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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A102773 a(n) = Sum_{i=0..n} binomial(n,i)^2*i!*4^i.

Original entry on oeis.org

1, 5, 49, 709, 13505, 318181, 8916145, 289283429, 10656031489, 439039941445, 19995858681521, 997184081617285, 54026137182982849, 3159127731435043109, 198258247783634075185, 13289190424904891606821, 947419111092028780186625
Offset: 0

Views

Author

Miklos Kristof, Mar 16 2005

Keywords

Crossrefs

Programs

  • Maple
    seq(sum('binomial(k,i)^2*i!*4^i', 'i'=0..k),k=0..30);
  • Mathematica
    f[n_] := Sum[k!*4^k*Binomial[n, k]^2, {k, 0, n}]; Table[ f[n], {n, 0, 16}] (* or *)
    Range[0, 16]! CoefficientList[ Series[1/(1 - 4x)*Exp[x/(1 - 4x)], {x, 0, 16}], x] (* Robert G. Wilson v, Mar 16 2005 *)
  • PARI
    a(n)=my(t=1); sum(i=1,n, t*=i; binomial(n,i)^2*t<<(2*i), 1) \\ Charles R Greathouse IV, Oct 23 2023

Formula

E.g.f.: (1/(1-4x))*exp(x/(1-4x)).
a(n) = (8*n-3)*a(n-1) - 16*(n-1)^2*a(n-2). - Vaclav Kotesovec, Oct 09 2013
a(n) ~ n^(n+1/4) * exp(sqrt(n)-n-1/8) * 4^n * (1 + 37/(96*sqrt(n))). - Vaclav Kotesovec, Oct 09 2013
Sum_{n>=0} a(n) * x^n / (n!)^2 = exp(4*x) * BesselI(0,2*sqrt(x)). - Ilya Gutkovskiy, Jul 17 2020

Extensions

More terms from Robert G. Wilson v, Mar 16 2005

A287065 Number of dominating sets on the n X n rook graph.

Original entry on oeis.org

1, 11, 421, 59747, 32260381, 67680006971, 559876911043381, 18412604442711949187, 2416403019417984915336061, 1267413006543912045144741284411, 2658304092145691708492995820522716981, 22300364428188338185156192161829091442585827
Offset: 1

Views

Author

Eric W. Weisstein, May 19 2017

Keywords

Comments

Number of {0,1} n X n matrices with no zero rows or no zero columns. - Geoffrey Critzer, Jan 15 2024

Crossrefs

Main diagonal of A287274.
Row sums of A368831.

Programs

  • Mathematica
    Table[(2^n - 1)^n + Sum[Binomial[n, i] Sum[(-1)^j (-1 + 2^(n - j))^i Binomial[n, j], {j, 0, n}], {i, n - 1}], {n, 20}] (* Eric W. Weisstein, May 27 2017 *)
  • PARI
    b(m,n)=sum(j=0, m, (-1)^j*binomial(m, j)*(2^(m - j) - 1)^n);
    a(n)=(2^n-1)^n + sum(i=1,n-1,b(n,i)*binomial(n,i)); \\ Andrew Howroyd, May 22 2017

Formula

a(n) = (2^n-1)^n + Sum_{i=1..n-1} binomial(n,i) * A183109(n,i). - Andrew Howroyd, May 22 2017

Extensions

a(6)-a(12) from Andrew Howroyd, May 22 2017

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)*binomial(n,k)*binomial(n+k,k) (n >= k >= 0).

Original entry on oeis.org

-1, -1, 2, -1, 6, -6, -1, 12, -30, 20, -1, 20, -90, 140, -70, -1, 30, -210, 560, -630, 252, -1, 42, -420, 1680, -3150, 2772, -924, -1, 56, -756, 4200, -11550, 16632, -12012, 3432, -1, 72, -1260, 9240, -34650, 72072, -84084, 51480, -12870, -1, 90, -1980, 18480, -90090, 252252, -420420, 411840, -218790, 48620, -1, 110, -2970, 34320, -210210, 756756, -1681680, 2333760, -1969110, 923780, -184756
Offset: 0

Views

Author

N. J. A. Sloane, Jan 17 2020

Keywords

Comments

This is Table I of Ser (1933), page 92.
From Petros Hadjicostas, Jul 09 2020: (Start)
Essentially Ser (1933) in his book (and in particular for Tables I-IV) finds triangular arrays that allow him to express the coefficients of various kinds of series in terms of the coefficients of other series.
He uses Newton's series (or some variation of it), factorial series, and inverse factorial series. Unfortunately, he uses unusual notation, and as a result it is difficult to understand what he is actually doing.
Rivoal (2008, 2009) essentially uses factorial series and transformations to other kinds of series to provide new proofs of the irrationality of log(2), zeta(2), and zeta(3). As a result, the triangular array T(n,k) appears in various parts of his papers.
We believe Table I (p. 92) in Ser (1933), regarding the numbers T(n,k), corresponds to four different formulas. We have deciphered the first two of them. (End)

Examples

			Triangle T(n,k) (with rows n >= 0 and columns k=0..n) begins:
  -1;
  -1,  2;
  -1,  6,   -6;
  -1, 12,  -30,   20;
  -1, 20,  -90,  140,    -70;
  -1, 30, -210,  560,   -630,   252;
  -1, 42, -420, 1680,  -3150,  2772,   -924;
  -1, 56, -756, 4200, -11550, 16632, -12012, 3432;
  ...
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92) is
  -1;
  -1/2, 1/2;
  -1/3, 1/2,   -1/6;
  -1/4, 9/20,  -1/4,  1/20;
  -1/5, 2/5,   -2/7,  1/10, -1/70;
  -1/6, 5/14, -25/84, 5/36, -1/28,  1/252;
  -1/7, 9/28, -25/84, 1/6,  -9/154,  1/84, -1/924;
   ... (End)
		

References

  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, pp. 92-93.

Crossrefs

A063007 is the same triangle without the minus signs, and has much more information.
Columns 1 and 2 are A002378 and A033487; the last three diagonals are A002544, A002457, A000984.

Programs

  • Magma
    /* As triangle: */ [[(-1)^(k+1) * Factorial(n+k) / (Factorial(k) * Factorial(k) * Factorial(n-k)): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jan 19 2020
    
  • Mathematica
    Table[CoefficientList[-Hypergeometric2F1[-n, n + 1, 1, x], x], {n, 0, 9}] // Flatten (* Georg Fischer, Jan 18 2020 after Peter Luschny in A063007 *)
  • SageMath
    def T(n,k): return (-1)^(k+1)*falling_factorial(n+k,2*k)/factorial(k)^2
    flatten([[T(n,k) for k in (0..n)] for n in (0..10)]) # Peter Luschny, Jul 09 2020

Formula

T(n,k) can also be written as (-1)^(k+1)*(n+k)!/(k!*k!*(n-k)!).
From Petros Hadjicostas, Jul 09 2020: (Start)
Ser's first formula from his Table I (p. 92) is the following:
Sum_{k=0..n} T(n,k)*k!/(x*(x+1)*...*(x+k)) = -(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, Sum_{k=0..n} T(n,k)/binomial(m+k, k) = 0 for m = 1..n.
Ser's second formula from his Table I appears also in Rivoal (2008, 2009) in a slightly different form:
Sum_{k=0..n} T(n,k)/(x + k) = (-1)^(n+1)*(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, for m = 1..n, Sum_{k=0..n} T(n,k)/(m + k) = 0. (End)
T(n,k) = (-1)^(k+1)*FallingFactorial(n+k,2*k)/(k!)^2. - Peter Luschny, Jul 09 2020
From Petros Hadjicostas, Jul 10 2020: (Start)
Peter Luschny's formula above is essentially the way the numbers T(n,k) appear in Eq. (7) on p. 86 of Ser's (1933) book. Eq. (7) is essentially equivalent to the first formula above (related to Table I on p. 92).
By inverting that formula (in some way), he gets
n!/(x*(x+1)*...*(x+n)) = Sum_{p=0..n} (-1)^p*(2*p+1)*f_p(n+1)*f_p(x), where f_p(x) = (x-1)*...*(x-p)/(x*(x+1)*...*(x+p)). This is equivalent to Eq. (8) on p. 86 of Ser's book.
The rational coefficients A(n,p) = (2*p+1)*f_p(n+1) = (2*p+1)*(n*(n-1)*...*(n+1-p))/((n+1)*...*(n+1+p)) appear in Table II on p. 92 of Ser's book.
If we consider the coefficients T(n,k) and (-1)^(p+1)*A(n,p) as infinite lower triangular matrices, then they are inverses of one another (see the example below). This means that, for m >= s,
Sum_{k=s..m} T(m,k)*(-1)^(s+1)*A(k,s) = I(s=m) = Sum_{k=s..m} (-1)^(k+1)*A(m,k)*T(k,s), where I(s=m) = 1, if s = m, and = 0, otherwise.
Without the (-1)^p, we get the formula
1/(x+n) = Sum_{p=0..n} (2*p+1)*f_p(n+1)*f_p(x),
which apparently is the inversion of the second of Ser's formulas (related to Table I on p. 92).
In all of the above formulas, an empty product is by definition 1, so f_0(x) = 1/x. (End)

Extensions

Thanks to Bob Selcoe, who noticed a typo in one of the entries, which, when corrected, led to an explicit formula for the whole of Ser's Table I.

A056953 Denominators of continued fraction for alternating factorial.

Original entry on oeis.org

1, 1, 2, 3, 7, 13, 34, 73, 209, 501, 1546, 4051, 13327, 37633, 130922, 394353, 1441729, 4596553, 17572114, 58941091, 234662231, 824073141, 3405357682, 12470162233, 53334454417, 202976401213, 896324308634, 3535017524403, 16083557845279, 65573803186921
Offset: 0

Views

Author

Aleksandar Petojevic, Sep 05 2000

Keywords

Comments

Starting (1, 2, 3, ...) with offset 0 = eigensequence of an infinite lower triangular matrix with 1's in the main diagonal and the natural numbers repeated in the subdiagonal. - Gary W. Adamson, Feb 14 2011
a(n) is the number of involutions of [n] such that every 2-cycle contains one odd and one even element; a(4) = 7: 1234, 1243, 1324, 2134, 2143, 4231, 4321. - Alois P. Heinz, Feb 14 2013

Crossrefs

Bisections are A000262 and A002720.
Cf. A124428, diagonals of A088699.

Programs

  • Magma
    [(&+[Factorial(k)*Binomial(Floor(n/2),k)*Binomial(Floor((n+1)/2),k): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, May 16 2018
  • Maple
    a:= proc(n) option remember; `if`(n<4, [1, 1, 2, 3][n+1],
          ((4*n-2)*a(n-2) +2*a(n-3) -(n-2)*(n-3)*a(n-4)) /4)
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Feb 14 2013
  • Mathematica
    Table[Sum[k!*Binomial[Floor[n/2], k]*Binomial[Floor[(n+1)/2], k] , {k,0,Floor[n/2]}], {n,0,30}] (* G. C. Greubel, May 16 2018 *)
  • PARI
    a(n)=sum(k=0,n\2,k!*binomial(n\2,k)*binomial((n+1)\2,k)) \\ Paul D. Hanna, Oct 31 2006
    

Formula

a(0)=1; a(1)=1; a(n) = a(n-1) + n*a(n-2)/2.
a(n) = Sum_{k=0..[n/2]} k!*C([n/2],k)*C([(n+1)/2],k). - Paul D. Hanna, Oct 31 2006
a(n) ~ n^(n/2 + 1/4) / (2^(n/2 + 3/4) * exp(n/2 - sqrt(2*n) + 1/2)) * (1 + (25 + 6*(-1)^n)/(24*sqrt(2*n)) + (397 + 156*(-1)^n)/(2304*n)). - Vaclav Kotesovec, Feb 22 2019

A105219 a(n) = Sum_{k=0..n} C(n,k)^2*(n-k)!*k^2.

Original entry on oeis.org

0, 1, 8, 63, 544, 5225, 55656, 653023, 8379008, 116780049, 1757211400, 28394129951, 490371506208, 9013522796473, 175679564492264, 3618800515187775, 78547755741723136, 1791704327280481313, 42846080320725932808, 1071798626271975328639, 27989931083161219661600
Offset: 0

Views

Author

Miklos Kristof, Apr 13 2005

Keywords

Comments

If the e.g.f. of n^2 is E(x) and a(n) = Sum_{k=0..n} C(n,k)^2*(n-k)!*k^2, then the e.g.f. of a(n) is E(x/(1-x))/(1-x). (Thanks to Vladeta Jovovic for help.)
a(n) is the total number of edges in all matchings of the labeled complete bipartite graph K_n,n. Cf. A144084 for other interpretations. - Geoffrey Critzer, Nov 17 2021

Examples

			b(n) = 0,1,4,9,16,25,36,49,64,...
a(3) = C(3,0)^2*3!*b(0) + C(3,1)^2*2!*b(1) + C(3,2)^2*1!*b(2) + C(3,3)^2*0!*b(3) = 1*6*0 + 9*2*1 + 9*1*4 + 1*1*9 = 0 + 18 + 36 + 9 = 63.
		

Crossrefs

Programs

  • Maple
    for n from 0 to 30 do b[n]:=n^2 od: seq(add(binomial(n,k)^2*(n-k)!*b[k], k=0..n), n=0..30);
    seq(`if`(n=0,0,simplify(n!*LaguerreL(n-1,2,-1))),n=0..17); # Peter Luschny, Apr 11 2015
  • Mathematica
    CoefficientList[Series[(x/(1-x)^2+x^2/(1-x)^3)*E^(x/(1-x)), {x, 0, 20}], x]* Table[n!, {n, 0, 20}] (* Vaclav Kotesovec, Oct 17 2012 *)

Formula

E.g.f.: (x/(1-x)^2+x^2/(1-x)^3)*exp(x/(1-x)).
a(n) = n^2*A002720(n-1) for n>=1 [Riordan]. - N. J. A. Sloane, Jan 10 2018
a(n) = (n+1)!*(2*L(n,-1)-L(n+1,-1)) where L(n,x) is the n-th Laguerre polynomial. - Peter Luschny, Jan 19 2012
Recurrence: a(n) = 2*(n+2)*a(n-1) - (n^2+4*n-4)*a(n-2) + 2*(n-2)*(n-1)*a(n-3). - Vaclav Kotesovec, Oct 17 2012
a(n) ~ exp(2*sqrt(n)-n-1/2)*n^(n+5/4)/sqrt(2)*(1-17/(48*sqrt(n))). - Vaclav Kotesovec, Oct 17 2012
a(n) = n!*L(n-1,2,-1) for n>=1 where L(n,b,x) is the n-th generalized Laguerre polynomial. - Peter Luschny, Apr 11 2015
a(n) = Sum_{k=0...n} A144084(n,k)*k. - Geoffrey Critzer, Nov 17 2021
a(n) = Sum_{k=0..n} (n-k) * A206703(n,k). - Alois P. Heinz, Feb 19 2022
a(n) = Sum_{k=1..n} k*k!*binomial(n,k)^2. - Ridouane Oudra, Jun 15 2025

A123510 Arises in the normal ordering of functions of a*(a+)*a, where a and a+ are the boson annihilation and creation operators, respectively.

Original entry on oeis.org

1, 6, 42, 340, 3135, 32466, 373156, 4713192, 64877805, 966466270, 15487707246, 265617899196, 4853435351947, 94114052406570, 1930026941433480, 41728495237790416, 948549349736725401, 22613209058160908982, 564104540143144909810, 14694713818659640322340
Offset: 0

Views

Author

Karol A. Penson, Oct 02 2006

Keywords

Crossrefs

Programs

  • Magma
    I:=[6,42]; [1] cat [n le 2 select I[n] else 2*(n+2)*Self(n-1) - (n^2 -1)*((n+2)/n)*Self(n-2): n in [1..30]]; // G. C. Greubel, May 16 2018
  • Mathematica
    max = 16; s = (1/(1-x)^3)*Exp[x/(1-x)]*LaguerreL[2, -x/(1-x)] + O[x]^(max+1); CoefficientList[s, x]*Range[0, max]! (* Jean-François Alcover, May 23 2016 *)
  • PARI
    m=30; v=concat([6,42], vector(m-2)); for(n=3, m, v[n]=2*(n+2)*v[n-1]-(n^2 - 1)*((n+2)/n)*v[n-2]); concat([1], v) \\ G. C. Greubel, May 16 2018
    

Formula

E.g.f.: (1/(1-x)^3)*exp(x/(1-x))*LaguerreL(2,-x/(1-x)), where LaguerreL(p,y) are the Laguerre polynomials.
From Vaclav Kotesovec, Nov 13 2017: (Start)
Recurrence: n*a(n) = 2*n*(n+2)*a(n-1) - (n-1)*(n+1)*(n+2)*a(n-2).
a(n) ~ exp(2*sqrt(n) - n - 1/2) * n^(n + 9/4) / 2^(3/2) * (1 + 31/(48*sqrt(n))).
(End)

Extensions

a(0)=1 prepended by G. C. Greubel, Oct 31 2017
More terms from G. C. Greubel, May 16 2018

A123511 Arises in the normal ordering of functions of a*(a+)*a, where a and a+ are the boson annihilation and creation operators, respectively.

Original entry on oeis.org

1, 8, 70, 680, 7315, 86576, 1119468, 15710640, 237885285, 3865865080, 67113398066, 1239550196248, 24267176759735, 501941612835040, 10936819334789720, 250370971426742496, 6007479214999260873
Offset: 0

Views

Author

Karol A. Penson, Oct 02 2006

Keywords

Crossrefs

Programs

  • Mathematica
    max = 16; s = (1/(1 - x)^4)*Exp[x/(1 - x)]*LaguerreL[3, -x/(1 - x)] + O[x]^(max + 1); CoefficientList[s, x]*Range[0, max]! (* Jean-François Alcover, May 23 2016 *)

Formula

E.g.f.: (1/(1-x)^4)*exp(x/(1-x))*LaguerreL(3,-x/(1-x)).
From Vaclav Kotesovec, Nov 13 2017: (Start)
Recurrence: n*a(n) = 2*n*(n+3)*a(n-1) - (n-1)*(n+2)*(n+3)*a(n-2).
a(n) ~ exp(2*sqrt(n)-n-1/2) * n^(n + 13/4) / (3*2^(3/2)) * (1 + 31/(48*sqrt(n))).
(End)

Extensions

a(0)=1 prepended by G. C. Greubel, Oct 31 2017

A123512 Arises in the normal ordering of functions of a*(a+)*a, where a and a+ are the boson annihilation and creation operators, respectively.

Original entry on oeis.org

1, 10, 105, 1190, 14630, 194796, 2798670, 43204260, 713655855, 12564061510, 234896893231, 4648313235930, 97068707038940, 2133251854548920, 49215687006553740, 1189262114277026856, 30037396074996304365
Offset: 0

Views

Author

Karol A. Penson, Oct 02 2006

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[ Series[(1/(1 - x)^5)*Exp[x/(1 - x)]LaguerreL[4, -x/(1 - x)], {x,0,16}], x]*Range[0, 16]! (* Robert G. Wilson v, Oct 03 2006 *)
  • PARI
    LaguerreL(n,v='x) = {
      my(x='x+O('x^(n+1)), t='t);
      subst(polcoeff(exp(-x*t/(1-x))/(1-x), n), 't, v);
    };
    N=17;x='x+O('x^N); Vec(serlaplace((1/(1-x)^5)*exp(x/(1-x))*LaguerreL(4,-x/(1-x)))) \\ Gheorghe Coserea, Oct 26 2017

Formula

E.g.f.: (1/(1-x)^5)*exp(x/(1-x))*LaguerreL(4,-x/(1-x)).
From Vaclav Kotesovec, Nov 13 2017: (Start)
Recurrence: n*a(n) = 2*n*(n+4)*a(n-1) - (n-1)*(n+3)*(n+4)*a(n-2).
a(n) ~ exp(2*sqrt(n)-n-1/2) * n^(n + 17/4) / (3*2^(7/2)) * (1 + 31/(48*sqrt(n))).
(End)

Extensions

a(0)=1 prepended by Gheorghe Coserea, Oct 26 2017

A206703 Triangular array read by rows. T(n,k) is the number of partial permutations (injective partial functions) of {1,2,...,n} that have exactly k elements in a cycle. The k elements are not necessarily in the same cycle. A fixed point is considered to be in a cycle.

Original entry on oeis.org

1, 1, 1, 3, 2, 2, 13, 9, 6, 6, 73, 52, 36, 24, 24, 501, 365, 260, 180, 120, 120, 4051, 3006, 2190, 1560, 1080, 720, 720, 37633, 28357, 21042, 15330, 10920, 7560, 5040, 5040, 394353, 301064, 226856, 168336, 122640, 87360, 60480, 40320, 40320
Offset: 0

Views

Author

Geoffrey Critzer, Feb 11 2012

Keywords

Examples

			     1;
     1,     1;
     3,     2,     2;
    13,     9,     6,     6;
    73,    52,    36,    24,    24;
   501,   365,   260,   180,   120,  120;
  4051,  3006,  2190,  1560,  1080,  720,   720;
  ...
		

References

  • Mohammad K. Azarian, On the Fixed Points of a Function and the Fixed Points of its Composite Functions, International Journal of Pure and Applied Mathematics, Vol. 46, No. 1, 2008, pp. 37-44. Mathematical Reviews, MR2433713 (2009c:65129), March 2009. Zentralblatt MATH, Zbl 1160.65015.
  • Mohammad K. Azarian, Fixed Points of a Quadratic Polynomial, Problem 841, College Mathematics Journal, Vol. 38, No. 1, January 2007, p. 60. Solution published in Vol. 39, No. 1, January 2008, pp. 66-67.

Crossrefs

Columns k = 0..1 give: A000262, A006152.
Main diagonal gives A000142.
Row sums give A002720.
T(2n,n) gives A088026.

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=0, 1, add((p-> p+x^j*
          coeff(p, x, 0))(b(n-j)*binomial(n-1, j-1)*j!), j=1..n))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
    seq(T(n), n=0..10);  # Alois P. Heinz, Feb 19 2022
  • Mathematica
    nn = 7; a = 1/(1 - x); ay = 1/(1 - y x); f[list_] := Select[list, # > 0 &]; Map[f, Range[0, nn]! CoefficientList[Series[Exp[a x] ay, {x, 0, nn}], {x, y}]] // Flatten

Formula

E.g.f.: exp(x/(1-x))/(1-y*x).
From Alois P. Heinz, Feb 19 2022: (Start)
Sum_{k=1..n} T(n,k) = A052852.
Sum_{k=0..n} k * T(n,k) = A103194(n).
Sum_{k=0..n} (n-k) * T(n,k) = A105219(n).
Sum_{k=0..n} (-1)^k * T(n,k) = A331725(n). (End)

A216294 Triangular array read by rows: T(n,k) is the number of partial permutations of {1,2,...,n} that have exactly k cycles, 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 3, 3, 1, 13, 14, 6, 1, 73, 84, 41, 10, 1, 501, 609, 325, 95, 15, 1, 4051, 5155, 2944, 965, 190, 21, 1, 37633, 49790, 30023, 10689, 2415, 343, 28, 1, 394353, 539616, 340402, 129220, 32179, 5348, 574, 36, 1, 4596553, 6478521, 4246842, 1698374, 455511, 84567, 10794, 906, 45, 1
Offset: 0

Views

Author

Geoffrey Critzer, Sep 04 2012

Keywords

Comments

A partial permutation on a set X is a bijection between two subsets of X.
Row sums are A002720.
First column (corresponding to k=0) is A000262.

Examples

			1;
1,     1;
3,     3,   1;
13,   14,   6,  1;
73,   84,  41, 10,  1;
501, 609, 325, 95, 15,  1;
		

Crossrefs

Programs

  • Maple
    gf := exp(x / (1 - x)) / (1 - x)^y:
    serx := series(gf, x, 10): poly := n -> simplify(coeff(serx, x, n)):
    seq(print(seq(n!*coeff(poly(n), y, k), k = 0..n)), n = 0..9); # Peter Luschny, Feb 23 2023
  • Mathematica
    nn=10;t=Sum[n^(n-1)x^n/n!,{n,1,nn}];f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[Series[Exp[ x/(1-x)]/(1-x)^y,{x,0,nn}],{x,y}]]//Flatten

Formula

E.g.f.: exp(x/(1-x))/(1-x)^y.
From Peter Bala, Aug 23 2013: (Start)
Exponential Riordan array [exp(x/(1-x)), log(1/(1-x))].
The row polynomials R(n,y), n > = 0, satisfy the 2nd order recurrence equation R(n,y) = (2*n + y - 1)*R(n-1,y) - (n - 1)*(n + y - 2)*R(n-2,y) with R(0,y) = 1 and R(1,y) = 1 + y.
Modulo variations in offset we have: R(n,0) = A000262, R(n,1) = A002720, R(n,2) = A000262, R(n,3) = A052852, R(n,4) = A062147, R(n,5) = A062266 and R(n,6) = A062192. In general, for fixed k, the sequence {R(n,k)}n>=1 gives the entries on a diagonal of the square array A088699. (End)
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