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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A136126 Triangle read by rows: T(n,k) is the number of permutations of {1,2,...,k+n} having excedance set {1,2,...,k} (the empty set for k=0), 0 <= k <= n-1.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 7, 1, 1, 15, 31, 15, 1, 1, 31, 115, 115, 31, 1, 1, 63, 391, 675, 391, 63, 1, 1, 127, 1267, 3451, 3451, 1267, 127, 1, 1, 255, 3991, 16275, 25231, 16275, 3991, 255, 1, 1, 511, 12355, 72955, 164731, 164731, 72955, 12355, 511, 1
Offset: 1

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Author

Emeric Deutsch, Jan 17 2008

Keywords

Comments

The excedance set of a permutation p in S_n is the set of indices i such that p(i) > i.
Columns 1,2,3,4 yield A000225, A091344, A091347, A091348, respectively. Row sums yield A136127.
T(a+b-1,b-1)*(-1)^(a+b-1) = Sum_{k=0..} F(a,b,k)*(-1)^k where F(a,b,k) is the number of connected subgraphs of K(a,b) (the complete bipartite graph) with k edges. F(n,n,k) is A255192(n,k). - Thomas Dybdahl Ahle, Feb 18 2015 [The sum starts with k=0, and F(n,n,k) is A255192(n,k), but there seems to be no A255192(n,0). Is there no upper k-summation limit? - Wolfdieter Lang, Mar 15 2015]
Comment from Don Knuth, Aug 25 2020, added by N. J. A. Sloane, Sep 07 2020: (Start)
This array also arises from a problem about {0,1}-matrices. Symmetric array read by antidiagonals: A(n,k) (n >= 1, k >= 0) = number of n X k matrices of 0's and 1's satisfying two conditions: (i) no column is entirely 0; (ii) no 0 has simultaneously a 1 above it and another 1 to its left.
Equivalently (see the Steingrímsson-Williams reference) A(n,k) is the number of permutations p_1...p_{n+k} on {1,...,n+k} for which p_1 >= 1, ..., p_n >= n, p_{n+1} < n+1,..., p_{n+k} < n+k. Then A(n,k) = A(k+1,n-1), for n >= 1 and k >= 0.
For example, the seven 2 X 2 matrices satisfying (i) and (ii) are
00 01 10 10 11 11 11
11 11 01 11 00 01 11
and the seven permutations of {1, 2, 3, 4} satisfying the other definition are
1423, 2413, 3412, 3421, 4213, 4312, 4321.
(End)

Examples

			T(4,2) = 7 because 3412, 4312, 2413, 2314, 2431, 3421 and 4321 are the only permutations of {1,2,3,4} with excedance set {1,2}.
Triangle starts:
  1;
  1,   1;
  1,   3,    1;
  1,   7,    7,     1;
  1,  15,   31,    15,     1;
  1,  31,  115,   115,    31,     1;
  1,  63,  391,   675,   391,    63,    1;
  1, 127, 1267,  3451,  3451,  1267,  127,   1;
  1, 255, 3991, 16275, 25231, 16275, 3991, 255, 1;
  ...
Formatted as a square array A(n,k) with 0 <= k <= n:
  1,   1,    1,     1,      1,        1,         1,          1, ... [A000012]
  1,   3,    7,    15,     31,       63,       127,        255, ... [A000225]
  1,   7,   31,   115,    391,     1267,      3991,      12355, ... [A091344]
  1,  15,  115,   675,   3451,    16275,     72955,     316275, ... [A091347]
  1,  31,  391,  3451,  25231,   164731,    999391,    5767051, ... [A091348]
  1,  63, 1267, 16275, 164731,  1441923,  11467387,   85314915, ...
  1, 127, 3991, 72955, 999391, 11467387, 116914351, 1096832395, ...
		

Crossrefs

Programs

  • Maple
    with(combinat): T:=proc(n,k) if k < n then add((-1)^(k+1-i)*factorial(i)*i^(n-1-k)* stirling2(k+1,i),i=1..k+1) else 0 end if end proc: for n to 10 do seq(T(n,k),k=0..n-1) end do; # yields sequence in triangular form
    # Alternatively as a square array:
    A := (n, k) -> add((-1)^(k-j)*j!*Stirling2(k+1,j+1)*(j+1)^(n+1), j=0..k);
    seq(print(seq(A(n, k), k=0..7)), n=0..6); # Peter Luschny, Mar 14 2018
    # Using the exponential generating function as given by Arakawa & Kaneko:
    gf := polylog(-t, 1-exp(-x))/(exp(x)-1):
    ser := series(gf, x, 12): c := n -> n!*coeff(ser, x, n):
    seq(lprint(seq(subs(t=k, c(n)), n=0..8)), k=0..8); # Peter Luschny, Apr 29 2021
    # Using recurrence relations:
    A := proc(n, k) option remember; local j; if n = 0 then return k^n fi;
    add(binomial(k+1, j+1)*A(n-1, k-j), j = 0..k) end:
    for n from 0 to 7 do lprint(seq(A(n, k), k=0..8)) od;  # Peter Luschny, Apr 19 2024
  • Mathematica
    T[n_, k_] := Sum[(-1)^(k + 1 - i)*i!*i^(n - 1 - k)*StirlingS2[k + 1, i], {i, 1, k + 1}];
    Table[T[n, k], {n, 1, 10}, {k, 0, n - 1}] // Flatten (* Jean-François Alcover, Nov 16 2017 *)
  • PARI
    {T(n,k)=polcoeff(polcoeff( x*y*sum(m=0, n, m!*x^m*prod(k=1, m, (1+y+k*x*y)/(1+(1+y)*k*x+k^2*x^2*y +x*O(x^n))) ), n,x),k,y)} \\ Paul D. Hanna, Feb 01 2013
    for(n=1, 10,for(k=1,n, print1(T(n,k), ", "));print(""))
    
  • PARI
    tabl(nn) = {default(seriesprecision, nn+1); pol = log(1/(1-(exp(x)-1)*(exp(y)-1))) + O(x^nn); for (n=1, nn-1, poly = n!*polcoeff(pol, n, x); for (k=1, n, print1(k!*polcoeff(poly, k, y), ", ");); print(););} \\ Michel Marcus, Apr 17 2015

Formula

T(n,k) = Sum_{i=1..k+1} (-1)^(k+1-i)*i!*i^(n-1-k)*Stirling2(k+1,i) (0 <= k <= n-1).
G.f.: A(x,y) = x*y*Sum_{n>=1} n! * x^n*Product_{k=1..n} (1 + y + k*x*y) / (1 + (1+y)*k*x + k^2*x^2*y). - Paul D. Hanna, Feb 01 2013
Central terms of triangle equals A092552. - Paul D. Hanna, Feb 01 2013
T(n,k-1) = Sum_{i=0..k, m=0..i} binomial(i,m)*(-1)^(k-m)*i^(n-k)*m^k (1 <= k <= n). - Thomas Dybdahl Ahle, Feb 18 2015
E.g.f.: log(1/(1-(exp(x)-1)*(exp(y)-1))). - Vladimir Kruchinin, Apr 17 2015
Let W(n,k) = k!*Stirling2(n+1, k+1) denote the Worpitzky numbers, then A(n,k) = Sum_{j=0..k} (-1)^(k-j)*W(k,j)*(j+1)^(n+1) enumerates the square array. - Peter Luschny, Mar 14 2018
Assume the missing first row (1,0,0,...) of the array which Ayyer and Bényi call the 'poly-Bernoulli numbers of type C'. Then T(n, k) = p_{n}(k) where p_{n}(x) = Sum_{k=0..n} (-1)^(n-k)*(k+1)^x*Sum_{j=0..n} E1(n,j)*binomial(n-j, n-k), and E1(n, k) are the Eulerian numbers of first order. This reflects the Worpitzky approach to the Bernoulli numbers. This formula can alternatively be written as: T(n, k) = Sum_{j=0..k} (-1)^(k-j)*(j+1)^n*A028246(k+1, j+1). - Peter Luschny, Apr 29 2021

Extensions

Definition corrected. Changed "T(n,k) is the number of permutations of {1,2,...,n}..." to "T(n,k) is the number of permutations of {1,2,...,k+n}..." - Karel Casteels (kcasteel(AT)sfu.ca), Feb 17 2010

A005461 Number of simplices in barycentric subdivision of n-simplex.

Original entry on oeis.org

1, 15, 180, 2100, 25200, 317520, 4233600, 59875200, 898128000, 14270256000, 239740300800, 4249941696000, 79332244992000, 1556132497920000, 32011868528640000, 689322235650048000, 15509750302126080000, 364022962973429760000, 8898339094906060800000
Offset: 1

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Author

Keywords

Examples

			G.f. = x + 15*x^2 + 180*x^3 + 2100*x^4 + 25200*x^5 + 317520*x^6 + ...
		

References

  • R. Austin, R. K. Guy, and R. Nowakowski, unpublished notes, circa 1987.
  • R. K. Guy, personal communication.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [Factorial(n-1)*StirlingSecond(n+3,n): n in [1..35]]; // G. C. Greubel, Nov 23 2022
  • Maple
    a:=n->sum((n-j)*n!/4!, j=3..n): seq(a(n), n=4..17); # Zerinvary Lajos, Apr 29 2007
  • Mathematica
    Table[(n(n+1)(n+3)!)/48,{n,20}] (* Harvey P. Dale, Mar 14 2012 *)
    a[ n_] := If[ n < 0, 0, n (n + 1) (n + 3)! / 48]; (* Michael Somos, May 27 2014 *)
  • Sage
    [factorial(m+1)*binomial(m-1,2)/24 for m in range(3, 19)] # Zerinvary Lajos, Jul 05 2008
    
  • Sage
    [binomial(n,4)*factorial (n-2)/2 for n in range(4, 18)] #  Zerinvary Lajos, Jul 07 2009
    

Formula

a(n) = n*(n + 1)*(n + 3)!/48.
Essentially Stirling numbers of second kind - see A028246.
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j) then a(n-3) = (-1)^n*f(n,4,-3), (n>=4). - Milan Janjic, Mar 01 2009
E.g.f.: t*(3*t + 2)/(2*(t - 1)^6). - Ran Pan, Jul 10 2016
a(n) ~ sqrt(Pi/2)*exp(-n)*n^(n+1/2)*(n^5/24 + 85*n^4/288 + 5065*n^3/6912 + 955841*n^2/1244160 + 3710929*n/11943936). - Ilya Gutkovskiy, Jul 10 2016
From Amiram Eldar, May 06 2022: (Start)
Sum_{n>=1} 1/a(n) = 16*(e + gamma - Ei(1)) - 64/3, where e = A001113, gamma = A001620, and Ei(1) = A091725.
Sum_{n>=1} (-1)^(n+1)/a(n) = 32*(gamma - Ei(-1)) - 16/e - 56/3, where Ei(-1) = -A099285. (End)
a(n) = (n-1)! * Stirling2(n+3, n). - G. C. Greubel, Nov 23 2022

Extensions

More terms from Harvey P. Dale, Mar 14 2012

A161739 The RSEG2 triangle.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 4, 1, 0, 0, 13, 10, 1, 0, -4, 30, 73, 20, 1, 0, 0, -14, 425, 273, 35, 1, 0, 120, -504, 1561, 3008, 798, 56, 1, 0, 0, 736, -2856, 25809, 14572, 1974, 84, 1, 0, -12096, 44640, -73520, 125580, 218769, 55060, 4326, 120, 1
Offset: 0

Views

Author

Johannes W. Meijer & Nico Baken (n.h.g.baken(AT)tudelft.nl), Jun 18 2009

Keywords

Comments

The EG2[2*m,n] matrix coefficients were introduced in A008955. We discovered that EG2[2m,n] = Sum_{k = 1..n} (-1)^(k+n)*t1(n-1,k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2 with t1(n,m) the central factorial numbers A008955 and eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function.
A different way to define these matrix coefficients is EG2[2*m,n] = (1/m)*Sum_{k = 0..m-1} ZETA(2*m-2*k, n-1)*EG2[2*k, n] with ZETA(2*m, n-1) = zeta(2*m) - Sum_{k = 1..n-1} (k)^(-2*m) and EG2[0, n] = 1, for m = 0, 1, 2, ..., and n = 1, 2, 3, ... .
We define the row sums of the EG2 matrix rs(2*m,p) = Sum_{n >= 1} (n^p)*EG2(2*m,n) for p = -2, -1, 0, 1, ... and m >= p+2. We discovered that rs(2*m,p=-2) = 2*eta(2*m+2) = (1 - 2^(1-(2*m+2)))*zeta(2*m+2). This formula is quite unlike the other rs(2*m,p) formulas, see the examples.
The series expansions of the row generators RGEG2(z,2*m) about z = 0 lead to the EG2[2*m,n] coefficients while the series expansions about z = 1 lead to the ZG1[2*m-1,n] coefficients, see the formulas.
The first Maple program gives the triangle coefficients. Adding the second program to the first one gives information about the row sums rs(2*m,p).
The a(n) formulas of the right hand columns are related to sequence A036283, see also A161740 and A161741.

Examples

			The first few expressions for the ZG1[2*m-1,p+1] coefficients are:
  ZG1[2*m-1, 1] = (zeta(2*m-1))/(1/2)
  ZG1[2*m-1, 2] = (zeta(2*m-3) - zeta(2*m-1))/1
  ZG1[2*m-1, 3] = (zeta(2*m-5) - 5*zeta(2*m-3) + 4*zeta(2*m-1))/6
  ZG1[2*m-1, 4] = (zeta(2*m-7) - 14*zeta(2*m-5) + 49*zeta(2*m-3) - 36*zeta(2*m-1))/72
The first few rs(2*m,p) are (m >= p+2)
  rs(2*m, p=0) = ZG1[2*m-1,1]
  rs(2*m, p=1) = ZG1[2*m-1,1] + ZG1[2*m-1,2]
  rs(2*m, p=2) = ZG1[2*m-1,1] + 3*ZG1[2*m-1,2] + 2*ZG1[2*m-1,3]
  rs(2*m, p=3) = ZG1[2*m-1,1] + 7*ZG1[2*m-1,2] + 12*ZG1[2*m-1,3] + 6*ZG1[2*m-1,4]
The first few rs(2*m,p) are (m >= p+2)
  rs(2*m, p=-1) = zeta(2*m+1)/(1/2)
  rs(2*m, p=0) = zeta(2*m-1)/(1/2)
  rs(2*m, p=1) = (zeta(2*m-1) + zeta(2*m-3))/1
  rs(2*m, p=2) = (zeta(2*m-1) + 4*zeta(2*m-3) + zeta(2*m-5))/3
  rs(2*m, p=3) = (0*zeta(2*m-1) + 13*zeta(2*m-3) + 10*zeta(2*m-5) + zeta(2*m-7))/12
The first few rows of the RSEG2 triangle are:
  [1]
  [0, 1]
  [0, 1, 1]
  [0, 1, 4, 1]
  [0, 0, 13, 10, 1]
  [0, -4, 30, 73, 20, 1]
		

Crossrefs

A000007, A129825, A161742 and A161743 are the first four left hand columns.
A000012, A000292, A107963, A161740 and A161741 are the first five right hand columns.
A010790 equals 2*r(n) and A054977 equals denom(r(n)).
A001710 equals numer(q(n)) and A141044 equals denom(q(n)).
A000142 equals the row sums.
A008955 is a central factorial number triangle.
A028246 is Worpitzky's triangle.

Programs

  • Maple
    nmax:=10; for n from 0 to nmax do A008955(n, 0) := 1 end do: for n from 0 to nmax do A008955(n, n) := (n!)^2 end do: for n from 1 to nmax do for m from 1 to n-1 do A008955(n, m) := A008955(n-1, m-1)*n^2 + A008955(n-1, m) end do: end do: for n from 1 to nmax do A028246(n, 1) := 1 od: for n from 1 to nmax do A028246(n, n) := (n-1)! od: for n from 3 to nmax do for m from 2 to n-1 do A028246(n, m) := m*A028246(n-1, m) + (m-1)*A028246(n-1, m-1) od: od: for i from 0 to nmax-2 do s(i) := ((i+1)!/2)*sum(A028246(i+1, k1+1)*(sum((-1)^(j)*A008955(k1, j)*2*x^(2*nmax-(2*k1+1-2*j)), j=0..k1)/ (k1!*(k1+1)!)), k1=0..i) od: a(0,0) := 1: for n from 1 to nmax-1 do for m from 0 to n do a(n,m) := coeff(s(n-1), x, 2*nmax-1-2*m+2) od: od: seq(seq(a(n, m), m=0..n), n=0..nmax-1); for n from 0 to nmax-1 do seq(a(n, m), m=0..n) od;
    m:=7: row := 2*m; rs(2*m, -2) := 2*eta(2*m+2); for p from -1 to m-2 do q(p+1) := (p+1)!/2 od: for p from -1 to m-2 do rs(2*m, p) := sum(a(p+1, k)*zeta(2*m+1-2*k), k=0..p+1)/q(p+1) od;

Formula

RGEG2(2*m,z) = Sum_{n >= 1} EG2[2*m,n]*z^(n-1) = Integral_{y = 0..oo}((2*y)^(2*m)/(2*m)!)* cosh(y)/(cosh(y)^2 - z)^(3/2) for m >= 0.
EG2[2*m,n] = Sum_{k = 1..n} (-1)^(k+n)* A008955(n-1, k-1)*2*eta(2*m-2*n+2*k)/((n-1)!)^2.
ZG1[2*m-1,p+1] = Sum_{j = 0..p} (-1)^j*A008955(p, j)*zeta(2*m-(2*p+1-2*j))/ r(p) with r(p)= p!*(p+1)!/2 and p >= 0.
rs(2*m,p) = Sum_{k = 0..p} A028246(p+1,k+1)*ZG1[2*m-1,k+1] and p >= 0; p <= m-2.
rs(2*m,p) = Sum_{k = 0..p+1} A161739(p+1,k)*zeta(2*m+1-2*k)/q(p+1) with q(p+1) = (p+1)!/2 and p >= -1; p <= m-2.
From Peter Bala, Mar 19 2022: (Start)
It appears that the k-th row polynomial (with indexing starting at k = 1) is given by R(k,n^2) = (k-1)!*Sum_{i = 0..n} (-1)^(n-i)*(i^k)* binomial(n,i)*binomial(n+i,i)/(n+i) for n >= 1.
For example, for k = 6, Maple's SumTools:-Summation procedure gives 5!*Sum_{i = 0..n} (-1)^(n-i)*(i^6)*binomial(n,i)*binomial(n+i,i)/(n+i) = -4*n^2 + 30*n^4 + 73*n^6 + 20*n^8 + n^10 = R(6,n^2). (End)

Extensions

Minor error corrected and edited by Johannes W. Meijer, Sep 22 2012

A248727 A046802(x,y) --> A046802(x,y+1), transform of e.g.f. for the graded number of positroids of the totally nonnegative Grassmannians G+(k,n); enumerates faces of the stellahedra.

Original entry on oeis.org

1, 2, 1, 5, 5, 1, 16, 24, 10, 1, 65, 130, 84, 19, 1, 326, 815, 720, 265, 36, 1, 1957, 5871, 6605, 3425, 803, 69, 1, 13700, 47950, 65646, 44240, 15106, 2394, 134, 1, 109601, 438404, 707840, 589106, 267134, 63896, 7094, 263, 1
Offset: 0

Views

Author

Tom Copeland, Oct 12 2014

Keywords

Comments

This is a transform of A046802 treating it as an array of h-vectors, so y is replaced by (y+1) in the e.g.f. for A046802.
An e.g.f. for the reversed row polynomials with signs is given by exp(a.(0;t)x) = [e^{(1+t)x} [1+t(1-e^(-x))]]^(-1) = 1 - (1+2t)x + (1+5t+5t^2)x^2/2! + ... . The reciprocal is an e.g.f. for the reversed face polynomials of the simplices A074909, i.e., exp(b.(0;t)x) = e^{(1+t)x} [1+t(1-e^(-x))] = 1 + (1+2t)x +(1+3t+3t^2) x^2/2! + ... , so the relations of A133314 apply between the two sets of polynomials. In particular, umbrally [a.(0;t)+b.(0;t)]^n vanishes except for n=0 for which it's unity, implying the two sets of Appell polynomials formed from the two bases, a_n(z;t) = (a.(0;t)+z)^n and b_n(z;t) = (b.(0;t) + z)^n, are an umbral compositional inverse pair, i.e., b_n(a.(x;t);t)= x^n = a_n(b.(x;t);t). Raising operators for these Appell polynomials are related to the polynomials of A028246, whose reverse polynomials are given by A123125 * A007318. Compare: A248727 = A007318 * A123125 * A007318 and A046802 = A007318 * A123125. See A074909 for definitions and related links. - Tom Copeland, Jan 21 2015
The o.g.f. for the umbral inverses is Og(x) = x / (1 - x b.(0;t)) = x / [(1-tx)(1-(1+t)x)] = x + (1+2t) x^2 + (1+3t+3t^2) x^3 + ... . Its compositional inverse is an o.g.f for signed A033282, the reverse f-polynomials for the simplicial duals of the Stasheff polytopes, or associahedra of type A, Oginv(x) =[1+(1+2t)x-sqrt[1+2(1+2t)x+x^2]] / (2t(1+t)x) = x - (1+2t) x^2 + (1+5t+5t^2) x^3 + ... . Contrast this with the o.g.f.s related to the corresponding h-polynomials in A046802. - Tom Copeland, Jan 24 2015
Face vectors, or coefficients of the face polynomials, of the stellahedra, or stellohedra. See p. 59 of Buchstaber and Panov. - Tom Copeland, Nov 08 2016
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016

Examples

			The triangle T(n, k) starts:
n\k    0     1     2     3     4    5   6  7 ...
1:     1
2:     2     1
3:     5     5     1
4:    16    24    10     1
5:    65   130    84    19     1
6:   326   815   720   265    36    1
7:  1957  5871  6605  3425   803   69   1
8: 13700 47950 65646 44240 15106 2394 134  1
... reformatted, _Wolfdieter Lang_, Mar 27 2015
		

Crossrefs

Programs

  • Mathematica
    (* t = A046802 *) 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_, j_] := Sum[Binomial[k, j]*t[n + 1, k + 1], {k, j, n}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 23 2015, after Tom Copeland *)

Formula

Let M(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)]} with M(n,0)=1. Then M(n,k)= A046802(n,k), and T(n,j)= sum(k=j,..,n, C(k,j)*M(n,k)) for j>0 with T(n,0)= 1 + sum(k=1,..,n, M(n,k)) for n>0 and T(0,0)=1.
E.g.f: y * exp[x*(y+1)]/[y+1-exp(x*y)].
Row sums are A007047. Row polynomials evaluated at -1 are unity. Row polynomials evaluated at -2 are A122045.
First column is A000522. Second column appears to be A036918/2 = (A001339-1)/2 = n*A000522(n)/2.
Second diagonal is A052944. (Changed from conjecture to fact on Nov 08 2016.)
The raising operator for the reverse row polynomials with row signs is R = x - (1+t) - t e^(-D) / [1 + t(1-e^(-D))] evaluated at x = 0, with D = d/dx. Also R = x - d/dD log[exp(a.(0;t)D], or R = - d/dz log[e^(-xz) exp(a.(0;t)z)] = - d/dz log[exp(a.(-x;t)z)] with the e.g.f. defined in the comments and z replaced by D. Note that t e ^(-D) / [1+t(1-e^(-D))] = t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... is an e.g.f. for the signed reverse row polynomials of A028246. - Tom Copeland, Jan 23 2015
Equals A007318*(padded A090582)*A007318*A097808 = A007318*(padded (A008292*A007318))*A007318*A097808 = A007318*A130850 = A007318*(mirror of A028246). Padded means in the same way that A097805 is padded A007318. - Tom Copeland, Nov 14 2016
Umbrally, the row polynomials are p_n(x) = (1 + q.(x))^n, where (q.(x))^k = q_k(x) are the row polynomials of A130850. - 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)] = x/((1+x)*exp(-x*y) - 1), the e.g.f. of A130850, so OP(x,d/dy) y^n evaluated at y = 1 is p_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 A046082, 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 A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of this entry (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

Extensions

Title expanded by Tom Copeland, Nov 08 2016

A129825 a(n) = n!*Bernoulli(n-1), n > 2; a(0)=0, a(1)=1, a(2)=1.

Original entry on oeis.org

0, 1, 1, 1, 0, -4, 0, 120, 0, -12096, 0, 3024000, 0, -1576143360, 0, 1525620096000, 0, -2522591034163200, 0, 6686974460694528000, 0, -27033456071346536448000, 0, 160078872315904478576640000, 0, -1342964491649083924630732800000, 0, 15522270327163593186886877184000000, 0
Offset: 0

Views

Author

Paul Curtz, Jun 03 2007

Keywords

Comments

Define "conjugated" Bernoulli numbers G(n) via G(0)=0, G(1)=B(0)=1, G(2)=-B(1)=1/2, G(n+1)=B(n), where B(n)=A027641(n)/A027642(n).
The sequence is then defined by a(n) = n!*G(n).
The first differences are 1, 0, 0, -1, -4, 4, 120, -120, -12096, ...
The 2nd differences are -1, 0, -1, -3, 8, 116, -240, -11976, 24192, 3011904, ...

Crossrefs

Equals second left hand column of A161739 (RSEG2 triangle).
Other left hand columns are A161742 and A161743.
Cf. A094310 [T(n,k) = n!/k], A008277 [S2(n,k); Stirling numbers of the second kind], A028246 [Worpitzky's triangle] and A008955 [CFN triangle].

Programs

  • Magma
    [n le 2 select Floor((n+1)/2) else Factorial(n)*Bernoulli(n-1): n in [0..40]]; // G. C. Greubel, Apr 26 2024
    
  • Maple
    A129825 := proc(n) if n <= 1 then n; elif n = 2 then 1; else n!*bernoulli(n-1) ; fi; end: # R. J. Mathar, May 21 2009
  • Mathematica
    a[n_] := n!*BernoulliB[n-1]; a[0]=0; a[2]=1; Table[a[n], {n, 0, 28}] (* Jean-François Alcover, Mar 04 2013 *)
  • SageMath
    [(n+1)//2 if n <3 else factorial(n)*bernoulli(n-1) for n in range(41)] # G. C. Greubel, Apr 26 2024

Formula

From Johannes W. Meijer, Jun 18 2009: (Start)
a(n) = Sum_{k=1..n} (-1)^(k+1)*(n!/k)*S2(n, k)*(k-1)!.
a(n) = Sum_{k=0..n-1} ((-1)^k/(k!*(k+1)!))*n!*A028246(n, k+1) *A008955(k, k). (End)
a(n) = A129814(n-1) for n > 2. - Georg Fischer, Oct 07 2018

Extensions

Edited by R. J. Mathar, May 21 2009

A005460 a(n) = (3*n+4)*(n+3)!/24.

Original entry on oeis.org

1, 7, 50, 390, 3360, 31920, 332640, 3780000, 46569600, 618710400, 8821612800, 134399865600, 2179457280000, 37486665216000, 681734237184000, 13071512982528000, 263564384219136000, 5575400435404800000, 123469776914964480000, 2856835183101419520000
Offset: 0

Views

Author

Keywords

Comments

Essentially Stirling numbers of second kind - third external diagonal of Worpitzky triangle A028246.

References

  • R. Austin, R. K. Guy, and R. Nowakowski, unpublished notes, circa 1987.
  • R. K. Guy, personal communication.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A028246.

Programs

  • Magma
    [(3*n+4)*Factorial(n+3)/24: n in [0..20]]; // Vincenzo Librandi, Oct 08 2011
    
  • Mathematica
    Table[StirlingS2[n+3, n+1]*n!, {n,0,20}]
  • PARI
    a(n)=(3*n+4)*(n+3)!/24 \\ Charles R Greathouse IV, Jun 30 2017
    
  • SageMath
    [factorial(n)*stirling_number2(n+3,n+1) for n in range(21)] # G. C. Greubel, Nov 22 2022

Formula

E.g.f.: (1+2*x)/(1-x)^5.
a(n) = S2(n+3, n+1)*n! = n!*A001296(n+1). - Olivier Gérard, Sep 13 2016

A130850 Triangle read by rows, 0 <= k <= n, T(n,k) = Sum_{j=0..n} A(n,j)*binomial(n-j,k) where A(n,j) are the Eulerian numbers A173018.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 12, 7, 1, 24, 60, 50, 15, 1, 120, 360, 390, 180, 31, 1, 720, 2520, 3360, 2100, 602, 63, 1, 5040, 20160, 31920, 25200, 10206, 1932, 127, 1, 40320, 181440, 332640, 317520, 166824, 46620, 6050, 255, 1, 362880, 1814400, 3780000, 4233600, 2739240, 1020600, 204630, 18660, 511, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 20 2007

Keywords

Comments

Old name was: Triangle T(n,k), 0<=k<=n, read by rows given by [1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,2,0,3,0,4,0,5,0,6,0,...] where DELTA is the operator defined in A084938.
Vandervelde (2018) refers to this as the Worpitzky number triangle - N. J. A. Sloane, Mar 27 2018 [Named after the German mathematician Julius Daniel Theodor Worpitzky (1835-1895). - Amiram Eldar, Jun 24 2021]
Triangle given by A123125*A007318 (as infinite lower triangular matrices), A123125 = Euler's triangle, A007318 = Pascal's triangle; A007318*A123125 gives A046802.
Taylor coefficients of Eulerian polynomials centered at 1. - Louis Zulli, Nov 28 2015
A signed refinement is A263634. - Tom Copeland, Nov 14 2016
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 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 this entry (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

Examples

			Triangle begins:
1
1      1
2      3       1
6      12      7       1
24     60      50      15      1
120    360     390     180     31      1
720    2520    3360    2100    602     63      1
5040   20160   31920   25200   10206   1932    127    1
40320  181440  332640  317520  166824  46620   6050   255   1
362880 1814400 3780000 4233600 2739240 1020600 204630 18660 511 1
...
		

Crossrefs

Programs

  • Mathematica
    Table[(n-k)!*StirlingS2[n+1, n-k+1], {n, 0, 10}, {k, 0, n}] (* G. C. Greubel, Nov 15 2015 *)
  • PARI
    t(n, k) = (n-k)!*stirling(n+1, n-k+1, 2);
    tabl(nn) = for (n=0, 10, for (k=0, n, print1(t(n,k),", ")); print()); \\ Michel Marcus, Nov 16 2015
  • Sage
    from sage.combinat.combinat import eulerian_number
    def A130850(n, k):
        return add(eulerian_number(n, j)*binomial(n-j, k) for j in (0..n))
    for n in (0..7): [A130850(n, k) for k in (0..n)] # Peter Luschny, May 21 2013
    

Formula

T(n,k) = (-1)^k*A075263(n,k).
T(n,k) = (n-k)!*A008278(n+1,k+1).
T(n,n-1) = 2^n - 1 for n > 0. - Derek Orr, Dec 31 2015
E.g.f.: x/(e^(-x*t)*(1+x)-1). - Tom Copeland, Nov 14 2016
Sum_{k=1..floor(n/2)} T(n,2k) = Sum_{k=0..floor(n/2)} T(n,2k+1) = A000670(n). - Jacob Sprittulla, Oct 03 2021

Extensions

New name from Peter Luschny, May 21 2013

A141618 Triangle read by rows: number of nilpotent partial transformations (of an n-element set) of height r (height(alpha) = |Im(alpha)|), 0 <= r < n.

Original entry on oeis.org

1, 1, 2, 1, 9, 6, 1, 28, 72, 24, 1, 75, 500, 600, 120, 1, 186, 2700, 7800, 5400, 720, 1, 441, 12642, 73500, 117600, 52920, 5040, 1, 1016, 54096, 571536, 1764000, 1787520, 564480, 40320, 1, 2295, 217800, 3916080, 21019824, 40007520, 27941760, 6531840, 362880, 1, 5110, 839700, 24555600, 214326000
Offset: 1

Views

Author

Abdullahi Umar, Aug 23 2008

Keywords

Comments

The sum of each row of the sequence (as a triangular array) is A000272. Second left-downward diagonal is A058877.
From Tom Copeland, Oct 26 2014: (Start)
With T(x,t) the e.g.f. for A055302 for the number of labeled rooted trees with n nodes and k leaves, the mirror of the row polynomials of this array are given by e^T(x,t) = exp[ t * x + (2t) * x^2/2! + (6t + 3t^2) * x^3/3! + ...] = 1 + t * x + (2t + t^2) * x^2/2! + (6t + 9t^2 + t^3) * x^3/3! + ... = 1 + Nr(x,t).
Equivalently, e^x-1 = Nr[Tinv(x,t),t] = t * N[t*Tinv(x,t),1/t], where N(x,t) is the e.g.f. of this array and Tinv(x,t) is the comp. inverse in x of T(x,t). Note that Nr(x,t) = t * N(x*t,1/t), and N(x,t) = t * Nr(x*t,1/t). Also, log[1 + Nr(x,t)]= x * [t + Nr(x,t)] = T(x,t).
E.g.f. is N(x,t)= t * {exp[T(x*t,1/t)] - 1}, and log[1 + N(x,t)/t] = T(x*t,1/t) = x + (2t) * x^2/2! + (3t + 6t^2) * x^3/3! + (4t + 36t^2 + 24t^3) * x^4/4! + ... = x + (t) * x^2 + (t + 2t^2) * x^3/2! + (t + 9t^2 + 6t^3) * x^4/3! + ... is the comp. inverse in x of x / [1 + t * (e^x - 1)].
The exp/log transforms (A036040/A127671) generally give associations between enumerations of sets of connected graphs/objects (in this case, trees) and sets of disconnected (or not necessarily connected) graphs/objects (in this case, bipartite graphs of the nilpotent transformations). The transforms also relate formal cumulants and moments so that Nr(x,t) is then the e.g.f. for the formal moments associated to the formal cumulants whose e.g.f. is T(x,t). (End)
T(n,k) is the number of parking functions of length n containing exactly k+1 distinct values in its image. - Alan Kappler, Jun 08 2024

Examples

			N(J(4,2)) = 6*6*2 = 72.
From _Peter Bala_, Oct 22 2008: (Start)
Triangle begins
n\k|..0.....1.....2.....3.....4....5
=====================================
.1.|..1
.2.|..1.....2
.3.|..1.....9.....6
.4.|..1....28....72....24
.5.|..1....75...500...600...120
.6.|..1...186..2700..7800..5400...720
...
(End)
		

Crossrefs

Programs

  • Maple
    A048993 := proc(n,k)
        combinat[stirling2](n,k) ;
    end proc:
    A141618 := proc(n,k)
        binomial(n,k)*k!*A048993(n,k+1) ;
    end proc:
  • Mathematica
    Flatten[CoefficientList[CoefficientList[InverseSeries[Series[Log[1 + x]/(1 + t*x),{x,0,9}]],x]*Table[n!, {n,0,9}],t]] (* Peter Luschny, Oct 24 2015, after Peter Bala *)
  • PARI
    A055302(n,k)=n!/k!*stirling(n-1, n-k,2);
    T(n,k)=A055302(n+1,n+1-k) / (n+1);
    for(n=1,10,for(k=1,n,print1(T(n,k),", "));print());
    \\ Joerg Arndt, Oct 27 2014

Formula

N(J(n,r)) = C(n,r)*S(n,r+1)*r! where S(n, r + 1) is a Stirling number of the second kind (given by A048993 with zeros removed); generating function = (x+1)^(n-1).
From Peter Bala, Oct 22 2008: (Start)
Define a functional I on formal power series of the form f(x) = 1 + ax + bx^2 + ... by the following iterative process. Define inductively f^(1)(x) = f(x) and f^(n+1)(x) = f(x*f^(n)(x)) for n >= 1. Then set I(f(x)) = lim_{n -> infinity} f^(n)(x) in the x-adic topology on the ring of formal power series; the operator I may also be defined by I(f(x)) := 1/x*series reversion of x/f(x).
Let f(x) = 1 + a*x + a*x^2/2! + a*x^3/3! + ... . Then the e.g.f. for this table is I(f(x)) = 1 + a*x +(a + 2*a^2)*x^2/2! + (a + 9*a^2 + 6*a^3)*x^3/3! + (a + 28*a^2 + 72*a^3 + 24*a^4)*x^4/4! + ... . Note, if we take f(x) = 1 + a*x + a*x^2 + a*x^3 + ... then I(f(x)) is the o.g.f. of the Narayana triangle A001263. (End)
A generator for this array is given by the inverse, g(x,t), of f(x,t)= x/(1 + t * (e^x-1)). Then A248927 gives h(x,t)= x / f(x,t) = 1 + t*(e^x-1)= 1 + t * (x + x^2/2! + x^3/3! + ...) and g(x,t)= x * (1 + t * x + (t + 2 t^2) * x^2/2! + (t + 9 t^2 + 6 t^3) * x^3/3! + ...), so by Bala's arguments A248927 is a refinement of A141618 with row sums A000272. The connection to Narayana numbers is reflected in the relation between A248927 and A134264. See A145271 for more relations that g(x,t) and f(x,t) must satisfy. - Tom Copeland, Oct 17 2014
T(n,k) = C(n,k-1) * A028246(n,k) = C(n,k-1) * A019538(n,k)/k = A055302(n+1,n+1-k) / (n+1). - Tom Copeland, Oct 25 2014
E.g.f. is the series reversion of log(1 + x)/(1 + t*x) with respect to x. Cf. A198204. - Peter Bala, Oct 21 2015

Extensions

More terms from Joerg Arndt, Oct 27 2014

A123227 Expansion of e.g.f.: 2*exp(2*x) / (3 - exp(2*x)).

Original entry on oeis.org

1, 3, 12, 66, 480, 4368, 47712, 608016, 8855040, 145083648, 2641216512, 52891055616, 1155444326400, 27344999497728, 696933753434112, 19031293222127616, 554336947975618560, 17155693983744196608, 562168282464340672512, 19444889661250162262016
Offset: 0

Views

Author

Philippe Deléham, Oct 06 2006

Keywords

Crossrefs

Programs

  • Maple
    a := n -> 2^(n+1)*polylog(-n, 1/3):
    seq(round(evalf(a(n),32)), n=0..19); # Peter Luschny, Nov 03 2015
    seq(expand(2^(n+1)*polylog(-n,1/3)), n=0..100); # Robert Israel, Nov 03 2015
  • Mathematica
    CoefficientList[Series[2*Exp[2*x]/(3-Exp[2*x]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Jun 24 2013 *)
    Round@Table[(-1)^(n+1) (LerchPhi[Sqrt[3], -n, 0] + LerchPhi[-Sqrt[3], -n, 0]), {n, 0, 20}] (* Vladimir Reshetnikov, Oct 31 2015 *)
  • PARI
    {a(n)=n!*polcoeff(2*exp(2*x+x*O(x^n))/(3 - exp(2*x+x*O(x^n))), n)} /* Paul D. Hanna */
    
  • PARI
    {a(n)=polcoeff(sum(m=0, n, 3^m*m!*x^m/prod(k=1, m, 1+2*k*x+x*O(x^n))), n)} /* Paul D. Hanna */
    
  • PARI
    {Stirling2(n, k)=if(k<0||k>n, 0, sum(i=0, k, (-1)^i*binomial(k, i)/k!*(k-i)^n))}
    {a(n)=sum(k=0, n, (-2)^(n-k)*3^k*Stirling2(n, k)*k!)} /* Paul D. Hanna */
    
  • PARI
    my(x='x+O('x^20)); Vec(serlaplace(2*exp(2*x)/(3-exp(2*x)))) \\ Joerg Arndt, May 06 2013
  • Sage
    @CachedFunction
    def BB(n, k, x):  # Modified Cardinal B-splines
        if n == 1: return 0 if (x < 0) or (x >= k) else 1
        return x*BB(n-1, k, x) + (n*k-x)*BB(n-1, k, x-k)
    def EulerianPolynomial(n, k, x):
        if n == 0: return 1
        return add(BB(n+1, k, k*m+1)*x^m for m in (0..n))
    def A123227(n) : return 3^n*EulerianPolynomial(n, 1, 1/3)
    [A123227(n) for n in (0..18)]  # Peter Luschny, May 04 2013
    

Formula

a(n) = abs(A009362(n+1)).
a(n-1) = Sum_{k=1..n} 2^(n-k)*A028246(n,k), n>=1.
a(n) = Sum_{k=0..n} 3^k*A123125(n,k).
From Paul D. Hanna, Nov 30 2011: (Start)
a(n) = 3*A122704(n) for n>0.
a(n) = Sum_{k=0..n} (-2)^(n-k) * 3^k * Stirling2(n,k) * k!.
O.g.f.: Sum_{n>=0} 3^n * n!*x^n / Product_{k=0..n} (1+2*k*x).
O.g.f.: 1/(1 - 3*x/(1-x/(1 - 6*x/(1-2*x/(1 - 9*x/(1-3*x/(1 - 12*x/(1-4*x/(1 - 15*x/(1-5*x/(1 - ...))))))))))), a continued fraction.
(End)
a(n) ~ n! * (2/log(3))^(n+1). - Vaclav Kotesovec, Jun 24 2013
a(n) = 2^n*log(3)*Integral_{x = 0..oo} (ceiling(x))^n * 3^(-x) dx. - Peter Bala, Feb 06 2015
a(n) = (-1)^(n+1)*(LerchPhi(sqrt(3), -n, 0) + LerchPhi(-sqrt(3), -n, 0)) = (-1)^(n+1)*(Li_{-n}(sqrt(3)) + Li_{-n}(-sqrt(3))) - 2*0^n, where Li_n(x) is the polylogarithm. - Vladimir Reshetnikov, Oct 31 2015
a(n) = 2^(n+1)*Li_{-n}(1/3). - Peter Luschny, Nov 03 2015
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Jul 05 2020

Extensions

Name changed and a(8) corrected by Paul D. Hanna, Nov 30 2011

A191302 Denominators in triangle that leads to the Bernoulli numbers.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 2, 3, 15, 2, 6, 3, 2, 1, 5, 105, 2, 6, 15, 15, 2, 3, 3, 105, 105, 2, 2, 5, 7, 35, 2, 3, 3, 21, 21, 231, 2, 6, 15, 15, 21, 21, 2, 1, 5, 15, 1, 77, 15015, 2, 6, 3, 35, 15, 33, 1155
Offset: 0

Views

Author

Paul Curtz, May 30 2011

Keywords

Comments

For the definition of the ASPEC array coefficients see the formulas; see also A029635 (Lucas triangle), A097207 and A191662 (k-dimensional square pyramidal numbers).
The antidiagonal row sums of the ASPEC array equal A042950(n) and A098011(n+3).
The coefficients of the T(n,m) array are defined in A190339. We define the coefficients of the SBD array with the aid of the T(n,n+1), see the formulas and the examples.
Multiplication of the coefficients in the rows of the ASPEC array with the coefficients in the columns of the SBD array leads to the coefficients of the BSPEC triangle, see the formulas. The BSPEC triangle can be looked upon as a spectrum for the Bernoulli numbers.
The row sums of the BSPEC triangle give the Bernoulli numbers A164555(n)/A027642(n).
For the numerators of the BSPEC triangle coefficients see A192456.

Examples

			The first few rows of the array ASPEC array:
  2, 1,  1,  1,   1,   1,    1,
  2, 3,  4,  5,   6,   7,    8,
  2, 5,  9, 14,  20,  27,   35,
  2, 7, 16, 30,  50,  77,  112,
  2, 9, 25, 55, 105, 182,  294,
The first few T(n,n+1) = T(n,n)/2 coefficients:
1/2, -1/6, 1/15, -4/105, 4/105, -16/231, 3056/15015, ...
The first few rows of the SBD array:
  1/2,   0,   0,     0
  1/2,   0,   0,     0
  1/2, -1/6,  0,     0
  1/2, -1/6,  0,     0
  1/2, -1/6, 1/15,   0
  1/2, -1/6, 1/15,   0
  1/2, -1/6, 1/15, -4/105
  1/2, -1/6, 1/15, -4/105
The first few rows of the BSPEC triangle:
  B(0) =   1   = 1/1
  B(1) =  1/2  = 1/2
  B(2) =  1/6  = 1/2 - 1/3
  B(3) =   0   = 1/2 - 1/2
  B(4) = -1/30 = 1/2 - 2/3 +  2/15
  B(5) =   0   = 1/2 - 5/6 +  1/3
  B(6) =  1/42 = 1/2 - 1/1 +  3/5  - 8/105
  B(7) =   0   = 1/2 - 7/6 + 14/15 - 4/15
		

Crossrefs

Cf. A028246 (Worpitzky), A085737/A085738 (Conway-Sloane) and A051714/A051715 (Akiyama-Tanigawa) for other triangles that lead to the Bernoulli numbers. - Johannes W. Meijer, Jul 02 2011

Programs

  • Maple
    nmax:=13: mmax:=nmax:
    A164555:=proc(n): if n=1 then 1 else numer(bernoulli(n)) fi: end:
    A027642:=proc(n): if n=1 then 2 else denom(bernoulli(n)) fi: end:
    for m from 0 to 2*mmax do T(0,m):=A164555(m)/A027642(m) od:
    for n from 1 to nmax do for m from 0 to 2*mmax do T(n,m):=T(n-1,m+1)-T(n-1,m) od: od:
    seq(T(n,n+1),n=0..nmax):
    for n from 0 to nmax do ASPEC(n,0):=2: for m from 1 to mmax do ASPEC(n,m):= (2*n+m)*binomial(n+m-1,m-1)/m od: od:
    for n from 0 to nmax do seq(ASPEC(n,m),m=0..mmax) od:
    for n from 0 to nmax do for m from 0 to 2*mmax do SBD(n,m):=0 od: od:
    for m from 0 to mmax do for n from 2*m to nmax do SBD(n,m):= T(m,m+1) od: od:
    for n from 0 to nmax do seq(SBD(n,m), m= 0..mmax/2) od:
    for n from 0 to nmax do BSPEC(n,2) := SBD(n,2)*ASPEC(2,n-4) od:
    for m from 0 to mmax do for n from 0 to nmax do BSPEC(n,m) := SBD(n,m)*ASPEC(m,n-2*m) od: od:
    for n from 0 to nmax do seq(BSPEC(n,m), m=0..mmax/2) od:
    seq(add(BSPEC(n, k), k=0..floor(n/2)) ,n=0..nmax):
    Tx:=0:
    for n from 0 to nmax do for m from 0 to floor(n/2) do a(Tx):= denom(BSPEC(n,m)): Tx:=Tx+1: od: od:
    seq(a(n),n=0..Tx-1); # Johannes W. Meijer, Jul 02 2011
  • Mathematica
    (* a=ASPEC, b=BSPEC *) nmax = 13; a[n_, 0] = 2; a[n_, m_] := (2n+m)*Binomial[n+m-1, m-1]/m; b[n_] := BernoulliB[n]; b[1]=1/2; bb = Table[b[n], {n, 0, nmax}]; diff = Table[ Differences[bb, n], {n, 1, nmax}]; dd = Diagonal[diff]; sbd[n_, m_] := If[n >= 2m, -dd[[m+1]], 0]; b[n_, m_] := sbd[n, m]*a[m, n-2m]; Table[b[n, m], {n, 0, nmax}, {m, 0, Floor[n/2]}] // Flatten // Denominator (* Jean-François Alcover_, Aug 09 2012 *)

Formula

ASPEC(n, 0) = 2 and ASPEC(n, m) = (2*n+m)*binomial(n+m-1, m-1)/m, n >= 0, m >= 1.
ASPEC(n, m) = ASPEC(n-1, m) + ASPEC(n, m-1), n >= 1, m >= 1, with ASPEC(n, 0) = 2, n >= 0, and ASPEC(0,m) = 1, m >= 1.
SBD(n, m) = T(m, m+1), n >= 2*m; see A190339 for the definition of the T(n, m).
BSPEC(n, m) = SBD(n, m)*ASPEC(m, n-2*m)
Sum_{k=0..floor(n/2)} BSPEC(n, k) = A164555(n)/A027642(n).

Extensions

Edited, Maple program and crossrefs added by Johannes W. Meijer, Jul 02 2011
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