(n,k) = binomial(n,k) u_{n+1-k}. 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. - Tom Copeland, Jul 15 2016
Also, P(n,t) = (1, 0, 0, 0, ...) [UP^(-1) * S]^n (0, 1, 0, ..)^T * t^n/n! in agreement with A139605. - Tom Copeland, Aug 27 2016
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 = -u5, b2 = 15 u2 u4 + 10 u3^2, b3 = -105 u2^2 u3, and b4 = 105 u2^4.
The relation between solutions of the inviscid Burgers' 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 * 105 u2^2 u3 + 1 * 15 u2 u4 + 1 * 10 u3^2 - 0 * u5 = 315 u2^4 - 210 u2^2 u3 + 15 u2 u4 + 10 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 + 10 u2 u3 - u4) + 6!/(3!*3!) * (3 u2^2 - 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 Bell polynomials of A036040 is presented in the blog entry "Formal group laws and binomial Sheffer sequences." - Tom Copeland, Feb 06 2018

Extensions

P(7,t) and P(8,t) data added by Tom Copeland, Jan 14 2016
Terms in rows 5-8 reordered by Andrei Zabolotskii, Feb 19 2024
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