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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A074143 a(1) = 1; a(n) = n * Sum_{k=1..n-1} a(k).

Original entry on oeis.org

1, 2, 9, 48, 300, 2160, 17640, 161280, 1632960, 18144000, 219542400, 2874009600, 40475635200, 610248038400, 9807557760000, 167382319104000, 3023343138816000, 57621363351552000, 1155628453883904000, 24329020081766400000, 536454892802949120000
Offset: 1

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Author

Amarnath Murthy, Aug 28 2002

Keywords

Comments

a(n) is also the number of elements of the alternating semigroup (A^c_n) for F(n, p) if p = n - 1 (cf. A001710). - Bakare Gatta Naimat, Jan 15 2016

Crossrefs

Programs

Formula

a(n) = n^2 * a(n-1)/(n-1) for n > 2.
a(n) = n*ceiling(n!/2) = n*A001710(n) = ceiling(A001563(n)/2). - Henry Bottomley, Nov 27 2002
a(n) = ((n+1)!-n!)/2 for n > 1. - Vladimir Joseph Stephan Orlovsky, Apr 03 2011
G.f.: (U(0) + x)/(2*x) where U(k) = 1 - 1/(k+1 - x*(k+1)^2*(k+2)/(x*(k+1)*(k+2) - 1/U(k+1))); (continued fraction). - Sergei N. Gladkovskii, Sep 27 2012
G.f.: 1/2 + Q(0), where Q(k)= 1 - 1/(k+2 - x*(k+2)^2*(k+3)/(x*(k+2)*(k+3)-1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 19 2013
a(n) = Sum_{j = 0..n} (-1)^(n-j)*binomial(n, j)*(j)^(n+1) / (n+1), n > 1, a(1) = 1. - Vladimir Kruchinin, Jun 01 2013
a(n) = numerator(n!/2*n). - Vincenzo Librandi, Apr 15 2014
a(n) is F(n;p) = n^2(n-1)!/2 if p = n-1 in A^c_n. For instance for n=4 and p=n-1: F(4; 4-1)= 4^2(4-1)!/2 = 16*6/2 = 48. - Bakare Gatta Naimat, Nov 18 2015
From Seiichi Manyama, Apr 27 2025: (Start)
E.g.f.: x/2 * (1 + 1/(1-x)^2).
a(n) = (n+2) * a(n-1) - (n-1) * a(n-2) for n > 3. (End)
From Amiram Eldar, May 04 2025: (Start)
Sum_{n>=1} 1/a(n) = 2*ExpIntegralEi(1) - 2*gamma - 1 = 2*A091725 - 2*A001620 - 1.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*gamma - 1 - 2*ExpIntegralEi(-1) = 2*A001620 - 1 + 2*A099285. (End)

Extensions

More terms from Henry Bottomley, Nov 27 2002

A143491 Unsigned 2-Stirling numbers of the first kind.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 24, 26, 9, 1, 120, 154, 71, 14, 1, 720, 1044, 580, 155, 20, 1, 5040, 8028, 5104, 1665, 295, 27, 1, 40320, 69264, 48860, 18424, 4025, 511, 35, 1, 362880, 663696, 509004, 214676, 54649, 8624, 826, 44, 1, 3628800, 6999840, 5753736, 2655764
Offset: 2

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Author

Peter Bala, Aug 20 2008

Keywords

Comments

Essentially the same as A136124 but with column numbers differing by one. See A049444 for a signed version of this array. The unsigned 2-Stirling numbers of the first kind count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the elements 1 and 2 belong to distinct cycles. This is the particular case r = 2 of the unsigned r-Stirling numbers of the first kind, which count the permutations of the set {1,2,...,n} into k disjoint cycles, with the restriction that the numbers 1, 2, ..., r belong to distinct cycles. The case r = 1 gives the usual unsigned Stirling numbers of the first kind, abs(A008275); for other cases see A143492 (r = 3) and A143493 (r = 4). The corresponding 2-Stirling numbers of the second kind can be found in A143494.
In general, the lower unitriangular array of unsigned r-Stirling numbers of the first kind (with suitable offsets in the row and column indexing) equals the matrix product St1 * P^(r-1), where St1 is the array of unsigned Stirling numbers of the first kind, abs(A008275) and P is Pascal's triangle, A007318. The theory of r-Stirling numbers of both kinds is developed in [Broder]. For details of the related r-Lah numbers see A143497.
This sequence also represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}. For a bijective proof of the relation between these numbers and the 2-Stirling numbers of the first kind see the Rotbart link. - Aviv Rotbart, May 05 2011
With offset n=0,k=0 : triangle T(n,k), read by rows, given by [2,1,3,2,4,3,5,4,6,5,...] DELTA [1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 29 2011
With offset n=0 and k=0, this is the Sheffer triangle (1/(1-x)^2,-log(1-x)) (in the umbral notation of S. Roman's book this would be called Sheffer for (exp(-2*t),1-exp(-t))). See the e.g.f. given below. Compare also with the e.g.f. for the signed version A049444. - Wolfdieter Lang, Oct 10 2011
Reversed rows correspond to the Betti numbers of the moduli space M(0,n) of smooth Riemann surfaces (see Murri link). - Tom Copeland, Sep 19 2012

Examples

			Triangle begins
  n\k|.....2.....3.....4.....5.....6.....7
  ========================================
  2..|.....1
  3..|.....2.....1
  4..|.....6.....5.....1
  5..|....24....26.....9.....1
  6..|...120...154....71....14.....1
  7..|...720..1044...580...155....20.....1
  ...
T(4,3) = 5. The permutations of {1,2,3,4} with 3 cycles such that 1 and 2 belong to different cycles are: (1)(2)(3 4), (1)(3)(24), (1)(4)(23), (2)(3)(14) and (2)(4)(13). The remaining possibility (3)(4)(12) is not allowed.
From _Aviv Rotbart_, May 05 2011: (Start)
Example of the alternating group permutations numbers:
Triangle begins
  n\k|.....0.....1.....2.....3.....4.....5.....6.....7
  ====================================================
  2..|.....1
  3..|.....1.....2
  4..|.....1.....5.....6
  5..|.....1.....9....26....24
  6..|.....1....14....71...154...120
  7..|.....1....20...155...580..1044..720
A(n,k) = number of permutations in An of length k, with respect to the generators set {(12)(ij)}. For example, A(2,0)=1 (only the identity is there), for A4, the generators are {(12)(13),(12)(14),(12,23),(12)(24),(12)(34)}, thus we have A(4,1)=5 (exactly 5 generators), the permutations of length 2 are:
   (12)(13)(12)(13) = (312)
   (12)(13)(12)(14) = (41)(23)
   (12)(13)(12)(24) = (432)(1)
   (12)(13)(12)(34) = (342)(1)
   (12)(23)(12)(24) = (13)(24)
   (12)(14)(12)(14) = (412)(3)
Namely, A(4,2)=6. Together with the identity [=(12)(12), of length 0. therefore A(4,0)=1] we have 12 permutations, comprising all A4 (4!/2=12). (End)
		

Crossrefs

Cf. A001705 - A001709 (column 3..7), A001710 (row sums), A008275, A049444 (signed version), A136124, A143492, A143493, A143494, A143497.
Cf. A094638.

Programs

  • Maple
    with combinat: T := (n, k) -> (n-2)! * add((n-j-1)*abs(stirling1(j,k-2))/j!,j = k-2..n-2): for n from 2 to 10 do seq(T(n, k), k = 2..n) end do;
  • Mathematica
    t[n_, k_] := (n-2)!*Sum[(n-j-1)*Abs[StirlingS1[j, k-2]]/j!, {j, k-2, n-2}]; Table[t[n, k], {n, 2, 11}, {k, 2, n}] // Flatten (* Jean-François Alcover, Apr 16 2013, after Maple *)

Formula

T(n,k) = (n-2)! * Sum_{j = k-2 .. n-2} (n-j-1)*|stirling1(j,k-2)|/j!.
Recurrence relation: T(n,k) = T(n-1,k-1) + (n-1)*T(n-1,k) for n > 2, with boundary conditions: T(n,1) = T(1,n) = 0, for all n; T(2,2) = 1; T(2,k) = 0 for k > 2.
Special cases: T(n,2) = (n-1)!; T(n,3) = (n-1)!*(1/2 + 1/3 + ... + 1/(n-1)).
T(n,k) = Sum_{2 <= i_1 < ... < i_(n-k) < n} (i_1*i_2*...*i_(n-k)). For example, T(6,4) = Sum_{2 <= i < j < 6} (i*j) = 2*3 + 2*4 + 2*5 + 3*4 + 3*5 + 4*5 = 71.
Row g.f.: Sum_{k = 2..n} T(n,k)*x^k = x^2*(x+2)*(x+3)*...*(x+n-1).
E.g.f. for column (k+2): Sum_{n>=k} T(n+2,k+2)*x^n/n! = (1/k!)*(1/(1-x)^2)*(log(1/(1-x)))^k.
E.g.f.: (1/(1-t))^(x+2) = Sum_{n>=0} Sum_{k = 0..n} T(n+2,k+2)*x^k*t^n/n! = 1 + (2+x)*t/1! + (6+5*x+x^2)*t^2/2! + ... .
This array is the matrix product St1 * P, where St1 denotes the lower triangular array of unsigned Stirling numbers of the first kind, abs(A008275) and P denotes Pascal's triangle, A007318. The row sums are n!/2 ( A001710 ). The alternating row sums are (n-2)!.
If we define f(n,i,a) = Sum_{k=0..n-i} binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a - j), then T(n-1,i) = |f(n,i,2)|, for n=1,2,...; i=0..n. - Milan Janjic, Dec 21 2008
From Gary W. Adamson, Jul 19 2011: (Start)
n-th row of the triangle = top row of M^(n-2), M = a reversed variant of the (1,2) Pascal triangle (Cf. A029635); as follows:
2, 1, 0, 0, 0, 0, ...
2, 3, 1, 0, 0, 0, ...
2, 5, 4, 1, 0, 0, ...
2, 7, 9, 5, 1, 0, ...
... (End)
The reversed, row polynomials of this entry multiplied by (1+x) are the row polynomials of A094638. E.g., (1+x)(1+5x+6x^2) = (1+6x+11x^2+6x^3). - Tom Copeland, Dec 11 2016

A187735 G.f.: Sum_{n>=0} (2*n+1)^n * x^n / (1 + (2*n+1)*x)^n.

Original entry on oeis.org

1, 3, 16, 120, 1152, 13440, 184320, 2903040, 51609600, 1021870080, 22295347200, 531372441600, 13733933875200, 382588157952000, 11426632984166400, 364223926370304000, 12340763622899712000, 442896294466289664000, 16783438527143608320000
Offset: 0

Views

Author

Paul D. Hanna, Jan 02 2013

Keywords

Comments

Compare g.f. to the identity (cf. A001710):
Sum_{n>=0} n^n * x^n / (1 + n*x)^n = 1 + Sum_{n>=1} (n+1)!/2 * x^n.
More generally,
if Sum_{n>=0} a(n)*x^n = Sum_{n>=0} (b*n+c)^n * x^n / (1 + (b*n+c)*x)^n,
then Sum_{n>=0} a(n)*x^n/n! = (2 - 2*(b-c)*x + b*(b-2*c)*x^2)/(2*(1-b*x)^2)
so that a(n) = (b*n + (b+2*c)) * b^(n-1) * n!/2 for n>0 with a(0)=1.

Examples

			G.f.: A(x) = 1 + 3*x + 16*x^2 + 120*x^3 + 1152*x^4 + 13440*x^5 +...
where
A(x) = 1 + 3*x/(1+3*x) + 5^2*x^2/(1+5*x)^2 + 7^3*x^3/(1+7*x)^3 + 9^4*x^4/(1+9*x)^4 + 11^5*x^5/(1+11*x)^5 +...
		

Crossrefs

Programs

  • Mathematica
    a[n_] := (n + 2)*2^(n - 1)*n!; Array[a, 20, 0] (* Amiram Eldar, Dec 23 2022 *)
  • PARI
    {a(n)=polcoeff( sum(m=0,n,((2*m+1)*x)^m / (1 + (2*m+1)*x +x*O(x^n))^m),n)}
    for(n=0,20,print1(a(n),", "))
    
  • PARI
    {a(n) = (n+2)*2^(n-1)*n!}

Formula

a(n) = (n+2) * 2^(n-1) * n!.
E.g.f.: (1-x)/(1-2*x)^2.
From Amiram Eldar, Dec 23 2022: (Start)
Sum_{n>=0} 1/a(n) = 8 - 4*sqrt(e).
Sum_{n>=0} (-1)^n/a(n) = 8 - 12/sqrt(e). (End)

A215771 Number T(n,k) of undirected labeled graphs on n nodes with exactly k cycle graphs as connected components; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 3, 7, 6, 1, 0, 12, 25, 25, 10, 1, 0, 60, 127, 120, 65, 15, 1, 0, 360, 777, 742, 420, 140, 21, 1, 0, 2520, 5547, 5446, 3157, 1190, 266, 28, 1, 0, 20160, 45216, 45559, 27342, 10857, 2898, 462, 36, 1, 0, 181440, 414144, 427275, 264925, 109935, 31899, 6300, 750, 45, 1
Offset: 0

Views

Author

Alois P. Heinz, Aug 23 2012

Keywords

Comments

Also the Bell transform of A001710. For the definition of the Bell transform see A264428 and the links given there. - Peter Luschny, Jan 21 2016

Examples

			T(4,1) = 3:  .1-2.  .1 2.  .1-2.
.            .| |.  .|X|.  . X .
.            .3-4.  .3 4.  .3-4.
.
T(4,2) = 7:  .1 2.  .1-2.  .1 2.  o1 2.  .1 2o  .1-2.  .1-2.
.            .| |.  .   .  . X .  . /|.  .|\ .  . \|.  .|/ .
.            .3 4.  .3-4.  .3 4.  .3-4.  .3-4.  o3 4.  .3 4o
.
T(4,3) = 6:  .1 2o  .1-2.  o1 2.  o1 2o  o1 2.  .1 2o
.            .|  .  .   .  .  |.  .   .  . / .  . \ .
.            .3 4o  o3 4o  o3 4.  .3-4.  .3 4o  o3 4.
.
T(4,4) = 1:  o1 2o
.            .   .
.            o3 4o
Triangle T(n,k) begins:
  1;
  0,   1;
  0,   1,   1;
  0,   1,   3,   1;
  0,   3,   7,   6,   1;
  0,  12,  25,  25,  10,   1;
  0,  60, 127, 120,  65,  15,  1;
  0, 360, 777, 742, 420, 140, 21,  1;
		

Crossrefs

Columns k=0-10 give: A000007, A001710(n-1) for n>0, A215772, A215763, A215764, A215765, A215766, A215767, A215768, A215769, A215770.
Diagonal and lower diagonals give: A000012, A000217, A001296, A215773, A215774.
Row sums give A002135.
T(2n,n) gives A253276.

Programs

  • Maple
    T:= proc(n, k) option remember; `if`(k<0 or k>n, 0, `if`(n=0, 1,
          add(binomial(n-1, i)*T(n-1-i, k-1)*ceil(i!/2), i=0..n-k)))
        end:
    seq(seq(T(n, k), k=0..n), n=0..12);
    # Alternatively, with the function BellMatrix defined in A264428:
    BellMatrix(n -> `if`(n<2, 1, n!/2), 8); # Peter Luschny, Jan 21 2016
  • Mathematica
    t[n_, k_] := t[n, k] = If[k < 0 || k > n, 0, If[n == 0, 1, Sum[Binomial[n-1, i]*t[n-1-i, k-1]*Ceiling[i!/2], {i, 0, n-k}]]]; Table[Table[t[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Dec 18 2013, translated from Maple *)
    rows = 10;
    t = Table[If[n<2, 1, n!/2], {n, 0, rows}];
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 0, rows}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)
  • Sage
    # uses[bell_matrix from A264428]
    bell_matrix(lambda n: factorial(n)//2 if n>=2 else 1, 8)

A306343 Number T(n,k) of defective (binary) heaps on n elements with k defects; triangle T(n,k), n>=0, 0<=k<=max(0,n-1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 3, 9, 9, 3, 8, 28, 48, 28, 8, 20, 90, 250, 250, 90, 20, 80, 360, 1200, 1760, 1200, 360, 80, 210, 1526, 5922, 12502, 12502, 5922, 1526, 210, 896, 7616, 34160, 82880, 111776, 82880, 34160, 7616, 896, 3360, 32460, 185460, 576060, 1017060, 1017060, 576060, 185460, 32460, 3360
Offset: 0

Views

Author

Alois P. Heinz, Feb 08 2019

Keywords

Comments

A defect in a defective heap is a parent-child pair not having the correct order.
T(n,k) is the number of permutations p of [n] having exactly k indices i in {1,...,n} such that p(i) > p(floor(i/2)).
T(n,0) counts perfect (binary) heaps on n elements (A056971).

Examples

			T(4,0) = 3: 4231, 4312, 4321.
T(4,1) = 9: 2413, 3124, 3214, 3241, 3412, 3421, 4123, 4132, 4213.
T(4,2) = 9: 1342, 1423, 1432, 2134, 2143, 2314, 2341, 2431, 3142.
T(4,3) = 3: 1234, 1243, 1324.
(The examples use max-heaps.)
Triangle T(n,k) begins:
    1;
    1;
    1,    1;
    2,    2,     2;
    3,    9,     9,     3;
    8,   28,    48,    28,      8;
   20,   90,   250,   250,     90,    20;
   80,  360,  1200,  1760,   1200,   360,    80;
  210, 1526,  5922, 12502,  12502,  5922,  1526,  210;
  896, 7616, 34160, 82880, 111776, 82880, 34160, 7616, 896;
  ...
		

Crossrefs

Row sums give A000142.
T(n,floor(n/2)) gives A306356.

Programs

  • Maple
    b:= proc(u, o) option remember; local n, g, l; n:= u+o;
          if n=0 then 1
        else g:= 2^ilog2(n); l:= min(g-1, n-g/2); expand(
             add(add(binomial(j-1, i)*binomial(n-j, l-i)*
             b(i, l-i)*b(j-1-i, n-l-j+i), i=0..min(j-1, l)), j=1..u)+
             add(add(binomial(j-1, i)*binomial(n-j, l-i)*
             b(l-i, i)*b(n-l-j+i, j-1-i), i=0..min(j-1, l)), j=1..o)*x)
          fi
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0)):
    seq(T(n), n=0..10);
  • Mathematica
    b[u_, o_] := b[u, o] = Module[{n = u + o, g, l},
         If[n == 0, 1, g := 2^Floor@Log[2, n]; l = Min[g-1, n-g/2]; Expand[
         Sum[Sum[ Binomial[j-1, i]* Binomial[n-j, l-i]*b[i, l-i]*
         b[j-1-i, n-l-j+i], {i, 0, Min[j-1, l]}], {j, 1, u}]+
         Sum[Sum[Binomial[j - 1, i]* Binomial[n-j, l-i]*b[l-i, i]*
         b[n-l-j+i, j-1-i], {i, 0, Min[j-1, l]}], {j, 1, o}]*x]]];
    T[n_] := CoefficientList[b[n, 0], x];
    T /@ Range[0, 10] // Flatten (* Jean-François Alcover, Feb 17 2021, after Alois P. Heinz *)

Formula

T(n,k) = T(n,n-1-k) for n > 0.
Sum_{k>=0} k * T(n,k) = A001286(n).
Sum_{k>=0} (k+1) * T(n,k) = A001710(n-1) for n > 0.
Sum_{k>=0} (k+2) * T(n,k) = A038720(n) for n > 0.
Sum_{k>=0} (k+3) * T(n,k) = A229039(n) for n > 0.
Sum_{k>=0} (k+4) * T(n,k) = A230056(n) for n > 0.

A111530 Row 3 of table A111528.

Original entry on oeis.org

1, 1, 5, 33, 261, 2361, 23805, 263313, 3161781, 40907241, 567074925, 8385483393, 131787520101, 2194406578521, 38605941817245, 715814473193073, 13956039627763221, 285509132504621001, 6116719419966460365
Offset: 0

Views

Author

Paul D. Hanna, Aug 06 2005

Keywords

Examples

			(1/3)*(log(1 + 3*x + 12*x^2 + 60*x^3 + ... + (n+2)!/2!)*x^n + ...)
= x + 5/2*x^2 + 33/3*x^3 + 261/4*x^4 + 2361/5*x^5 + ...
		

Crossrefs

Cf: A111528 (table), A003319 (row 1), A111529 (row 2), A111531 (row 4), A111532 (row 5), A111533 (row 6), A111534 (diagonal).

Programs

  • Mathematica
    T[n_, k_] := T[n, k] = Which[n<0 || k<0, 0, k==0 || k==1, 1, n==0, k!, True, (T[n-1, k+1]-T[n-1, k])/n - Sum[T[n, j]*T[n-1, k-j], {j, 1, k-1}]];
    a[n_] := T[3, n];
    Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Aug 09 2018 *)
  • PARI
    {a(n)=if(n<0,0,if(n==0,1, (n/3)*polcoeff(log(sum(m=0,n,(m+2)!/2!*x^m) + x*O(x^n)),n)))} \\ fixed by Vaclav Kotesovec, Jul 27 2015

Formula

G.f.: (1/3)*log(Sum_{n>=0} (n+2)!/2!*x^n) = Sum_{n>=1} a(n)*x^n/n.
G.f.: A(x) = 1/(1 + 3*x - 4*x/(1 + 4*x - 5*x/(1 + 5*x - ... (continued fraction).
a(n) = Sum_{k=0..n} 3^(n-k)*A089949(n,k). - Philippe Deléham, Oct 16 2006
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(k+1)/(x*(k-1/2) + 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: W(0), where W(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1+R)/( x*(k+1+R) - 1/W(k+1) ))); R=3 is Row R of table A111528 (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) ~ n! * n^3/6 * (1 - 4/n^2 - 15/n^3 - 99/n^4 - 882/n^5 - 9531/n^6 - 119493/n^7 - 1693008/n^8 - 26638245/n^9 - 459682047/n^10). - Vaclav Kotesovec, Jul 27 2015
From Peter Bala, May 24 2017: (Start)
O.g.f. A(x) = ( Sum_{n >= 0} (n+3)!/3!*x^n ) / ( Sum_{n >= 0} (n+2)!/2!*x^n ).
1/(1 - 3*x*A(x)) = Sum_{n >= 0} (n+2)!/2!*x^n. Cf. A001710.
A(x)/(1 - 3*x*A(x)) = Sum_{n >= 0} (n+3)!/3!*x^n. Cf. A001715.
A(x) satisfies the Riccati equation x^2*A'(x) + 3*x*A^2(x) - (1 + 2*x)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - 6*x/(1 - ... - n*x/(1 - (n+3)*x/(1 - ... ))))))))), by Stokes 1982.
A(x) = 1/(1 + 3*x - 4*x/(1 - x/(1 - 5*x/(1 - 2*x/(1 - 6*x/(1 - 3*x/(1 - ... - (n + 3)*x/(1 - n*x/(1 - ... ))))))))). (End)

A144739 7-factorial numbers A114799(7*n+3): Partial products of A017017(k) = 7*k+3, a(0) = 1.

Original entry on oeis.org

1, 3, 30, 510, 12240, 379440, 14418720, 648842400, 33739804800, 1990648483200, 131382799891200, 9590944392057600, 767275551364608000, 66752972968720896000, 6274779459059764224000, 633752725365036186624000, 68445294339423908155392000, 7871208849033749437870080000
Offset: 0

Views

Author

Philippe Deléham, Sep 20 2008

Keywords

Examples

			a(0)=1, a(1)=3, a(2)=3*10=30, a(3)=3*10*17=510, a(4)=3*10*17*24=12240, ...
		

Crossrefs

Programs

  • GAP
    List([0..20], n-> Product([0..n-1], k-> 7*k+3) ); # G. C. Greubel, Aug 19 2019
  • Magma
    [ 1 ] cat [ &*[ (7*k+3): k in [0..n] ]: n in [0..20] ]; // Klaus Brockhaus, Nov 10 2008
    
  • Maple
    a:= n-> product(7*j+3, j=0..n-1); seq(a(n), n=0..20); # G. C. Greubel, Aug 19 2019
  • Mathematica
    Table[7^n*Pochhammer[3/7, n], {n,0,20}] (* G. C. Greubel, Aug 19 2019 *)
  • PARI
    a(n)=prod(i=1,n,7*i-4) \\ Charles R Greathouse IV, Jul 02 2013
    
  • Sage
    [product(7*k+3 for k in (0..n-1)) for n in (0..20)] # G. C. Greubel, Aug 19 2019
    

Formula

a(n) = Sum_{k=0..n} A132393(n,k)*3^k*7^(n-k).
G.f.: 1/(1-3*x/(1-7*x/(1-10*x/(1-14*x/(1-17*x/(1-21*x/(1-24*x/(1-... (continued fraction). - Philippe Deléham, Jan 08 2012
a(n) = (-4)^n*Sum_{k=0..n} (7/4)^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
From Ilya Gutkovskiy, Mar 23 2017: (Start)
E.g.f.: 1/(1 - 7*x)^(3/7).
a(n) ~ sqrt(2*Pi)*7^n*n^n/(exp(n)*n^(1/14)*Gamma(3/7)). (End)
a(n) = A114799(7*n-4). - M. F. Hasler, Feb 23 2018
D-finite with recurrence: a(n) +(-7*n+4)*a(n-1)=0. - R. J. Mathar, Feb 21 2020
Sum_{n>=0} 1/a(n) = 1 + (e/7^4)^(1/7)*(Gamma(3/7) - Gamma(3/7, 1/7)). - Amiram Eldar, Dec 19 2022

A162990 Triangle of polynomial coefficients related to 3F2([1,n+1,n+1],[n+2,n+2],z).

Original entry on oeis.org

4, 36, 9, 576, 144, 64, 14400, 3600, 1600, 900, 518400, 129600, 57600, 32400, 20736, 25401600, 6350400, 2822400, 1587600, 1016064, 705600, 1625702400, 406425600, 180633600, 101606400, 65028096, 45158400, 33177600, 131681894400
Offset: 1

Views

Author

Johannes W. Meijer, Jul 21 2009

Keywords

Comments

The hypergeometric function 3F2([1,n+1,n+1],[n+2,n+2],z) = (n+1)^2*Li2(z)/z^(n+1) - MN(z;n)/(n!^2*z^n) for n >= 1, with Li2(z) the dilogarithm. The polynomial coefficients of MN(z;n) lead to the triangle given above.
We observe that 3F2([1,1,1],[2,2],z) = Li2(z)/z and that 3F2([1,0,0],[1,1],z) = 1.
The generating function for the EG1[3,n] coefficients of the EG1 matrix, see A162005, is GFEG1(z;m=2) = 1/(1-z)*(3*zeta(3)/2-2*z*log(2)* 3F2([1,1,1],[2,2],z) + sum((2^(1-2*n)* factorial(2*n-1)*z^(n+1)*3F2([1,n+1,n+1],[n+2,n+2],z))/(factorial(n+1)^2), n=1..infinity)).
The zeros of the MN(z;n) polynomials for larger values of n get ever closer to the unit circle and resemble the full moon, hence we propose to call the MN(z;n) the moon polynomials.

Examples

			The first few rows of the triangle are:
  [4]
  [36, 9]
  [576, 144, 64]
  [14400, 3600, 1600, 900]
The first few MN(z;n) polynomials are:
  MN(z;n=1) = 4
  MN(z;n=2) = 36 + 9*z
  MN(z;n=3) = 576 + 144*z + 64*z^2
  MN(z;n=4) = 14400 + 3600*z + 1600*z^2 + 900*z^3
		

References

  • Lewin, L., Polylogarithms and Associated Functions. New York, North-Holland, 1981.

Crossrefs

A162995 is a scaled version of this triangle.
A001819(n)*(n+1)^2 equals the row sums for n>=1.
A162991 and A162992 equal the first and second right hand columns.
A001048, A052747, A052759, A052778, A052794 are related to the square root of the first five right hand columns.
A001044, A162993 and A162994 equal the first, second and third left hand columns.
A000142, A001710, A002301, A133799, A129923, A001715 are related to the square root of the first six left hand columns.
A027451(n+1) equals the denominators of M(z, n)/(n!)^2.
A129202(n)/A129203(n) = (n+1)^2*Li2(z=1)/(Pi^2) = (n+1)^2/6.
Cf. A002378 and A035287.

Programs

  • Maple
    a := proc(n, m): ((n+1)!/m)^2 end: seq(seq(a(n, m), m=1..n), n=1..7); # Johannes W. Meijer, revised Nov 29 2012
  • Mathematica
    Table[((n+1)!/m)^2, {n, 10}, {m, n}] (* Paolo Xausa, Mar 30 2024 *)

Formula

a(n,m) = ((n+1)!/m)^2 for n >= 1 and 1 <= m <= n.

A051431 a(n) = (n+10)!/10!.

Original entry on oeis.org

1, 11, 132, 1716, 24024, 360360, 5765760, 98017920, 1764322560, 33522128640, 670442572800, 14079294028800, 309744468633600, 7124122778572800, 170978946685747200, 4274473667143680000, 111136315345735680000, 3000680514334863360000, 84019054401376174080000
Offset: 0

Views

Author

Keywords

Comments

The p=10 member of the p-family of sequences {(n+p-1)!/p!}, n >= 1.
The asymptotic expansion of the higher-order exponential integral E(x,m=1,n=11) ~ exp(-x)/x*(1 - 11/x + 132/x^2 - 1716/x^3 + 24024/x^4 - 360360/x^5 + 5765760/x^6 - ...) leads to the sequence given above. See A163931 and A130534 for more information. - Johannes W. Meijer, Oct 20 2009

Crossrefs

Programs

  • Haskell
    a051431 = (flip div 3628800) . a000142 . (+ 10)
    -- Reinhard Zumkeller, Aug 31 2014
  • Magma
    [Factorial(n+10)/3628800: n in [0..25]]; // Vincenzo Librandi, Jul 20 2011
    
  • Mathematica
    a[n_] := (n + 10)!/10!; Array[a, 20, 0] (* Amiram Eldar, Jan 15 2023 *)

Formula

a(n) = (n+10)!/10!
E.g.f.: 1/(1-x)^11.
a(n) = A173333(n+10,10). - Reinhard Zumkeller, Feb 19 2010
a(n) = A245334(n+10,n) / 11. - Reinhard Zumkeller, Aug 31 2014
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=0} 1/a(n) = 3628800*e - 9864100.
Sum_{n>=0} (-1)^n/a(n) = 3628800/e - 1334960. (End)

A051578 a(n) = (2*n+4)!!/4!!, related to A000165 (even double factorials).

Original entry on oeis.org

1, 6, 48, 480, 5760, 80640, 1290240, 23224320, 464486400, 10218700800, 245248819200, 6376469299200, 178541140377600, 5356234211328000, 171399494762496000, 5827582821924864000, 209792981589295104000, 7972133300393213952000, 318885332015728558080000
Offset: 0

Views

Author

Keywords

Comments

Row m=4 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.

Crossrefs

Cf. A000165, A001147(n+1), A002866(n+1), A051577 (rows m=0..3), A051579, A051580, A051581, A051582, A051583.
Cf. A052587 (essentially the same).

Programs

  • GAP
    List([0..20], n-> 2^(n-1)*Factorial(n+2) ); # G. C. Greubel, Nov 11 2019
  • Magma
    [2^(n-1)*Factorial(n+2): n in [0..20]]; // G. C. Greubel, Nov 11 2019
    
  • Maple
    a:= proc(n) option remember; `if`(n=0, 1, 2*(n+2)*a(n-1)) end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Apr 29 2019
    seq(2^(n-1)*(n+2)!, n=0..20); # G. C. Greubel, Nov 11 2019
  • Mathematica
    Table[2^(n-1)(n+2)!, {n,0,20}] (* Jean-François Alcover, Oct 05 2019 *)
    Table[(2n+4)!!/8,{n,0,20}] (* Harvey P. Dale, Apr 06 2023 *)
  • PARI
    vector(21, n, 2^(n-2)*(n+1)! ) \\ G. C. Greubel, Nov 11 2019
    
  • PARI
    apply( {A051578(n)=(n+2)!<<(n-1)}, [0..18]) \\ M. F. Hasler, Nov 10 2024
    
  • Sage
    [2^(n-1)*factorial(n+2) for n in (0..20)] # G. C. Greubel, Nov 11 2019
    

Formula

a(n) = (2*n+4)!!/4!!.
E.g.f.: 1/(1-2*x)^3.
a(n) ~ 2^(-1/2)*Pi^(1/2)*n^(5/2)*2^n*e^-n*n^n*{1 + 37/12*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
a(n) = (n+2)!*2^(n-1). - Zerinvary Lajos, Sep 23 2006. [corrected by Gary Detlefs, Apr 29 2019]
a(n) = 2^n*A001710(n+2). - R. J. Mathar, Feb 22 2008
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 6)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 6*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 6*x/(1 - 2*x/(1 - 8*x/(1 - 4*x/(1 - 10*x/(1 - 6*x/(1 - ... - (2*n + 4)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 6*x/(1 - 8*x/(1 - 2*x/(1 - 10*x/(1 - 4*x/(1 - 12*x/(1 - 6*x/(1 - ... - (2*n + 6)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 8*sqrt(e) - 12.
Sum_{n>=0} (-1)^n/a(n) = 8/sqrt(e) - 4. (End)
a(n) = A052587(n+2) for n > 0. - M. F. Hasler, Nov 10 2024
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