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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A036038 Triangle of multinomial coefficients.

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 6, 12, 24, 1, 5, 10, 20, 30, 60, 120, 1, 6, 15, 20, 30, 60, 90, 120, 180, 360, 720, 1, 7, 21, 35, 42, 105, 140, 210, 210, 420, 630, 840, 1260, 2520, 5040, 1, 8, 28, 56, 70, 56, 168, 280, 420, 560, 336, 840, 1120, 1680, 2520, 1680, 3360, 5040, 6720
Offset: 1

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Author

Keywords

Comments

The number of terms in the n-th row is the number of partitions of n, A000041(n). - Amarnath Murthy, Sep 21 2002
For each n, the partitions are ordered according to A-St: first by length and then lexicographically (arranging the parts in nondecreasing order), which is different from the usual practice of ordering all partitions lexicographically. - T. D. Noe, Nov 03 2006
For this ordering of the partitions, for n >= 1, see the remarks and the C. F. Hindenburg link given in A036036. - Wolfdieter Lang, Jun 15 2012
The relation (n+1) * A134264(n+1) = A248120(n+1) / a(n) where the arithmetic is performed for matching partitions in each row n connects the combinatorial interpretations of this array to some topological and algebraic constructs of the two other entries. Also, these seem (cf. MOPS reference, Table 2) to be the coefficients of the Jack polynomial J(x;k,alpha=0). - Tom Copeland, Nov 24 2014
The conjecture on the Jack polynomials of zero order is true as evident from equation a) on p. 80 of the Stanley reference, suggested to me by Steve Kass. The conventions for denoting the more general Jack polynomials J(n,alpha) vary. Using Stanley's convention, these Jack polynomials are the umbral extensions of the multinomial expansion of (s_1*x_1 + s_2*x_2 + ... + s_(n+1)*x_(n+1))^n in which the subscripts of the (s_k)^j in the symmetric monomial expansions are finally ignored and the exponent dropped to give s_j(alpha) = j-th row polynomial of A094638 or |A008276| in ascending powers of alpha. (The MOPS table has some inconsistency between n = 3 and n = 4.) - Tom Copeland, Nov 26 2016

Examples

			1;
1, 2;
1, 3,  6;
1, 4,  6, 12, 24;
1, 5, 10, 20, 30, 60, 120;
1, 6, 15, 20, 30, 60,  90, 120, 180, 360, 720;
		

References

  • Abramowitz and Stegun, Handbook, p. 831, column labeled "M_1".

Crossrefs

Cf. A036036-A036040. Different from A078760. Row sums give A005651.
Cf. A183610 is a table of sums of powers of terms in rows.
Cf. A134264 and A248120.
Cf. A096162 for connections to A130561.

Programs

  • Maple
    nmax:=7: with(combinat): for n from 1 to nmax do P(n):=sort(partition(n)): for r from 1 to numbpart(n) do B(r):=P(n)[r] od: for m from 1 to numbpart(n) do s:=0: j:=0: while sA036038(n, m) := n!/ (mul((t!)^q(t), t=1..n)); od: od: seq(seq(A036038(n, m), m=1..numbpart(n)), n=1..nmax); # Johannes W. Meijer, Jul 14 2016
  • Mathematica
    Flatten[Table[Apply[Multinomial, Reverse[Sort[IntegerPartitions[i],  Length[ #1]>Length[ #2]&]], {1}], {i,9}]] (* T. D. Noe, Nov 03 2006 *)
  • Sage
    def ASPartitions(n, k):
        Q = [p.to_list() for p in Partitions(n, length=k)]
        for q in Q: q.reverse()
        return sorted(Q)
    def A036038_row(n):
        return [multinomial(p) for k in (0..n) for p in ASPartitions(n, k)]
    for n in (1..10): print(A036038_row(n))
    # Peter Luschny, Dec 18 2016, corrected Apr 30 2022

Formula

The n-th row is the expansion of (x_1 + x_2 + ... + x_(n+1))^n in the basis of the monomial symmetric polynomials (m.s.p.). E.g., (x_1 + x_2 + x_3 + x_4)^3 = m[3](x_1,..,x_4) + 3*m[1,2](x_1,..,x_4) + 6*m[1,1,1](x_1,..,x_4) = (Sum_{i=1..4} x_i^3) + 3*(Sum_{i,j=1..4;i != j} x_i^2 x_j) + 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k). The number of indeterminates can be increased indefinitely, extending each m.s.p., yet the expansion coefficients remain the same. In each m.s.p., unique combinations of exponents and subscripts appear only once with a coefficient of unity. Umbral reduction by replacing x_k^j with x_j in the expansions gives the partition polynomials of A248120. - Tom Copeland, Nov 25 2016
From Tom Copeland, Nov 26 2016: (Start)
As an example of the umbral connection to the Jack polynomials: J(3,alpha) = (Sum_{i=1..4} x_i^3)*s_3(alpha) + 3*(Sum_{i,j=1..4;i!=j} x_i^2 x_j)*s_2(alpha)*s_1(alpha)+ 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k)*s_1(alpha)*s_1(alpha)*s_1(alpha) = (Sum_{i=1..4} x_i^3)*(1+alpha)*(1+2*alpha)+ 3*(sum_{i,j=1..4;i!=j} x_i^2 x_j)*(1+alpha) + 6*(Sum_{i,j,k=1..4;i < j < k} x_i x_j x_k).
See the Copeland link for more relations between the multinomial coefficients and the Jack symmetric functions. (End)

Extensions

More terms from David W. Wilson and Wouter Meeussen

A001790 Numerators in expansion of 1/sqrt(1-x).

Original entry on oeis.org

1, 1, 3, 5, 35, 63, 231, 429, 6435, 12155, 46189, 88179, 676039, 1300075, 5014575, 9694845, 300540195, 583401555, 2268783825, 4418157975, 34461632205, 67282234305, 263012370465, 514589420475, 8061900920775, 15801325804719
Offset: 0

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Author

Keywords

Comments

Also numerator of e(n-1,n-1) (see Maple line).
Leading coefficient of normalized Legendre polynomial.
Common denominator of expansions of powers of x in terms of Legendre polynomials P_n(x).
Also the numerator of binomial(2*n,n)/2^n. - T. D. Noe, Nov 29 2005
This sequence gives the numerators of the Maclaurin series of the Lorentz factor (see Wikipedia link) of 1/sqrt(1-b^2) = dt/dtau where b=u/c is the velocity in terms of the speed of light c, u is the velocity as observed in the reference frame where time t is measured and tau is the proper time. - Stephen Crowley, Apr 03 2007
Truncations of rational expressions like those given by the numerator operator are artifacts in integer formulas and have many disadvantages. A pure integer formula follows. Let n$ denote the swinging factorial and sigma(n) = number of '1's in the base-2 representation of floor(n/2). Then a(n) = (2*n)$ / sigma(2*n) = A056040(2*n) / A060632(2*n+1). Simply said: this sequence is the odd part of the swinging factorial at even indices. - Peter Luschny, Aug 01 2009
It appears that a(n) = A060818(n)*A001147(n)/A000142(n). - James R. Buddenhagen, Jan 20 2010
The convolution of sequence binomial(2*n,n)/4^n with itself is the constant sequence with all terms = 1.
a(n) equals the denominator of Hypergeometric2F1[1/2, n, 1 + n, -1] (see Mathematica code below). - John M. Campbell, Jul 04 2011
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2-2*x+2)^n dx. - Leonid Bedratyuk, Nov 17 2012
a(n) = numerator of the mean value of cos(x)^(2*n) from x = 0 to 2*Pi. - Jean-François Alcover, Mar 21 2013
Constant terms for normalized Legendre polynomials. - Tom Copeland, Feb 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums:
a(n)/A060818(n) = (-2)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A060818(k) = -i.
Sum_{k>=0} (-1)^k*a(k)/A060818(k) = 1/sqrt(3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A060818(k) = -1/sqrt(3).
a(n)/A046161(n) = (-1)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} (-1)^k*a(k)/A046161(k) = 1/sqrt(2).
Sum_{k>=0} (-1)^(k+1)*a(k)/A046161(k) = -1/sqrt(2). (End)
a(n) = numerator of (1/Pi)*Integral_{x=-oo..+oo} 1/(x^2+1)^n dx. (n=1 is the Cauchy distribution.) - Harry Garst, May 26 2017
Let R(n, d) = (Product_{j prime to d} Pochhammer(j / d, n)) / n!. Then the numerators of R(n, 2) give this sequence and the denominators are A046161. For d = 3 see A273194/A344402. - Peter Luschny, May 20 2021
Using WolframAlpha, it appears a(n) gives the numerator in the residues of f(z) = 2z choose z at odd negative half integers. E.g., the residues of f(z) at z = -1/2, -3/2, -5/2 are 1/(2*Pi), 1/(16*Pi), and 3/(256*Pi) respectively. - Nicholas Juricic, Mar 31 2022
a(n) is the numerator of (1/Pi) * Integral_{x=-oo..+oo} sech(x)^(2*n+1) dx. The corresponding denominator is A046161. - Mohammed Yaseen, Jul 29 2023
a(n) is the numerator of (1/Pi) * Integral_{x=0..Pi/2} sin(x)^(2*n) dx. The corresponding denominator is A101926(n). - Mohammed Yaseen, Sep 19 2023

Examples

			1, 1, 3/2, 5/2, 35/8, 63/8, 231/16, 429/16, 6435/128, 12155/128, 46189/256, ...
binomial(2*n,n)/4^n => 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, 6435/32768, ...
		

References

  • P. J. Davis, Interpolation and Approximation, Dover Publications, 1975, p. 372.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 2nd ed. New York: Wiley, 1968; Chap. III, Eq. 4.1.
  • 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).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:6 at page 51.
  • J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 102.

Crossrefs

Cf. A060818 (denominator of binomial(2*n,n)/2^n), A061549 (denominators).
Cf. A123854 (denominators).
Cf. A161198 (triangle of coefficients for (1-x)^((-1-2*n)/2)).
Cf. A163590 (odd part of the swinging factorial).
Cf. A001405.
First column and diagonal 1 of triangle A100258.
Bisection of A036069.
Bisections give A061548 and A063079.
Inverse Moebius transform of A180403/A046161.
Numerators of [x^n]( (1-x)^(p/2) ): A161202 (p=5), A161200 (p=3), A002596 (p=1), this sequence (p=-1), A001803 (p=-3), A161199 (p=-5), A161201 (p=-7).

Programs

  • Magma
    A001790:= func< n | Numerator((n+1)*Catalan(n)/4^n) >;
    [A001790(n): n in [0..40]]; // G. C. Greubel, Sep 23 2024
  • Maple
    e := proc(l,m) local k; add(2^(k-2*m)*binomial(2*m-2*k,m-k)*binomial(m+k,m)*binomial(k,l),k=l..m); end;
    # From Peter Luschny, Aug 01 2009: (Start)
    swing := proc(n) option remember; if n = 0 then 1 elif irem(n, 2) = 1 then swing(n-1)*n else 4*swing(n-1)/n fi end:
    sigma := n -> 2^(add(i,i=convert(iquo(n,2),base,2))):
    a := n -> swing(2*n)/sigma(2*n); # (End)
    A001790 := proc(n) binomial(2*n, n)/4^n ; numer(%) ; end proc : # R. J. Mathar, Jan 18 2013
  • Mathematica
    Numerator[ CoefficientList[ Series[1/Sqrt[(1 - x)], {x, 0, 25}], x]]
    Table[Denominator[Hypergeometric2F1[1/2, n, 1 + n, -1]], {n, 0, 34}]   (* John M. Campbell, Jul 04 2011 *)
    Numerator[Table[(-2)^n*Sqrt[Pi]/(Gamma[1/2 - n]*Gamma[1 + n]),{n,0,20}]] (* Ralf Steiner, Apr 07 2017 *)
    Numerator[Table[Binomial[2n,n]/2^n, {n, 0, 25}]] (* Vaclav Kotesovec, Apr 07 2017 *)
    Table[Numerator@LegendreP[2 n, 0]*(-1)^n, {n, 0, 25}] (* Andres Cicuttin, Jan 22 2018 *)
    A = {1}; Do[A = Append[A, 2^IntegerExponent[n, 2]*(2*n - 1)*A[[n]]/n], {n, 1, 25}]; Print[A] (* John Lawrence, Jul 17 2020 *)
  • PARI
    {a(n) = if( n<0, 0, polcoeff( pollegendre(n), n) * 2^valuation((n\2*2)!, 2))};
    
  • PARI
    a(n)=binomial(2*n,n)>>hammingweight(n); \\ Gleb Koloskov, Sep 26 2021
    
  • Sage
    # uses[A000120]
    @CachedFunction
    def swing(n):
        if n == 0: return 1
        return swing(n-1)*n if is_odd(n) else 4*swing(n-1)/n
    A001790 = lambda n: swing(2*n)/2^A000120(2*n)
    [A001790(n) for n in (0..25)]  # Peter Luschny, Nov 19 2012
    

Formula

a(n) = numerator( binomial(2*n,n)/4^n ) (cf. A046161).
a(n) = A000984(n)/A001316(n) where A001316(n) is the highest power of 2 dividing C(2*n, n) = A000984(n). - Benoit Cloitre, Jan 27 2002
a(n) = denominator of (2^n/binomial(2*n,n)). - Artur Jasinski, Nov 26 2011
a(n) = numerator(L(n)), with rational L(n):=binomial(2*n,n)/2^n. L(n) is the leading coefficient of the Legendre polynomial P_n(x).
L(n) = (2*n-1)!!/n! with the double factorials (2*n-1)!! = A001147(n), n >= 0.
Numerator in (1-2t)^(-1/2) = 1 + t + (3/2)t^2 + (5/2)t^3 + (35/8)t^4 + (63/8)t^5 + (231/16)t^6 + (429/16)t^7 + ... = 1 + t + 3*t^2/2! + 15*t^3/3! + 105*t^4/4! + 945*t^5/5! + ... = e.g.f. for double factorials A001147 (cf. A094638). - Tom Copeland, Dec 04 2013
From Ralf Steiner, Apr 08 2017: (Start)
a(n)/A061549(n) = (-1/4)^n*sqrt(Pi)/(Gamma(1/2 - n)*Gamma(1 + n)).
Sum_{k>=0} a(k)/A061549(k) = 2/sqrt(3).
Sum_{k>=0} (-1)^k*a(k)/A061549(k) = 2/sqrt(5).
Sum_{k>=0} (-1)^(k+1)*a(k)/A061549(k) = -2/sqrt(5).
a(n)/A123854(n) = (-1/2)^n*sqrt(Pi)/(gamma(1/2 - n)*gamma(1 + n)).
Sum_{k>=0} a(k)/A123854(k) = sqrt(2).
Sum_{k>=0} (-1)^k*a(k)/A123854(k) = sqrt(2/3).
Sum_{k>=0} (-1)^(k+1)*a(k)/A123854(k) = -sqrt(2/3). (End)
a(n) = 2^A007814(n)*(2*n-1)*a(n-1)/n. - John Lawrence, Jul 17 2020
Sum_{k>=0} A086117(k+3)/a(k+2) = Pi. - Antonio Graciá Llorente, Aug 31 2024
a(n) = A001803(n)/(2*n+1). - G. C. Greubel, Sep 23 2024

A008544 Triple factorial numbers: Product_{k=0..n-1} (3*k+2).

Original entry on oeis.org

1, 2, 10, 80, 880, 12320, 209440, 4188800, 96342400, 2504902400, 72642169600, 2324549427200, 81359229952000, 3091650738176000, 126757680265216000, 5577337931669504000, 262134882788466688000
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

a(n-1), n >= 1, enumerates increasing plane (aka ordered) trees with n vertices (one of them a root labeled 1) where each vertex with outdegree r >= 0 comes in r+1 types (like an (r+1)-ary vertex). See the increasing tree comments under A004747. - Wolfdieter Lang, Oct 12 2007
An example for the case of 3 vertices is shown below. For the enumeration of non-plane trees of this type see A029768. - Peter Bala, Aug 30 2011
a(n) is the product of the positive integers k <= 3*n that have k modulo 3 = 2. - Peter Luschny, Jun 23 2011
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
Partial products of A016789. - Reinhard Zumkeller, Sep 20 2013
The Mathar conjecture is true. Generally from the factorial form, the last term is the "extra" product beyond the prior term, from k=n-1 and 3k+2 evaluates to 3*(n-1)+2 = 3n-1, yielding a(n) = a(n-1)*(3n-1) (eqn1). Similarly, a(n) = a(n-2)*(3n-1)*(3(n-2)+2) = a(n-2)*(3n-1)*(3n-4) (eqn2) and a(n) = a(n-3)*(3n-1)*(3n-4)*(3*(n-2)+2) = a(n-3)*(3n-1)*(3n-4)*(3n-7) (eqn3). We equate (eqn2) and (eqn3) to get a(n-2)*(3n-1)*(3n-4) = a(n-3)*(3n-1)*(3n-4)*(3n-7) or a(n-2)+(7-3n)*a(n-3) = 0 (eqn4). From (eqn1) we have a(n)+(1-3n)*a(n-1) = 0 (eqn5). Combining (eqn4) and (eqn5) yields a(n)+(1-3n)*a(n-1)+a(n-2)+(7-3n)*a(n-3) = 0. - Bill McEachen, Jan 01 2016
a(n-1), n>=1, is the dimension of the n-th component of the operad encoding the multilinearization of the following identity in nonassociative algebras: s*(a,a,b)-(s+t)*(a,b,a)+t*(b,a,a)=0, for any given pair of scalars (s,t). Here (a,b,c) is the associator (ab)c-a(bc). This is proved in the referenced article on associator dependent algebras by Bremner and me. - Vladimir Dotsenko, Mar 22 2022

Examples

			a(2) = 10 from the described trees with 3 vertices: there are three trees with a root vertex (label 1) with outdegree r=2 (like the three 3-stars each with one different ray missing) and the four trees with a root (r=1 and label 1) a vertex with (r=1) and a leaf (r=0). Assigning labels 2 and 3 yields 2*3+4=10 such trees.
a(2) = 10. The 10 possible plane increasing trees on 3 vertices, where vertices of outdegree 1 come in 2 colors (denoted a or b) and vertices of outdegree 2 come in 3 colors (a, b or c), are:
.
   1a    1b    1a    1b        1a       1b       1c
   |     |     |     |        / \      / \      / \
   2a    2b    2b    2a      2   3    2   3    2   3
   |     |     |     |
   3     3     3     3         1a       1b       1c
                              / \      / \      / \
                             3   2    3   2    3   2
		

Crossrefs

a(n) = A004747(n+1, 1) (first column of triangle). Cf. A051141.
Cf. A225470, A290596 (first columns).
Subsequence of A007661.

Programs

  • Haskell
    a008544 n = a008544_list !! n
    a008544_list = scanl (*) 1 a016789_list
    -- Reinhard Zumkeller, Sep 20 2013
    
  • Magma
    [Round((Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6) )/ Sqrt(3)*3^n/4^(n-1)): n in [1..20]]; // Vincenzo Librandi, Feb 21 2015
    
  • Magma
    [Round(3^n*Gamma(n+2/3)/Gamma(2/3)): n in [0..20]]; // G. C. Greubel, Mar 31 2019
  • Maple
    a := n -> mul(3*k-1, k = 1..n);
    A008544 := n -> mul(k, k = select(k-> k mod 3 = 2, [$1 .. 3*n])): seq(A008544(n), n = 0 .. 16); # Peter Luschny, Jun 23 2011
  • Mathematica
    k = 3; b[1]=2; b[n_]:= b[n] = b[n-1]+k; a[0]=1; a[1]=2; a[n_]:= a[n] = a[n-1]*b[n]; Table[a[n], {n,0,20}] (* Roger L. Bagula, Sep 17 2008 *)
    Product[3 k + 2, {k, 0, # - 1}] & /@ Range[0, 16] (* Michael De Vlieger, Jan 02 2016 *)
    Table[3^n*Pochhammer[2/3, n], {n,0,20}] (* G. C. Greubel, Mar 31 2019 *)
  • Maxima
    a(n):=((n)!*sum(binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k),k,floor(n/2),n)); /* Vladimir Kruchinin, Sep 28 2013 */
    
  • PARI
    a(n) = prod(k=0,n-1, 3*k+2 );
    
  • PARI
    vector(20, n, n--; round(3^n*gamma(n+2/3)/gamma(2/3))) \\ G. C. Greubel, Mar 31 2019
    
  • Sage
    @CachedFunction
    def A008544(n): return 1 if n == 0 else (3*n-1)*A008544(n-1)
    [A008544(n) for n in (0..16)]  # Peter Luschny, May 20 2013
    
  • Sage
    [3^n*rising_factorial(2/3, n) for n in (0..20)] # G. C. Greubel, Mar 31 2019
    

Formula

a(n) = Product_{k=0..n-1} (3*k+2) = A007661(3*n-1) (with A007661(-1) = 1).
E.g.f.: (1-3*x)^(-2/3).
a(n) = 2*A034000(n), n >= 1, a(0) = 1.
a(n) ~ 2^(1/2)*Pi^(1/2)*Gamma(2/3)^-1*n^(1/6)*3^n*e^-n*n^n*{1 - 1/36*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 22 2001
a(n) = (Gamma(2*n-5/3)/Gamma(n-5/6)*Gamma(2/3)/Gamma(5/6))/sqrt(3)*3^n/4^(n-1). - Jeremy L. Martin, Mar 31 2002 (typo fixed by Vincenzo Librandi, Feb 21 2015)
From Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003: (Start)
a(n) = A084939(n)/A000142(n)*A000079(n).
a(n) = 3^n*Pochhammer(2/3, n) = 3^n*Gamma(n+2/3)/Gamma(2/3). (End)
Let T = A094638 and c(t) = column vector(1, t, t^2, t^3, t^4, t^5,...), then A008544 = unsigned [ T * c(-3) ] and the list partition transform A133314 of [1,T * c(-3)] gives [1,T * c(3)] with all odd terms negated, which equals a signed version of A007559; i.e., LPT[(1,signed A008544)] = signed A007559. Also LPT[A007559] = (1,-A008544) and e.g.f. [1,T * c(t)] = (1-x*t)^(-1/t) for t = 3 or -3. Analogous results hold for the double factorial, quadruple factorial and so on. - Tom Copeland, Dec 22 2007
G.f.: 1/(1-2x/(1-3x/(1-5x/(1-6x/(1-8x/(1-9x/(1-11x/(1-12x/(1-...))))))))) (continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-1)^n*Sum_{k=0..n} 3^k*s(n+1,n+1-k), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
G.f.: 1/Q(0) where Q(k) = 1 - x*(3*k+2)/(1 - x*(3*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 20 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k+2)/(x*(3*k+2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 25 2013
D-finite with recurrence: a(n) = (9*(n-2)*(n-1)+2)*a(n-2) + 4*a(n-1), n>=2. - Ivan N. Ianakiev, Aug 09 2013
a(n) = n!*Sum_{k=floor(n/2)..n} binomial(k,n-k)*binomial(n+k,k)*3^(-n+k)*(-1)^(n-k). - Vladimir Kruchinin, Sep 28 2013
Recurrence equation: a(n) = 3*a(n-1) + (3*n - 4)^2*a(n-2) with a(0) = 1 and a(1) = 2. A024396 satisfies the same recurrence (but with different initial conditions). This observation leads to a continued fraction expansion for the constant A193534 due to Euler. - Peter Bala, Feb 20 2015
a(n) = A225470(n, 0), n >= 0. - Wolfdieter Lang, May 29 2017
G.f.: Hypergeometric2F0(1, 2/3; -; 3*x). - G. C. Greubel, Mar 31 2019
D-finite with recurrence: a(n) + (-3*n+1)*a(n-1)=0. - R. J. Mathar, Jan 17 2020
G.f.: 1/(1-2*x-6*x^2/(1-8*x-30*x^2/(1-14*x-72*x^2/(1-20*x-132*x^2/(1-...))))) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 28 2020
G.f.: 1/G(0), where G(k) = 1 - (6*k+2)*x - 3*(k+1)*(3*k+2)*x^2/G(k+1). - Nikolaos Pantelidis, Feb 28 2020
Sum_{n>=0} 1/a(n) = 1 + (e/3)^(1/3) * (Gamma(2/3) - Gamma(2/3, 1/3)). - Amiram Eldar, Mar 01 2022

A130534 Triangle T(n,k), 0 <= k <= n, read by rows, giving coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in increasing powers of x. T(n,k) is also the unsigned Stirling number |s(n+1, k+1)|, denoting the number of permutations on n+1 elements that contain exactly k+1 cycles.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 6, 1, 24, 50, 35, 10, 1, 120, 274, 225, 85, 15, 1, 720, 1764, 1624, 735, 175, 21, 1, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

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Author

Philippe Deléham, Aug 09 2007

Keywords

Comments

This triangle is an unsigned version of the triangle of Stirling numbers of the first kind, A008275, which is the main entry for these numbers. - N. J. A. Sloane, Jan 25 2011
Or, triangle T(n,k), 0 <= k <= n, read by rows given by [1,1,2,2,3,3,4,4,5,5,6,6,...] DELTA [1,0,1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938.
Reversal of A094638.
Equals A132393*A007318, as infinite lower triangular matrices. - Philippe Deléham, Nov 13 2007
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the exponential integrals E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + n*(n+1)/x^2 - n*(n+1)*(n+2)/x^3 + ...), see Abramowitz and Stegun. This formula follows from the general formula for the asymptotic expansion, see A163932. We rewrite E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) and observe that the T(n,m) are the polynomials coefficients in the denominators. Looking at the a(n,m) formula of A028421, A163932 and A163934, and shifting the offset given above to 1, we can write T(n-1,m-1) = a(n,m) = (-1)^(n+m)*Stirling1(n,m), see the Maple program.
The asymptotic expansion leads for values of n from one to eleven to known sequences, see the cross-references. With these sequences one can form the triangles A008279 (right-hand columns) and A094587 (left-hand columns).
See A163936 for information about the o.g.f.s. of the right-hand columns of this triangle.
(End)
The number of elements greater than i to the left of i in a permutation gives the i-th element of the inversion vector. (Skiena-Pemmaraju 2003, p. 69.) T(n,k) is the number of n-permutations that have exactly k 0's in their inversion vector. See evidence in Mathematica code below. - Geoffrey Critzer, May 07 2010
T(n,k) counts the rooted trees with k+1 trunks in forests of "naturally grown" rooted trees with n+2 nodes. This corresponds to sums of coefficients of iterated derivatives representing vectors, Lie derivatives, or infinitesimal generators for flow fields and formal group laws. Cf. links in A139605. - Tom Copeland, Mar 23 2014
A refinement is A036039. - Tom Copeland, Mar 30 2014
From Tom Copeland, Apr 05 2014: (Start)
With initial n=1 and row polynomials of T as p(n,x)=x(x+1)...(x+n-1), the powers of x correspond to the number of trunks of the rooted trees of the "naturally-grown" forest referred to above. With each trunk allowed m colors, p(n,m) gives the number of such non-plane colored trees for the forest with each tree having n+1 vertices.
p(2,m) = m + m^2 = A002378(m) = 2*A000217(m) = 2*(first subdiag of |A238363|).
p(3,m) = 2m + 3m^2 + m^3 = A007531(m+2) = 3*A007290(m+2) = 3*(second subdiag A238363).
p(4,m) = 6m + 11m^2 + 6m^3 + m^4 = A052762(m+3) = 4*A033487(m) = 4*(third subdiag).
From the Joni et al. link, p(n,m) also represents the disposition of n distinguishable flags on m distinguishable flagpoles.
The chromatic polynomial for the complete graph K_n is the falling factorial, which encodes the colorings of the n vertices of K_n and gives a shifted version of p(n,m).
E.g.f. for the row polynomials: (1-y)^(-x).
(End)
A relation to derivatives of the determinant |V(n)| of the n X n Vandermonde matrix V(n) in the indeterminates c(1) thru c(n):
|V(n)| = Product_{1<=jTom Copeland, Apr 10 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A094587 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well defined). See the Example section. (End)
For the relation of this rising factorial to the moments of Viennot's Laguerre stories, see the Hetyei link, p. 4. - Tom Copeland, Oct 01 2015
Can also be seen as the Bell transform of n! without column 0 (and shifted enumeration). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle  T(n,k) begins:
n\k         0        1        2       3       4      5      6     7    8  9 10
n=0:        1
n=1:        1        1
n=2:        2        3        1
n=3:        6       11        6       1
n=4:       24       50       35      10       1
n=5:      120      274      225      85      15      1
n=6:      720     1764     1624     735     175     21      1
n=7:     5040    13068    13132    6769    1960    322     28     1
n=8:    40320   109584   118124   67284   22449   4536    546    36    1
n=9:   362880  1026576  1172700  723680  269325  63273   9450   870   45  1
n=10: 3628800 10628640 12753576 8409500 3416930 902055 157773 18150 1320 55  1
[Reformatted and extended by _Wolfdieter Lang_, Feb 05 2013]
T(3,2) = 6 because there are 6 permutations of {1,2,3,4} that have exactly 2 0's in their inversion vector: {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {2, 1, 3, 4},{2, 3, 1, 4}, {2, 3, 4, 1}. The respective inversion vectors are {0, 0, 1}, {0, 1, 0}, {0, 2, 0}, {1, 0, 0}, {2, 0, 0}, {3, 0, 0}. - _Geoffrey Critzer_, May 07 2010
T(3,1)=11 since there are exactly 11 permutations of {1,2,3,4} with exactly 2 cycles, namely, (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(143), (12)(34), (13)(24), and (14)(23). - _Dennis P. Walsh_, Jan 25 2011
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
  / 1          \/1        \/1        \      / 1           \
  | 1  1       ||0 1      ||0 1      |      | 1  1        |
  | 2  2  1    ||0 1 1    ||0 0 1    |... = | 2  3  1     |
  | 6  6  3 1  ||0 2 2 1  ||0 0 1 1  |      | 6 11  6  1  |
  |24 24 12 4 1||0 6 6 3 1||0 0 2 2 1|      |24 50 35 10 1|
  |...         ||...      ||...      |      |...          |
(End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 93-94.
  • Sriram Pemmaraju and Steven Skiena, Computational Discrete Mathematics, Cambridge University Press, 2003, pp. 69-71. [Geoffrey Critzer, May 07 2010]

Crossrefs

See A008275, which is the main entry for these numbers; A094638 (reversed rows).
From Johannes W. Meijer, Oct 07 2009: (Start)
Row sums equal A000142.
The asymptotic expansions lead to A000142 (n=1), A000142(n=2; minus a(0)), A001710 (n=3), A001715 (n=4), A001720 (n=5), A001725 (n=6), A001730 (n=7), A049388 (n=8), A049389 (n=9), A049398 (n=10), A051431 (n=11), A008279 and A094587.
Cf. A163931 (E(x,m,n)), A028421 (m=2), A163932 (m=3), A163934 (m=4), A163936.
(End)
Cf. A136662.

Programs

  • Haskell
    a130534 n k = a130534_tabl !! n !! k
    a130534_row n = a130534_tabl !! n
    a130534_tabl = map (map abs) a008275_tabl
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    with(combinat): A130534 := proc(n,m): (-1)^(n+m)*stirling1(n+1,m+1) end proc: seq(seq(A130534(n,m), m=0..n), n=0..10); # Johannes W. Meijer, Oct 07 2009, revised Sep 11 2012
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0, ..) as column 0 (and shifts the enumeration).
    BellMatrix(n -> n!, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    Table[Table[ Length[Select[Map[ToInversionVector, Permutations[m]], Count[ #, 0] == n &]], {n, 0, m - 1}], {m, 0, 8}] // Grid (* Geoffrey Critzer, May 07 2010 *)
    rows = 10;
    t = Range[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 *)

Formula

T(0,0) = 1, T(n,k) = 0 if k > n or if n < 0, T(n,k) = T(n-1,k-1) + n*T(n-1,k). T(n,0) = n! = A000142(n). T(2*n,n) = A129505(n+1). Sum_{k=0..n} T(n,k) = (n+1)! = A000142(n+1). Sum_{k=0..n} T(n,k)^2 = A047796(n+1). T(n,k) = |Stirling1(n+1,k+1)|, see A008275. (x+1)(x+2)...(x+n) = Sum_{k=0..n} T(n,k)*x^k. [Corrected by Arie Bos, Jul 11 2008]
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. - Philippe Deléham, Nov 13 2007
For k=1..n, let A={a_1,a_2,...,a_k} denote a size-k subset of {1,2,...,n}. Then T(n,n-k) = Sum(Product_{i=1..k} a_i) where the sum is over all subsets A. For example, T(4,1)=50 since 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50. - Dennis P. Walsh, Jan 25 2011
The preceding formula means T(n,k) = sigma_{n-k}(1,2,3,..,n) with the (n-k)-th elementary symmetric function sigma with the indeterminates chosen as 1,2,...,n. See the Oct 24 2011 comment in A094638 with sigma called there a. - Wolfdieter Lang, Feb 06 2013
From Gary W. Adamson, Jul 08 2011: (Start)
n-th row of the triangle = top row of M^n, where M is the production matrix:
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 4, 6, 4, 1;
... (End)
Exponential Riordan array [1/(1 - x), log(1/(1 - x))]. Recurrence: T(n+1,k+1) = Sum_{i=0..n-k} (n + 1)!/(n + 1 - i)!*T(n-i,k). - Peter Bala, Jul 21 2014

A035342 The convolution matrix of the double factorial of odd numbers (A001147).

Original entry on oeis.org

1, 3, 1, 15, 9, 1, 105, 87, 18, 1, 945, 975, 285, 30, 1, 10395, 12645, 4680, 705, 45, 1, 135135, 187425, 82845, 15960, 1470, 63, 1, 2027025, 3133935, 1595790, 370125, 43890, 2730, 84, 1, 34459425, 58437855, 33453945, 8998290
Offset: 1

Views

Author

Keywords

Comments

Previous name was: A triangle of numbers related to the triangle A035324; generalization of Stirling numbers of second kind A008277 and Lah numbers A008297.
If one replaces in the recurrence the '2' by '0', resp. '1', one obtains the Lah-number, resp. Stirling-number of 2nd kind, triangle A008297, resp. A008277.
The product of two lower triangular Jabotinsky matrices (see A039692 for the Knuth 1992 reference) is again such a Jabotinsky matrix: J(n,m) = Sum_{j=m..n} J1(n,j)*J2(j,m). The e.g.f.s of the first columns of these triangular matrices are composed in the reversed order: f(x)=f2(f1(x)). With f1(x)=-(log(1-2*x))/2 for J1(n,m)=|A039683(n,m)| and f2(x)=exp(x)-1 for J2(n,m)=A008277(n,m) one has therefore f2(f1(x))=1/sqrt(1-2*x) - 1 = f(x) for J(n,m)=a(n,m). This proves the matrix product given below. The m-th column of a Jabotinsky matrix J(n,m) has e.g.f. (f(x)^m)/m!, m>=1.
a(n,m) gives the number of forests with m rooted ordered trees with n non-root vertices labeled in an organic way. Organic labeling means that the vertex labels along the (unique) path from the root with label 0 to any leaf (non-root vertex of degree 1) is increasing. Proof: first for m=1 then for m>=2 using the recurrence relation for a(n,m) given below. - Wolfdieter Lang, Aug 07 2007
Also the Bell transform of A001147(n+1) (adding 1,0,0,... as column 0). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 19 2016

Examples

			Matrix begins:
    1;
    3,   1;
   15,   9,   1;
  105,  87,  18,   1;
  945, 975, 285,  30,   1;
  ...
Combinatoric meaning of a(3,2)=9: The nine increasing path sequences for the three rooted ordered trees with leaves labeled with 1,2,3 and the root labels 0 are: {(0,3),[(0,1),(0,2)]}; {(0,3),[(0,2),(0,1)]}; {(0,3),(0,1,2)}; {(0,1),[(0,3),(0,2)]}; [(0,1),[(0,2),(0,3)]]; [(0,2),[(0,1),(0,3)]]; {(0,2),[(0,3),(0,1)]}; {(0,1),(0,2,3)}; {(0,2),(0,1,3)}.
		

Crossrefs

The column sequences are A001147, A035101, A035119, ...
Row sums: A049118(n), n >= 1.

Programs

  • Haskell
    a035342 n k = a035342_tabl !! (n-1) !! (k-1)
    a035342_row n = a035342_tabl !! (n-1)
    a035342_tabl = map fst $ iterate (\(xs, i) -> (zipWith (+)
       ([0] ++ xs) $ zipWith (*) [i..] (xs ++ [0]), i + 2)) ([1], 3)
    -- Reinhard Zumkeller, Mar 12 2014
    
  • Maple
    T := (n,k) -> 2^(k-n)*hypergeom([k-n,k+1],[k-2*n+1],2)*GAMMA(2*n-k)/
    (GAMMA(k)*GAMMA(n-k+1)); for n from 1 to 9 do seq(simplify(T(n,k)),k=1..n) od; # Peter Luschny, Mar 31 2015
    T := (n, k) -> local j; 2^n*add((-1)^(k-j)*binomial(k, j)*pochhammer(j/2, n), j = 1..k)/k!: for n from 1 to 6 do seq(T(n, k), k=1..n) od;  # Peter Luschny, Mar 04 2024
  • Mathematica
    a[n_, k_] := 2^(n+k)*n!/(4^n*n*k!)*Sum[(j+k)*2^(j)*Binomial[j + k - 1, k-1]*Binomial[2*n - j - k - 1, n-1], {j, 0, n-k}]; Flatten[Table[a[n,k], {n, 1, 9}, {k, 1, n}] ] [[1 ;; 40]] (* Jean-François Alcover, Jun 01 2011, after Vladimir Kruchinin *)
  • Maxima
    a(n,k):=2^(n+k)*n!/(4^n*n*k!)*sum((j+k)*2^(j)*binomial(j+k-1,k-1)*binomial(2*n-j-k-1,n-1),j,0,n-k); /* Vladimir Kruchinin, Mar 30 2011 */
    
  • Sage
    # uses[bell_matrix from A264428]
    # Adds a column 1,0,0,0, ... at the left side of the triangle.
    print(bell_matrix(lambda n: A001147(n+1), 9)) # Peter Luschny, Jan 19 2016

Formula

a(n, m) = Sum_{j=m..n} |A039683(n, j)|*S2(j, m) (matrix product), with S2(j, m) := A008277(j, m) (Stirling2 triangle). Priv. comm. to Wolfdieter Lang by E. Neuwirth, Feb 15 2001; see also the 2001 Neuwirth reference. See the comment on products of Jabotinsky matrices.
a(n, m) = n!*A035324(n, m)/(m!*2^(n-m)), n >= m >= 1; a(n+1, m)= (2*n+m)*a(n, m)+a(n, m-1); a(n, m) := 0, n
E.g.f. of m-th column: ((x*c(x/2)/sqrt(1-2*x))^m)/m!, where c(x) = g.f. for Catalan numbers A000108.
From Vladimir Kruchinin, Mar 30 2011: (Start)
G.f. (1/sqrt(1-2*x) - 1)^k = Sum_{n>=k} (k!/n!)*a(n,k)*x^n.
a(n,k) = 2^(n+k) * n! / (4^n*n*k!) * Sum_{j=0..n-k} (j+k) * 2^(j) * binomial(j+k-1,k-1) * binomial(2*n-j-k-1,n-1). (End)
From Peter Bala, Nov 25 2011: (Start)
E.g.f.: G(x,t) = exp(t*A(x)) = 1 + t*x + (3*t + t^2)*x^2/2! + (15*t + 9*t^2 + t^3)*x^3/3! + ..., where A(x) = -1 + 1/sqrt(1-2*x) satisfies the autonomous differential equation A'(x) = (1+A(x))^3.
The generating function G(x,t) satisfies the partial differential equation t*(dG/dt+G) = (1-2*x)*dG/dx, from which follows the recurrence given above.
The row polynomials are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+x)^3*d/dx. Cf. A008277 (D = (1+x)*d/dx), A105278 (D = (1+x)^2*d/dx), A035469 (D = (1+x)^4*d/dx) and A049029 (D = (1+x)^5*d/dx). (End)
The n-th row polynomial R(n,x) is given by the Dobinski-type formula R(n,x) = exp(-x)*Sum_{k>=1} k*(k+2)*...*(k+2*n-2)*x^k/k!. - Peter Bala, Jun 22 2014
T(n,k) = 2^(k-n)*hypergeom([k-n,k+1],[k-2*n+1],2)*Gamma(2*n-k)/(Gamma(k)*Gamma(n-k+1)). - Peter Luschny, Mar 31 2015
T(n,k) = 2^n*Sum_{j=1..k} ((-1)^(k-j)*binomial(k, j)*Pochhammer(j/2, n)) / k!. - Peter Luschny, Mar 04 2024

Extensions

Simpler name from Peter Luschny, Mar 31 2015

A374848 Obverse convolution A000045**A000045; see Comments.

Original entry on oeis.org

0, 1, 2, 16, 162, 3600, 147456, 12320100, 2058386904, 701841817600, 488286500625000, 696425232679321600, 2038348954317776486400, 12259459134020160144810000, 151596002479762016373851690400, 3855806813438155578522841251840000
Offset: 0

Author

Clark Kimberling, Jul 31 2024

Keywords

Comments

The obverse convolution of sequences
s = (s(0), s(1), ...) and t = (t(0), t(1), ...)
is introduced here as the sequence s**t given by
s**t(n) = (s(0)+t(n)) * (s(1)+t(n-1)) * ... * (s(n)+t(0)).
Swapping * and + in the representation s(0)*t(n) + s(1)*t(n-1) + ... + s(n)*t(0)
of ordinary convolution yields s**t.
If x is an indeterminate or real (or complex) variable, then for every sequence t of real (or complex) numbers, s**t is a sequence of polynomials p(n) in x, and the zeros of p(n) are the numbers -t(0), -t(1), ..., -t(n).
Following are abbreviations in the guide below for triples (s, t, s**t):
F = (0,1,1,2,3,5,...) = A000045, Fibonacci numbers
L = (2,1,3,4,7,11,...) = A000032, Lucas numbers
P = (2,3,5,7,11,...) = A000040, primes
T = (1,3,6,10,15,...) = A000217, triangular numbers
C = (1,2,6,20,70, ...) = A000984, central binomial coefficients
LW = (1,3,4,6,8,9,...) = A000201, lower Wythoff sequence
UW = (2,5,7,10,13,...) = A001950, upper Wythoff sequence
[ ] = floor
In the guide below, sequences s**t are identified with index numbers Axxxxxx; in some cases, s**t and Axxxxxx differ in one or two initial terms.
Table 1. s = A000012 = (1,1,1,1...) = (1);
t = A000012; 1 s**t = A000079; 2^(n+1)
t = A000027; n s**t = A000142; (n+1)!
t = A000040, P s**t = A054640
t = A000040, P (1/3) s**t = A374852
t = A000079, 2^n s**t = A028361
t = A000079, 2^n (1/3) s**t = A028362
t = A000045, F s**t = A082480
t = A000032, L s**t = A374890
t = A000201, LW s**t = A374860
t = A001950, UW s**t = A374864
t = A005408, 2*n+1 s**t = A000165, 2^n*n!
t = A016777, 3*n+1 s**t = A008544
t = A016789, 3*n+2 s**t = A032031
t = A000142, n! s**t = A217757
t = A000051, 2^n+1 s**t = A139486
t = A000225, 2^n-1 s**t = A006125
t = A032766, [3*n/2] s**t = A111394
t = A034472, 3^n+1 s**t = A153280
t = A024023, 3^n-1 s**t = A047656
t = A000217, T s**t = A128814
t = A000984, C s**t = A374891
t = A279019, n^2-n s**t = A130032
t = A004526, 1+[n/2] s**t = A010551
t = A002264, 1+[n/3] s**t = A264557
t = A002265, 1+[n/4] s**t = A264635
Sequences (c)**L, for c=2..4: A374656 to A374661
Sequences (c)**F, for c=2..6: A374662, A374662, A374982 to A374855
The obverse convolutions listed in Table 1 are, trivially, divisibility sequences. Likewise, if s = (-1,-1,-1,...) instead of s = (1,1,1,...), then s**t is a divisibility sequence for every choice of t; e.g. if s = (-1,-1,-1,...) and t = A279019, then s**t = A130031.
Table 2. s = A000027 = (0,1,2,3,4,5,...) = (n);
t = A000027, n s**t = A007778, n^(n+1)
t = A000290, n^2 s**t = A374881
t = A000040, P s**t = A374853
t = A000045, F s**t = A374857
t = A000032, L s**t = A374858
t = A000079, 2^n s**t = A374859
t = A000201, LW s**t = A374861
t = A005408, 2*n+1 s**t = A000407, (2*n+1)! / n!
t = A016777, 3*n+1 s**t = A113551
t = A016789, 3*n+2 s**t = A374866
t = A000142, n! s**t = A374871
t = A032766, [3*n/2] s**t = A374879
t = A000217, T s**t = A374892
t = A000984, C s**t = A374893
t = A038608, n*(-1)^n s**t = A374894
Table 3. s = A000290 = (0,1,4,9,16,...) = (n^2);
t = A000290, n^2 s**t = A323540
t = A002522, n^2+1 s**t = A374884
t = A000217, T s**t = A374885
t = A000578, n^3 s**t = A374886
t = A000079, 2^n s**t = A374887
t = A000225, 2^n-1 s**t = A374888
t = A005408, 2*n+1 s**t = A374889
t = A000045, F s**t = A374890
Table 4. s = t;
s = t = A000012, 1 s**s = A000079; 2^(n+1)
s = t = A000027, n s**s = A007778, n^(n+1)
s = t = A000290, n^2 s**s = A323540
s = t = A000045, F s**s = this sequence
s = t = A000032, L s**s = A374850
s = t = A000079, 2^n s**s = A369673
s = t = A000244, 3^n s**s = A369674
s = t = A000040, P s**s = A374851
s = t = A000201, LW s**s = A374862
s = t = A005408, 2*n+1 s**s = A062971
s = t = A016777, 3*n+1 s**s = A374877
s = t = A016789, 3*n+2 s**s = A374878
s = t = A032766, [3*n/2] s**s = A374880
s = t = A000217, T s**s = A375050
s = t = A005563, n^2-1 s**s = A375051
s = t = A279019, n^2-n s**s = A375056
s = t = A002398, n^2+n s**s = A375058
s = t = A002061, n^2+n+1 s**s = A375059
If n = 2*k+1, then s**s(n) is a square; specifically,
s**s(n) = ((s(0)+s(n))*(s(1)+s(n-1))*...*(s(k)+s(k+1)))^2.
If n = 2*k, then s**s(n) has the form 2*s(k)*m^2, where m is an integer.
Table 5. Others
s = A000201, LW t = A001950, UW s**t = A374863
s = A000045, F t = A000032, L s**t = A374865
s = A005843, 2*n t = A005408, 2*n+1 s**t = A085528, (2*n+1)^(n+1)
s = A016777, 3*n+1 t = A016789, 3*n+2 s**t = A091482
s = A005408, 2*n+1 t = A000045, F s**t = A374867
s = A005408, 2*n+1 t = A000032, L s**t = A374868
s = A005408, 2*n+1 t = A000079, 2^n s**t = A374869
s = A000027, n t = A000142, n! s**t = A374871
s = A005408, 2*n+1 t = A000142, n! s**t = A374872
s = A000079, 2^n t = A000142, n! s**t = A374874
s = A000142, n! t = A000045, F s**t = A374875
s = A000142, n! t = A000032, L s**t = A374876
s = A005408, 2*n+1 t = A016777, 3*n+1 s**t = A352601
s = A005408, 2*n+1 t = A016789, 3*n+2 s**t = A064352
Table 6. Arrays of coefficients of s(x)**t(x), where s(x) and t(x) are polynomials
s(x) t(x) s(x)**t(x)
n x A132393
n^2 x A269944
x+1 x+1 A038220
x+2 x+2 A038244
x x+3 A038220
nx x+1 A094638
1 x^2+x+1 A336996
n^2 x x+1 A375041
n^2 x 2x+1 A375042
n^2 x x+2 A375043
2^n x x+1 A375044
2^n 2x+1 A375045
2^n x+2 A375046
x+1 F(n) A375047
x+1 x+F(n) A375048
x+F(n) x+F(n) A375049

Examples

			a(0) = 0 + 0 = 0
a(1) = (0+1) * (1+0) = 1
a(2) = (0+1) * (1+1) * (1+0) = 2
a(3) = (0+2) * (1+1) * (1+1) * (2+0) = 16
As noted above, a(2*k+1) is a square for k>=0. The first 5 squares are 1, 16, 3600, 12320100, 701841817600, with corresponding square roots 1, 4, 60, 3510, 837760.
If n = 2*k, then s**s(n) has the form 2*F(k)*m^2, where m is an integer and F(k) is the k-th Fibonacci number; e.g., a(6) = 2*F(3)*(192)^2.
		

Crossrefs

Programs

  • Maple
    a:= n-> (F-> mul(F(n-j)+F(j), j=0..n))(combinat[fibonacci]):
    seq(a(n), n=0..15);  # Alois P. Heinz, Aug 02 2024
  • Mathematica
    s[n_] := Fibonacci[n]; t[n_] := Fibonacci[n];
    u[n_] := Product[s[k] + t[n - k], {k, 0, n}];
    Table[u[n], {n, 0, 20}]
  • PARI
    a(n)=prod(k=0, n, fibonacci(k) + fibonacci(n-k)) \\ Andrew Howroyd, Jul 31 2024

Formula

a(n) ~ c * phi^(3*n^2/4 + n) / 5^((n+1)/2), where c = QPochhammer(-1, 1/phi^2)^2/2 if n is even and c = phi^(1/4) * QPochhammer(-phi, 1/phi^2)^2 / (phi + 1)^2 if n is odd, and phi = A001622 is the golden ratio. - Vaclav Kotesovec, Aug 01 2024

A001715 a(n) = n!/6.

Original entry on oeis.org

1, 4, 20, 120, 840, 6720, 60480, 604800, 6652800, 79833600, 1037836800, 14529715200, 217945728000, 3487131648000, 59281238016000, 1067062284288000, 20274183401472000, 405483668029440000, 8515157028618240000, 187333454629601280000, 4308669456480829440000
Offset: 3

Keywords

Comments

The numbers (4, 20, 120, 840, 6720, ...) arise from the divisor values in the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ... *(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) (which covers the following sequences: A000578, A000537, A024166, A101094, A101097, A101102). - Alexander R. Povolotsky, May 17 2008
a(n) is also the number of decreasing 3-cycles in the decomposition of permutations as product of disjoint cycles, a(3)=1, a(4)=4, a(5)=20. - Wenjin Woan, Dec 21 2008
Equals eigensequence of triangle A130128 reflected. - Gary W. Adamson, Dec 23 2008
a(n) is the number of n-permutations having 1, 2, and 3 in three distinct cycles. - Geoffrey Critzer, Apr 26 2009
From Johannes W. Meijer, Oct 20 2009: (Start)
The asymptotic expansion of the higher order exponential integral E(x,m=1,n=4) ~ exp(-x)/x*(1 - 4/x + 20/x^2 - 120/x^3 + 840/x^4 - 6720/x^5 + 60480/x^6 - 604800/x^7 + ...) leads to the sequence given above. See A163931 and A130534 for more information.
(End)

References

  • 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).

Programs

Formula

a(n) = A049352(n-2, 1) (first column of triangle).
E.g.f. if offset 0: 1/(1-x)^4.
a(n) = A173333(n,3). - Reinhard Zumkeller, Feb 19 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(k+4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: W(0), where W(k) = 1 - x*(k+4)/( x*(k+4) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) = A245334(n,n-3) / 4. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, May 22 2017: (Start)
The o.g.f. A(x) satisfies the Riccati equation x^2*A'(x) + (4*x - 1)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - 4*x/(1 - x/(1 - 5*x/(1 - 2*x/(1 - 6*x/(1 - 3*x/(1 - ... - (n + 3)*x/(1 - n*x/(1 - ... ))))))))) (apply Stokes, 1982).
A(x) = 1/(1 - 3*x - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - 6*x/(1 - ... - n*x/(1 - (n+3)*x/(1 - ... ))))))))). (End)
H(x) = (1 - (1 + x)^(-3)) / 3 = x - 4 x^2/2! + 20 x^3/3! - ... is an e.g.f. of the signed sequence (n!/4!), which is the compositional inverse of G(x) = (1 - 3*x)^(-1/3) - 1, an e.g.f. for A007559. Cf. A094638, A001710 (for n!/2!), and A001720 (for n!/4!). Cf. columns of A094587, A173333, and A213936 and rows of A138533.- Tom Copeland, Dec 27 2019
E.g.f.: x^3 / (3! * (1 - x)). - Ilya Gutkovskiy, Jul 09 2021
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=3} 1/a(n) = 6*e - 15.
Sum_{n>=3} (-1)^(n+1)/a(n) = 3 - 6/e. (End)

Extensions

More terms from Harvey P. Dale, Aug 12 2012

A001720 a(n) = n!/24.

Original entry on oeis.org

1, 5, 30, 210, 1680, 15120, 151200, 1663200, 19958400, 259459200, 3632428800, 54486432000, 871782912000, 14820309504000, 266765571072000, 5068545850368000, 101370917007360000, 2128789257154560000, 46833363657400320000, 1077167364120207360000
Offset: 4

Keywords

Comments

The asymptotic expansion of the higher-order exponential integral E(x,m=1,n=5) ~ exp(-x)/x*(1 - 5/x + 30/x^2 - 210/x^3 + 1680/x^4 - 15120/x^5 + 151200/x^6 - 1663200/x^7 + ...) leads to this sequence. See A163931 and A130534 for more information. - Johannes W. Meijer, Oct 20 2009

References

  • 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).

Programs

Formula

a(n)= A049353(n-3, 1) (first column of triangle).
E.g.f. if offset 0: 1/(1-x)^5.
a(n) = A173333(n,4). - Reinhard Zumkeller, Feb 19 2010
a(n) = A245334(n,n-4) / 5. - Reinhard Zumkeller, Aug 31 2014
G(x) = (1 - (1 + x)^(-4)) / 4 = x - 5 x^2/2! + 30 x^3/3! - ..., an e.g.f. for this signed sequence (for n!/4!), is the compositional inverse of H(x) = (1 - 4*x)^(-1/4) - 1 = x + 5 x^2/2! + 45 x^3/3! + ..., an e.g.f. for A007696. Cf. A094638, A001710 (for n!/2!), and A001715 (for n!/3!). Cf. columns of A094587, A173333, and A213936 and rows of A138533. - Tom Copeland, Dec 27 2019
E.g.f.: x^4 / (4! * (1 - x)). - Ilya Gutkovskiy, Jul 09 2021
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=4} 1/a(n) = 24*e - 64.
Sum_{n>=4} (-1)^n/a(n) = 24/e - 8. (End)

A105278 Triangle read by rows: T(n,k) = binomial(n,k)*(n-1)!/(k-1)!.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 24, 36, 12, 1, 120, 240, 120, 20, 1, 720, 1800, 1200, 300, 30, 1, 5040, 15120, 12600, 4200, 630, 42, 1, 40320, 141120, 141120, 58800, 11760, 1176, 56, 1, 362880, 1451520, 1693440, 846720, 211680, 28224, 2016, 72, 1, 3628800, 16329600
Offset: 1

Author

Miklos Kristof, Apr 25 2005

Keywords

Comments

T(n,k) is the number of partially ordered sets (posets) on n elements that consist entirely of k chains. For example, T(4, 3)=12 since there are exactly 12 posets on {a,b,c,d} that consist entirely of 3 chains. Letting ab denote a<=b and using a slash "/" to separate chains, the 12 posets can be given by a/b/cd, a/b/dc, a/c/bd, a/c/db, a/d/bc, a/d/cb, b/c/ad, b/c/da, b/d/ac, b/d/ca, c/d/ab and c/d/ba, where the listing of the chains is arbitrary (e.g., a/b/cd = a/cd/b =...cd/b/a). - Dennis P. Walsh, Feb 22 2007
Also the matrix product |S1|.S2 of Stirling numbers of both kinds.
This Lah triangle is a lower triangular matrix of the Jabotinsky type. See the column e.g.f. and the D. E. Knuth reference given in A008297. - Wolfdieter Lang, Jun 29 2007
The infinitesimal matrix generator of this matrix is given in A132710. See A111596 for an interpretation in terms of circular binary words and generalized factorials. - Tom Copeland, Nov 22 2007
Three combinatorial interpretations: T(n,k) is (1) the number of ways to split [n] = {1,...,n} into a collection of k nonempty lists ("partitions into sets of lists"), (2) the number of ways to split [n] into an ordered collection of n+1-k nonempty sets that are noncrossing ("partitions into lists of noncrossing sets"), (3) the number of Dyck n-paths with n+1-k peaks labeled 1,2,...,n+1-k in some order. - David Callan, Jul 25 2008
Given matrices A and B with A(n,k) = T(n,k)*a(n-k) and B(n,k) = T(n,k)*b(n-k), then A*B = D where D(n,k) = T(n,k)*[a(.)+b(.)]^(n-k), umbrally. - Tom Copeland, Aug 21 2008
An e.g.f. for the row polynomials of A(n,k) = T(n,k)*a(n-k) is exp[a(.)* D_x * x^2] exp(x*t) = exp(x*t) exp[(.)!*Lag(.,-x*t,1)*a(.)*x], umbrally, where [(.)! Lag(.,x,1)]^n = n! Lag(n,x,1) is a normalized Laguerre polynomial of order 1. - Tom Copeland, Aug 29 2008
Triangle of coefficients from the Bell polynomial of the second kind for f = 1/(1-x). B(n,k){x1,x2,x3,...} = B(n,k){1/(1-x)^2,...,(j-1)!/(1-x)^j,...} = T(n,k)/(1-x)^(n+k). - Vladimir Kruchinin, Mar 04 2011
The triangle, with the row and column offset taken as 0, is the generalized Riordan array (exp(x), x) with respect to the sequence n!*(n+1)! as defined by Wang and Wang (the generalized Riordan array (exp(x), x) with respect to the sequence n! is Pascal's triangle A007318, and with respect to the sequence n!^2 is A021009 unsigned). - Peter Bala, Aug 15 2013
For a relation to loop integrals in QCD, see p. 33 of Gopakumar and Gross and Blaizot and Nowak. - Tom Copeland, Jan 18 2016
Also the Bell transform of (n+1)!. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016
Also the number of k-dimensional flats of the n-dimensional Shi arrangement. - Shuhei Tsujie, Apr 26 2019
The numbers T(n,k) appear as coefficients when expanding the rising factorials (x)^k = x(x+1)...(x+k-1) in the basis of falling factorials (x)k = x(x-1)...(x-k+1). Specifically, (x)^n = Sum{k=1..n} T(n,k) (x)k. - _Jeremy L. Martin, Apr 21 2021

Examples

			T(1,1) = C(1,1)*0!/0! = 1,
T(2,1) = C(2,1)*1!/0! = 2,
T(2,2) = C(2,2)*1!/1! = 1,
T(3,1) = C(3,1)*2!/0! = 6,
T(3,2) = C(3,2)*2!/1! = 6,
T(3,3) = C(3,3)*2!/2! = 1,
Sheffer a-sequence recurrence: T(6,2)= 1800 = (6/3)*120 + 6*240.
B(n,k) =
   1/(1-x)^2;
   2/(1-x)^3,  1/(1-x)^4;
   6/(1-x)^4,  6/(1-x)^5,  1/(1-x)^6;
  24/(1-x)^5, 36/(1-x)^6, 12/(1-x)^7, 1/(1-x)^8;
The triangle T(n,k) begins:
  n\k      1       2       3      4      5     6    7  8  9 ...
  1:       1
  2:       2       1
  3:       6       6       1
  4:      24      36      12      1
  5:     120     240     120     20      1
  6:     720    1800    1200    300     30     1
  7:    5040   15120   12600   4200    630    42    1
  8:   40320  141120  141120  58800  11760  1176   56  1
  9:  362880 1451520 1693440 846720 211680 28224 2016 72  1
  ...
Row n=10: [3628800, 16329600, 21772800, 12700800, 3810240, 635040, 60480, 3240, 90, 1]. - _Wolfdieter Lang_, Feb 01 2013
From _Peter Bala_, Feb 24 2025: (Start)
The array factorizes as an infinite product (read from right to left):
  /  1                \        /1             \^m /1           \^m /1           \^m
  |  2    1            |      | 0   1          |  |0  1         |  |1  1         |
  |  6    6   1        | = ...| 0   0   1      |  |0  1  1      |  |0  2  1      |
  | 24   36  12   1    |      | 0   0   1  1   |  |0  0  2  1   |  |0  0  3  1   |
  |120  240 120  20   1|      | 0   0   0  2  1|  |0  0  0  3  1|  |0  0  0  4  1|
  |...                 |      |...             |  |...          |  |...          |
where m = 2. Cf. A008277 (m = 1), A035342 (m = 3), A035469 (m = 4), A049029 (m = 5) A049385 (m = 6), A092082 (m = 7), A132056 (m = 8), A223511 - A223522 (m = 9 through 20), A001497 (m = -1), A004747 (m = -2), A000369 (m = -3), A011801 (m = -4), A013988 (m = -5). (End)
		

Crossrefs

Triangle of Lah numbers (A008297) unsigned.
Cf. A111596 (signed triangle with extra n=0 row and m=0 column).
Cf. A130561 (for a natural refinement).
Cf. A094638 (for differential operator representation).
Cf. A248045 (central terms), A002868 (row maxima).
Cf, A059110.
Cf. A089231 (triangle with mirrored rows).
Cf. A271703 (triangle with extra n=0 row and m=0 column).

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->Binomial(n,k)*Factorial(n-1)/Factorial(k-1)))); # Muniru A Asiru, Jul 25 2018
  • Haskell
    a105278 n k = a105278_tabl !! (n-1) !! (k-1)
    a105278_row n = a105278_tabl !! (n-1)
    a105278_tabl = [1] : f [1] 2 where
       f xs i = ys : f ys (i + 1) where
         ys = zipWith (+) ([0] ++ xs) (zipWith (*) [i, i + 1 ..] (xs ++ [0]))
    -- Reinhard Zumkeller, Sep 30 2014, Mar 18 2013
    
  • Magma
    /* As triangle */ [[Binomial(n,k)*Factorial(n-1)/Factorial(k-1): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 31 2014
    
  • Maple
    The triangle: for n from 1 to 13 do seq(binomial(n,k)*(n-1)!/(k-1)!,k=1..n) od;
    the sequence: seq(seq(binomial(n,k)*(n-1)!/(k-1)!,k=1..n),n=1..13);
    # The function BellMatrix is defined in A264428.
    # Adds (1, 0, 0, 0, ...) as column 0.
    BellMatrix(n -> (n+1)!, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    nn = 9; a = x/(1 - x); f[list_] := Select[list, # > 0 &]; Flatten[Map[f, Drop[Range[0, nn]! CoefficientList[Series[Exp[y a], {x, 0, nn}], {x, y}], 1]]] (* Geoffrey Critzer, Dec 11 2011 *)
    nn = 9; Flatten[Table[(j - k)! Binomial[j, k] Binomial[j - 1, k - 1], {j, nn}, {k, j}]] (* Jan Mangaldan, Mar 15 2013 *)
    rows = 10;
    t = Range[rows]!;
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 23 2018, after Peter Luschny *)
    T[n_, n_] := 1; T[n_, k_] /;0Oliver Seipel, Dec 06 2024 *)
  • Perl
    use ntheory ":all"; say join ", ", map { my $n=$; map { stirling($n,$,3) } 1..$n; } 1..9; # Dana Jacobsen, Mar 16 2017
    

Formula

T(n,k) = Sum_{m=n..k} |S1(n,m)|*S2(m,k), k>=n>=1, with Stirling triangles S2(n,m):=A048993 and S1(n,m):=A048994.
T(n,k) = C(n,k)*(n-1)!/(k-1)!.
Sum_{k=1..n} T(n,k) = A000262(n).
n*Sum_{k=1..n} T(n,k) = A103194(n) = Sum_{k=1..n} T(n,k)*k^2.
E.g.f. column k: (x^(k-1)/(1-x)^(k+1))/(k-1)!, k>=1.
Recurrence from Sheffer (here Jabotinsky) a-sequence [1,1,0,...] (see the W. Lang link under A006232): T(n,k)=(n/k)*T(n-1,m-1) + n*T(n-1,m). - Wolfdieter Lang, Jun 29 2007
The e.g.f. is, umbrally, exp[(.)!* L(.,-t,1)*x] = exp[t*x/(1-x)]/(1-x)^2 where L(n,t,1) = Sum_{k=0..n} T(n+1,k+1)*(-t)^k = Sum_{k=0..n} binomial(n+1,k+1)* (-t)^k / k! is the associated Laguerre polynomial of order 1. - Tom Copeland, Nov 17 2007
For this Lah triangle, the n-th row polynomial is given umbrally by
n! C(B.(x)+1+n,n) = (-1)^n C(-B.(x)-2,n), where C(x,n)=x!/(n!(x-n)!),
the binomial coefficient, and B_n(x)= exp(-x)(xd/dx)^n exp(x), the n-th Bell / Touchard / exponential polynomial (cf. A008277). E.g.,
2! C(-B.(-x)-2,2) = (-B.(x)-2)(-B.(x)-3) = B_2(x) + 5*B_1(x) + 6 = 6 + 6x + x^2.
n! C(B.(x)+1+n,n) = n! e^(-x) Sum_{j>=0} C(j+1+n,n)x^j/j! is a corresponding Dobinski relation. See the Copeland link for the relation to inverse Mellin transform. - Tom Copeland, Nov 21 2011
The row polynomials are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A008277 (D = (1+x)*d/dx), A035342 (D = (1+x)^3*d/dx), A035469 (D = (1+x)^4*d/dx) and A049029 (D = (1+x)^5*d/dx). - Peter Bala, Nov 25 2011
T(n,k) = Sum_{i=k..n} A130534(n-1,i-1)*A008277(i,k). - Reinhard Zumkeller, Mar 18 2013
Let E(x) = Sum_{n >= 0} x^n/(n!*(n+1)!). Then a generating function is exp(t)*E(x*t) = 1 + (2 + x)*t + (6 + 6*x + x^2)*t^2/(2!*3!) + (24 + 36*x + 12*x^2 + x^3)*t^3/(3!*4!) + ... . - Peter Bala, Aug 15 2013
P_n(x) = L_n(1+x) = n!*Lag_n(-(1+x);1), where P_n(x) are the row polynomials of A059110; L_n(x), the Lah polynomials of A105278; and Lag_n(x;1), the Laguerre polynomials of order 1. These relations follow from the relation between the iterated operator (x^2 D)^n and ((1+x)^2 D)^n with D = d/dx. - Tom Copeland, Jul 23 2018
Dividing each n-th diagonal by n!, where the main diagonal is n=1, generates the Narayana matrix A001263. - Tom Copeland, Sep 23 2020
T(n,k) = A089231(n,n-k). - Ron L.J. van den Burg, Dec 12 2021
T(n,k) = T(n-1,k-1) + (n+k-1)*T(n-1,k). - Bérénice Delcroix-Oger, Jun 25 2025

Extensions

Stirling comments and e.g.f.s from Wolfdieter Lang, Apr 11 2007

A049029 Triangle read by rows, the Bell transform of the quartic factorial numbers A007696(n+1) without column 0.

Original entry on oeis.org

1, 5, 1, 45, 15, 1, 585, 255, 30, 1, 9945, 5175, 825, 50, 1, 208845, 123795, 24150, 2025, 75, 1, 5221125, 3427515, 775845, 80850, 4200, 105, 1, 151412625, 108046575, 27478710, 3363045, 219450, 7770, 140, 1, 4996616625, 3824996175, 1069801425
Offset: 1

Keywords

Comments

Previous name was: Triangle of numbers related to triangle A048882; generalization of Stirling numbers of second kind A008277, Lah-numbers A008297, ...
a(n,m) enumerates unordered n-vertex m-forests composed of m plane increasing quintic (5-ary) trees. Proof based on the a(n,m) recurrence. See also the F. Bergeron et al. reference, especially Table 1, first row and Example 1 for the e.g.f. for m=1. - Wolfdieter Lang, Sep 14 2007
Also the Bell transform of A007696(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 28 2016

Examples

			Triangle starts:
{1};
{5,1};
{45,15,1};
{585,255,30,1};
{9945,5175,825,50,1};
...
		

Crossrefs

a(n, m) := S2(5, n, m) is the fifth triangle of numbers in the sequence S2(1, n, m) := A008277(n, m) (Stirling 2nd kind), S2(2, n, m) := A008297(n, m) (Lah), S2(3, n, m) := A035342(n, m), S2(4, n, m) := A035469(n, m). a(n, 1)= A007696(n). A007559(n).
Cf. A048882, A007696. Row sums: A049120(n), n >= 1.

Programs

Formula

a(n, m) = n!*A048882(n, m)/(m!*4^(n-m)); a(n+1, m) = (4*n+m)*a(n, m)+ a(n, m-1), n >= m >= 1; a(n, m) := 0, n
a(n, m) = sum(|A051142(n, j)|*S2(j, m), j=m..n) (matrix product), with S2(j, m) := A008277(j, m) (Stirling2 triangle). Priv. comm. to W. Lang by E. Neuwirth, Feb 15 2001; see also the 2001 Neuwirth reference. See the general comment on products of Jabotinsky matrices given under A035342.
From Peter Bala, Nov 25 2011: (Start)
E.g.f.: G(x,t) = exp(t*A(x)) = 1+t*x+(5*t+t^2)*x^2/2!+(45*t+15*t^2+t^3)*x^3/3!+..., where A(x) = -1+(1-4*x)^(-1/4) satisfies the autonomous differential equation A'(x) = (1+A(x))^5.
The generating function G(x,t) satisfies the partial differential equation t*(dG/dt+G) = (1-4*x)*dG/dx, from which follows the recurrence given above.
The row polynomials are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+x)^5*d/dx. Cf. A008277 (D = (1+x)*d/dx), A105278 (D = (1+x)^2*d/dx), A035342 (D = (1+x)^3*d/dx) and A035469 (D = (1+x)^4*d/dx).
(End)

Extensions

New name from Peter Luschny, Jan 30 2016
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