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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A001563 a(n) = n*n! = (n+1)! - n!.

Original entry on oeis.org

0, 1, 4, 18, 96, 600, 4320, 35280, 322560, 3265920, 36288000, 439084800, 5748019200, 80951270400, 1220496076800, 19615115520000, 334764638208000, 6046686277632000, 115242726703104000, 2311256907767808000, 48658040163532800000, 1072909785605898240000
Offset: 0

Views

Author

Keywords

Comments

A similar sequence, with the initial 0 replaced by 1, namely A094258, is defined by the recurrence a(2) = 1, a(n) = a(n-1)*(n-1)^2/(n-2). - Andrey Ryshevich (ryshevich(AT)notes.idlab.net), May 21 2002
Denominators in power series expansion of E_1(x) + gamma + log(x), x > 0. - Michael Somos, Dec 11 2002
If all the permutations of any length k are arranged in lexicographic order, the n-th term in this sequence (n <= k) gives the index of the permutation that rotates the last n elements one position to the right. E.g., there are 24 permutations of 4 items. In lexicographic order they are (0,1,2,3), (0,1,3,2), (0,2,1,3), ... (3,2,0,1), (3,2,1,0). Permutation 0 is (0,1,2,3), which rotates the last 1 element, i.e., it makes no change. Permutation 1 is (0,1,3,2), which rotates the last 2 elements. Permutation 4 is (0,3,1,2), which rotates the last 3 elements. Permutation 18 is (3,0,1,2), which rotates the last 4 elements. The same numbers work for permutations of any length. - Henry H. Rich (glasss(AT)bellsouth.net), Sep 27 2003
Stirling transform of a(n+1)=[4,18,96,600,...] is A083140(n+1)=[4,22,154,...]. - Michael Somos, Mar 04 2004
From Michael Somos, Apr 27 2012: (Start)
Stirling transform of a(n)=[1,4,18,96,...] is A069321(n)=[1,5,31,233,...].
Partial sums of a(n)=[0,1,4,18,...] is A033312(n+1)=[0,1,5,23,...].
Binomial transform of A000166(n+1)=[0,1,2,9,...] is a(n)=[0,1,4,18,...].
Binomial transform of A000255(n+1)=[1,3,11,53,...] is a(n+1)=[1,4,18,96,...].
Binomial transform of a(n)=[0,1,4,18,...] is A093964(n)=[0,1,6,33,...].
Partial sums of A001564(n)=[1,3,4,14,...] is a(n+1)=[1,4,18,96,...].
(End)
Number of small descents in all permutations of [n+1]. A small descent in a permutation (x_1,x_2,...,x_n) is a position i such that x_i - x_(i+1) =1. Example: a(2)=4 because there are 4 small descents in the permutations 123, 13\2, 2\13, 231, 312, 3\2\1 of {1,2,3} (shown by \). a(n)=Sum_{k=0..n-1}k*A123513(n,k). - Emeric Deutsch, Oct 02 2006
Equivalently, in the notation of David, Kendall and Barton, p. 263, this is the total number of consecutive ascending pairs in all permutations on n+1 letters (cf. A010027). - N. J. A. Sloane, Apr 12 2014
a(n-1) is the number of permutations of n in which n is not fixed; equivalently, the number of permutations of the positive integers in which n is the largest element that is not fixed. - Franklin T. Adams-Watters, Nov 29 2006
Number of factors in a determinant when writing down all multiplication permutations. - Mats Granvik, Sep 12 2008
a(n) is also the sum of the positions of the left-to-right maxima in all permutations of [n]. Example: a(3)=18 because the positions of the left-to-right maxima in the permutations 123,132,213,231,312 and 321 of [3] are 123, 12, 13, 12, 1 and 1, respectively and 1+2+3+1+2+1+3+1+2+1+1=18. - Emeric Deutsch, Sep 21 2008
Equals eigensequence of triangle A002024 ("n appears n times"). - Gary W. Adamson, Dec 29 2008
Preface the series with another 1: (1, 1, 4, 18, ...); then the next term = dot product of the latter with "n occurs n times". Example: 96 = (1, 1, 4, 8) dot (4, 4, 4, 4) = (4 + 4 + 16 + 72). - Gary W. Adamson, Apr 17 2009
Row lengths of the triangle in A030298. - Reinhard Zumkeller, Mar 29 2012
a(n) is also the number of minimum (n-)distinguishing labelings of the star graph S_{n+1} on n+1 nodes. - Eric W. Weisstein, Oct 14 2014
When the numbers denote finite permutations (as row numbers of A055089) these are the circular shifts to the right, i.e., a(n) is the permutation with the cycle notation (0 1 ... n-1 n). Compare array A051683 for circular shifts to the right in a broader sense. Compare sequence A007489 for circular shifts to the left. - Tilman Piesk, Apr 29 2017
a(n-1) is the number of permutations on n elements with no cycles of length n. - Dennis P. Walsh, Oct 02 2017
The number of pandigital numbers in base n+1, such that each digit appears exactly once. For example, there are a(9) = 9*9! = 3265920 pandigital numbers in base 10 (A050278). - Amiram Eldar, Apr 13 2020

Examples

			E_1(x) + gamma + log(x) = x/1 - x^2/4 + x^3/18 - x^4/96 + ..., x > 0. - _Michael Somos_, Dec 11 2002
G.f. = x + 4*x^2 + 18*x^3 + 96*x^4 + 600*x^5 + 4320*x^6 + 35280*x^7 + 322560*x^8 + ...
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 218.
  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 336.
  • F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 263.
  • 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 37, equation 37:6:1 at page 354.

Crossrefs

Cf. A163931 (E(x,m,n)), A002775 (n^2*n!), A091363 (n^3*n!), A091364 (n^4*n!).
Cf. sequences with formula (n + k)*n! listed in A282466.
Row sums of A185105, A322383, A322384, A094485.

Programs

  • GAP
    List([0..20], n-> n*Factorial(n) ); # G. C. Greubel, Dec 30 2019
  • Haskell
    a001563 n = a001563_list !! n
    a001563_list = zipWith (-) (tail a000142_list) a000142_list
    -- Reinhard Zumkeller, Aug 05 2013
    
  • Magma
    [Factorial(n+1)-Factorial(n): n in [0..20]]; // Vincenzo Librandi, Aug 08 2014
    
  • Maple
    A001563 := n->n*n!;
  • Mathematica
    Table[n!n,{n,0,25}] (* Harvey P. Dale, Oct 03 2011 *)
  • PARI
    {a(n) = if( n<0, 0, n * n!)} /* Michael Somos, Dec 11 2002 */
    
  • Sage
    [n*factorial(n) for n in (0..20)] # G. C. Greubel, Dec 30 2019
    

Formula

From Michael Somos, Dec 11 2002: (Start)
E.g.f.: x / (1 - x)^2.
a(n) = -A021009(n, 1), n >= 0. (End)
The coefficient of y^(n-1) in expansion of (y+n!)^n, n >= 1, gives the sequence 1, 4, 18, 96, 600, 4320, 35280, ... - Artur Jasinski, Oct 22 2007
Integral representation as n-th moment of a function on a positive half-axis: a(n) = Integral_{x=0..oo} x^n*(x*(x-1)*exp(-x)) dx, for n>=0. This representation may not be unique. - Karol A. Penson, Sep 27 2001
a(0)=0, a(n) = n*a(n-1) + n!. - Benoit Cloitre, Feb 16 2003
a(0) = 0, a(n) = (n - 1) * (1 + Sum_{i=1..n-1} a(i)) for i > 0. - Gerald McGarvey, Jun 11 2004
Arises in the denominators of the following identities: Sum_{n>=1} 1/(n*(n+1)*(n+2)) = 1/4, Sum_{n>=1} 1/(n*(n+1)*(n+2)*(n+3)) = 1/18, Sum_{n>=1} 1/(n*(n+1)*(n+2)*(n+3)*(n+4)) = 1/96, etc. The general expression is Sum_{n>=k} 1/C(n, k) = k/(k-1). - Dick Boland, Jun 06 2005 [And the general expression implies that Sum_{n>=1} 1/(n*(n+1)*...*(n+k-1)) = (Sum_{n>=k} 1/C(n, k))/k! = 1/((k-1)*(k-1)!) = 1/a(k-1), k >= 2. - Jianing Song, May 07 2023]
a(n) = Sum_{m=2..n+1} |Stirling1(n+1, m)|, n >= 1 and a(0):=0, where Stirling1(n, m) = A048994(n, m), n >= m = 0.
a(n) = 1/(Sum_{k>=0} k!/(n+k+1)!), n > 0. - Vladeta Jovovic, Sep 13 2006
a(n) = Sum_{k=1..n(n+1)/2} k*A143946(n,k). - Emeric Deutsch, Sep 21 2008
The reciprocals of a(n) are the lead coefficients in the factored form of the polynomials obtained by summing the binomial coefficients with a fixed lower term up to n as the upper term, divided by the term index, for n >= 1: Sum_{k = i..n} C(k, i)/k = (1/a(n))*n*(n-1)*..*(n-i+1). The first few such polynomials are Sum_{k = 1..n} C(k, 1)/k = (1/1)*n, Sum_{k = 2..n} C(k, 2)/k = (1/4)*n*(n-1), Sum_{k = 3..n} C(k, 3)/k = (1/18)*n*(n-1)*(n-2), Sum_{k = 4..n} C(k, 4)/k = (1/96)*n*(n-1)*(n-2)*(n-3), etc. - Peter Breznay (breznayp(AT)uwgb.edu), Sep 28 2008
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)* Stirling2(j,i)*x^(k-j) then a(n) = (-1)^(n-1)*f(n,1,-2), (n >= 1). - Milan Janjic, Mar 01 2009
Sum_{n>=1} (-1)^(n+1)/a(n) = 0.796599599... [Jolley eq. 289]
G.f.: 2*x*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
G.f.: W(0)*(1-sqrt(x)) - 1, where W(k) = 1 + sqrt(x)/( 1 - sqrt(x)*(k+2)/(sqrt(x)*(k+2) + 1/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 18 2013
G.f.: T(0)/x - 1/x, where T(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 - (1-x-2*x*k)*(1-3*x-2*x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 17 2013
G.f.: Q(0)*(1-x)/x - 1/x, where Q(k) = 1 - x*(k+1)/( x*(k+1) - 1/(1 - x*(k+1)/( x*(k+1) - 1/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013
D-finite with recurrence: a(n) +(-n-2)*a(n-1) +(n-1)*a(n-2)=0. - R. J. Mathar, Jan 14 2020
a(n) = (-1)^(n+1)*(n+1)*Sum_{k=1..n} A094485(n,k)*Bernoulli(k). The inverse of the Worpitzky representation of the Bernoulli numbers. - Peter Luschny, May 28 2020
From Amiram Eldar, Aug 04 2020: (Start)
Sum_{n>=1} 1/a(n) = Ei(1) - gamma = A229837.
Sum_{n>=1} (-1)^(n+1)/a(n) = gamma - Ei(-1) = A239069. (End)
a(n) = Gamma(n)*A000290(n) for n > 0. - Jacob Szlachetka, Jan 01 2022

A270236 Triangle T(n,p) read by rows: the number of occurrences of p in the restricted growth functions of length n.

Original entry on oeis.org

1, 3, 1, 9, 5, 1, 30, 21, 8, 1, 112, 88, 47, 12, 1, 463, 387, 253, 97, 17, 1, 2095, 1816, 1345, 675, 184, 23, 1, 10279, 9123, 7304, 4418, 1641, 324, 30, 1, 54267, 48971, 41193, 28396, 13276, 3645, 536, 38, 1, 306298, 279855, 243152, 183615, 102244, 36223, 7473, 842, 47, 1
Offset: 1

Views

Author

R. J. Mathar, Mar 13 2016

Keywords

Comments

The RG functions used here are defined by f(1)=1, f(j) <= 1+max_{i
T(n,p) is the number of elements in the p-th subset in all set partitions of [n]. - Joerg Arndt, Mar 14 2016

Examples

			The two restricted growth functions of length 2 are (1,1) and (1,2). The 1 appears 3 times and the 2 once, so T(2,1)=3 and T(2,2)=1.
1;
3,1;
9,5,1;
30,21,8,1;
112,88,47,12,1;
463,387,253,97,17,1;
2095,1816,1345,675,184,23,1;
10279,9123,7304,4418,1641,324,30,1;
54267,48971,41193,28396,13276,3645,536,38,1;
306298,279855,243152,183615,102244,36223,7473,842,47,1;
1838320,1695902,1506521,1211936,770989,334751,90223,14303,1267,57,1;
11677867,10856879,9799547,8237223,5795889,2965654,995191,207186,25820, 1839,68,1;
		

Crossrefs

Cf. A070071 (row sums).
T(2n+1,n+1) gives A270529.

Programs

  • Maple
    b:= proc(n, m) option remember; `if`(n=0, [1, 0], add((p->
          [p[1], p[2]+p[1]*x^j])(b(n-1, max(m, j))), j=1..m+1))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, 0)[2]):
    seq(T(n), n=0..12);  # Alois P. Heinz, Mar 14 2016
  • Mathematica
    b[n_, m_] := b[n, m] = If[n == 0, {1, 0}, Sum[Function[p, {p[[1]], p[[2]] + p[[1]]*x^j}][b[n-1, Max[m, j]]], {j, 1, m+1}]];
    T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n, 0][[2]] ]; Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Apr 07 2016, after Alois P. Heinz *)

Formula

T(n,n) = 1.
Conjecture: T(n,n-1) = 2+n*(n-1)/2 for n>1.
Conjecture: T(n+1,n-1) = 2+n*(n+1)*(3*n^2-5*n+26)/24 for n>1.
Sum_{k=1..n} k * T(n,k) = A346772(n). - Alois P. Heinz, Aug 03 2021

A322383 Number T(n,k) of entries in the k-th cycles of all permutations of [n] when cycles are ordered by increasing lengths (and increasing smallest elements); triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 3, 1, 10, 7, 1, 45, 37, 13, 1, 236, 241, 101, 21, 1, 1505, 1661, 896, 226, 31, 1, 10914, 13301, 7967, 2612, 442, 43, 1, 90601, 117209, 78205, 29261, 6441, 785, 57, 1, 837304, 1150297, 827521, 346453, 88909, 14065, 1297, 73, 1, 8610129, 12314329, 9507454, 4338214, 1253104, 234646, 28006, 2026, 91, 1
Offset: 1

Author

Alois P. Heinz, Dec 05 2018

Keywords

Examples

			The 6 permutations of {1,2,3} are:
  (1)     (2)   (3)
  (1)     (2,3)
  (2)     (1,3)
  (3)     (1,2)
  (1,2,3)
  (1,3,2)
so there are 10 elements in the first cycles, 7 in the second cycles and only 1 in the third cycles.
Triangle T(n,k) begins:
      1;
      3,      1;
     10,      7,     1;
     45,     37,    13,     1;
    236,    241,   101,    21,    1;
   1505,   1661,   896,   226,   31,   1;
  10914,  13301,  7967,  2612,  442,  43,  1;
  90601, 117209, 78205, 29261, 6441, 785, 57, 1;
  ...
		

Crossrefs

Row sums give A001563.

Programs

  • Maple
    b:= proc(n, l) option remember; `if`(n=0, add(l[i]*
          x^i, i=1..nops(l)), add(binomial(n-1, j-1)*
          b(n-j, sort([l[], j]))*(j-1)!, j=1..n))
        end:
    T:= n-> (p-> (seq(coeff(p, x, i), i=1..n)))(b(n, [])):
    seq(T(n), n=1..12);
  • Mathematica
    b[n_, l_] := b[n, l] = If[n == 0, l.x^Range[Length[l]], Sum[Binomial[n - 1, j - 1] b[n - j, Sort[Append[l, j]]] (j - 1)!, {j, 1, n}]];
    T[n_] := Rest @ CoefficientList[b[n, {}], x];
    Array[T, 12] // Flatten (* Jean-François Alcover, Mar 03 2020, after Alois P. Heinz *)

A322384 Number T(n,k) of entries in the k-th cycles of all permutations of [n] when cycles are ordered by decreasing lengths (and increasing smallest elements); triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 3, 1, 13, 4, 1, 67, 21, 7, 1, 411, 131, 46, 11, 1, 2911, 950, 341, 101, 16, 1, 23563, 7694, 2871, 932, 197, 22, 1, 213543, 70343, 26797, 9185, 2311, 351, 29, 1, 2149927, 709015, 275353, 98317, 27568, 5119, 583, 37, 1, 23759791, 7867174, 3090544, 1141614, 343909, 73639, 10366, 916, 46, 1
Offset: 1

Author

Alois P. Heinz, Dec 05 2018

Keywords

Examples

			The 6 permutations of {1,2,3} are:
  (1)     (2) (3)
  (1,2)   (3)
  (1,3)   (2)
  (2,3)   (1)
  (1,2,3)
  (1,3,2)
so there are 13 elements in the first cycles, 4 in the second cycles and only 1 in the third cycles.
Triangle T(n,k) begins:
       1;
       3,     1;
      13,     4,     1;
      67,    21,     7,    1;
     411,   131,    46,   11,    1;
    2911,   950,   341,  101,   16,   1;
   23563,  7694,  2871,  932,  197,  22,  1;
  213543, 70343, 26797, 9185, 2311, 351, 29, 1;
  ...
		

Crossrefs

Row sums give A001563.
T(2n,n) gives A332928.

Programs

  • Maple
    b:= proc(n, l) option remember; `if`(n=0, add(l[-i]*
          x^i, i=1..nops(l)), add(binomial(n-1, j-1)*
          b(n-j, sort([l[], j]))*(j-1)!, j=1..n))
        end:
    T:= n-> (p-> (seq(coeff(p, x, i), i=1..n)))(b(n, [])):
    seq(T(n), n=1..12);
  • Mathematica
    b[n_, l_] := b[n, l] = If[n == 0, Sum[l[[-i]]*x^i, {i, 1, Length[l]}], Sum[Binomial[n-1, j-1]*b[n-j, Sort[Append[l, j]]]*(j-1)!, {j, 1, n}]];
    T[n_] := CoefficientList[b[n, {}], x] // Rest;
    Array[T, 12] // Flatten  (* Jean-François Alcover, Feb 26 2020, after Alois P. Heinz *)

A138772 Number of entries in the second cycles of all permutations of {1,2,...,n}; each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.

Original entry on oeis.org

0, 1, 5, 27, 168, 1200, 9720, 88200, 887040, 9797760, 117936000, 1536796800, 21555072000, 323805081600, 5187108326400, 88268019840000, 1590132031488000, 30233431388160000, 605024315191296000, 12711912992722944000, 279783730940313600000, 6437458713635389440000
Offset: 1

Author

Emeric Deutsch, Apr 10 2008

Keywords

Examples

			a(3) = 5 because the number of entries in the second cycles of (1)(2)(3), (1)(23), (132), (12)(3), (123) and (13)(2) is 1+2+0+1+0+1=5.
		

Crossrefs

Cf. A138771.
Column k=2 of A185105.

Programs

  • GAP
    List([1..30], n-> (n-1)*(n+2)*Factorial(n-1)/4); # G. C. Greubel, Jul 07 2019
  • Magma
    [(n-1)*(n+2)*Factorial(n-1)/4: n in [1..30]]; // G. C. Greubel, Jul 07 2019
    
  • Maple
    seq((1/4)*factorial(n-1)*(n-1)*(n+2), n = 1 .. 30);
  • Mathematica
    Table[((1/4)(n-1)!(n-1)(n+2)),{n,1,30}] (* Vincenzo Librandi, May 14 2012 *)
  • PARI
    vector(30, n, (n-1)*(n+2)*(n-1)!/4) \\ G. C. Greubel, Jul 07 2019
    
  • Sage
    [(n-1)*(n+2)*factorial(n-1)/4 for n in (1..30)] # G. C. Greubel, Jul 07 2019
    

Formula

a(n) = (1/4)*(n-1)!*(n-1)*(n+2).
a(n) = (n+1)*a(n-1) + (n-2)!.
a(n) = (n-1)*a(n-1) + n!/2.
a(n) = Sum_{k=0..n-1} k*A138771(n,k).
E.g.f. if offset 0: x*(2-x)/(2*(1-x)^3). Such e.g.f. computations resulted from e-mail exchange with Gary Detlefs. - Wolfdieter Lang, May 27 2010
a(n) = A000254(n-1) + A159324(n-1). - Gary Detlefs, May 13 2012
a(n) = n! * Sum_{i=1..n} (Sum_{j=1..i} (j/i)). - Pedro Caceres, Apr 19 2019
E.g.f.: ( x*(2-x)/(1-x)^2 + 2*log(1-x) )/4. - G. C. Greubel, Jul 07 2019
D-finite with recurrence a(n) +(-n-1)*a(n-1) -2*a(n-2) +2*(n-3)*a(n-3)=0. - R. J. Mathar, Jul 26 2022

A285439 Sum T(n,k) of the entries in the k-th cycles of all permutations of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 4, 2, 21, 12, 3, 132, 76, 28, 4, 960, 545, 235, 55, 5, 7920, 4422, 2064, 612, 96, 6, 73080, 40194, 19607, 6692, 1386, 154, 7, 745920, 405072, 202792, 75944, 18736, 2816, 232, 8, 8346240, 4484808, 2280834, 911637, 254061, 46422, 5256, 333, 9
Offset: 1

Author

Alois P. Heinz, Apr 19 2017

Keywords

Comments

Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.

Examples

			T(3,1) = 21 because the sum of the entries in the first cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 6+6+3+4+1+1 = 21.
Triangle T(n,k) begins:
       1;
       4,      2;
      21,     12,      3;
     132,     76,     28,     4;
     960,    545,    235,    55,     5;
    7920,   4422,   2064,   612,    96,    6;
   73080,  40194,  19607,  6692,  1386,  154,   7;
  745920, 405072, 202792, 75944, 18736, 2816, 232, 8;
  ...
		

Crossrefs

Columns k=1-2 give: A284816, A285489.
Row sums give A000142 * A000217 = A180119.
Main diagonal and first lower diagonal give: A000027, A006000 (for n>0).

Programs

  • Maple
    T:= proc(h) option remember; local b; b:=
          proc(n, l) option remember; `if`(n=0, [mul((i-1)!, i=l), 0],
            (p-> p+[0, (h-n+1)*p[1]*x^(nops(l)+1)])(b(n-1, [l[], 1]))+
             add((p-> p+[0, (h-n+1)*p[1]*x^j])(
             b(n-1, subsop(j=l[j]+1, l))), j=1..nops(l)))
          end: (p-> seq(coeff(p, x, i), i=1..n))(b(h, [])[2])
        end:
    seq(T(n), n=1..10);
  • Mathematica
    T[h_] := T[h] = Module[{b}, b[n_, l_] := b[n, l] = If[n == 0, {Product[(i - 1)!, {i, l}], 0}, # + {0, (h - n + 1)*#[[1]]*x^(Length[l] + 1)}&[b[n - 1, Append[l, 1]]] + Sum[# + {0, (h-n+1)*#[[1]]*x^j}&[b[n - 1, ReplacePart[ l, j -> l[[j]] + 1]]], {j, 1, Length[l]}]]; Table[Coefficient[#, x, i], {i, 1, n}]&[b[h, {}][[2]]]];
    Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, May 25 2018, translated from Maple *)

Formula

Sum_{k=1..n} k * T(n,k) = n^2 * n! = A002775(n).

A284816 Sum of entries in the first cycles of all permutations of [n].

Original entry on oeis.org

1, 4, 21, 132, 960, 7920, 73080, 745920, 8346240, 101606400, 1337212800, 18920563200, 286442956800, 4620449433600, 79114299264000, 1433211107328000, 27387931963392000, 550604138692608000, 11617107089043456000, 256671161862635520000, 5926549291918295040000
Offset: 1

Author

Alois P. Heinz, Apr 15 2017

Keywords

Comments

Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.
Also, the number of colorings of n+1 given balls, two thereof identical, using n given colors (each color is used). - Ivaylo Kortezov, Jan 27 2024

Examples

			a(3) = 21 because the sum of the entries in the first cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 6+6+3+4+1+1 = 21.
		

Crossrefs

Column k=1 of A285439.

Programs

  • Maple
    a:= n-> n!*(n*(n+1)-(n-1)*(n+2)/2)/2:
    seq(a(n), n=1..25);
    # second Maple program:
    a:= proc(n) option remember; `if`(n<2, n,
           (n^2+n+2)*n*a(n-1)/(n^2-n+2))
        end:
    seq(a(n), n=1..25);

Formula

a(n) = n!*(n*(n+1) - (n-1)*(n+2)/2)/2.
E.g.f.: -x*(x^2-2*x+2)/(2*(x-1)^3).
a(n) = (n^2+n+2)*n*a(n-1)/(n^2-n+2) for n > 1, a(n) = n for n < 2.
a(n) = n*A006595(n-1). - Ivaylo Kortezov, Feb 02 2024

A285231 Number of entries in the third cycles of all permutations of [n].

Original entry on oeis.org

1, 8, 59, 463, 3978, 37566, 388728, 4385592, 53653680, 708126480, 10034314560, 152001161280, 2451821339520, 41964428419200, 759698874547200, 14505012898790400, 291323663566387200, 6140173922952652800, 135515391451776000000, 3125606951427609600000
Offset: 3

Author

Alois P. Heinz, Apr 15 2017

Keywords

Comments

Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.

Crossrefs

Column k=3 of A185105.
Cf. A159324.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<4, [0$3, 1][n+1],
          ((2*n^3-7*n^2+3*n+4)*a(n-1)-(n-2)^3*(n+1)*a(n-2))/(n*(n-3)))
        end:
    seq(a(n), n=0..25);
  • Mathematica
    a[3] = 1; a[4] = 8;
    a[n_] := a[n] = ((2n^3-7n^2+3n+4) a[n-1] - (n-2)^3 (n+1) a[n-2])/(n(n-3));
    Table[a[n], {n, 3, 25}] (* Jean-François Alcover, May 30 2018, from Maple *)

Formula

a(n) = A185105(n,3) = A159324(n-1)/2.
a(n) ~ n!*n/8. - Vaclav Kotesovec, Apr 25 2017

A285239 Number of entries in the n-th cycles of all permutations of [2n].

Original entry on oeis.org

3, 27, 463, 12217, 441383, 20338679, 1141073295, 75473055841, 5748862140283, 495446888127507, 47648289796265871, 5057570671179281161, 587173799850231036207, 74005641366738437835967, 10062023872139208015273375, 1467822867614662009540883265
Offset: 1

Author

Alois P. Heinz, Apr 15 2017

Keywords

Comments

Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.
All terms are odd.

Crossrefs

Cf. A185105.

Programs

  • Maple
    b:= proc(n, i) option remember; expand(`if`(n=0, 1,
          add((p-> p+`if`(i=1, coeff(p, x, 0)*j*x, 0))((j-1)!
          *b(n-j, max(0, i-1)))*binomial(n-1, j-1), j=1..n)))
        end:
    a:= n-> coeff(b(2*n, n), x, 1):
    seq(a(n), n=1..20);
  • Mathematica
    b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, Sum[Function[p, p + If[i == 1, Coefficient[p, x, 0] j x, 0]][(j - 1)! b[n - j, Max[0, i - 1]]] Binomial[ n - 1, j - 1], {j, 1, n}]]];
    a[n_] := Coefficient[b[2n, n], x, 1];
    Array[a, 20] (* Jean-François Alcover, Jun 01 2018, from Maple *)

Formula

a(n) = A185105(2n,n).
a(n) ~ 2^(3*n-1) * c^(2*n) * n^(n - 1/2) / (sqrt(Pi*(c-1)) * (2*c-1)^n * exp(n)), where c = -LambertW(-1,-exp(-1/2)/2) = 1.7564312086261696769827376166... - Vaclav Kotesovec, Apr 15 2017, updated Mar 10 2020

A285232 Number of entries in the fourth cycles of all permutations of [n].

Original entry on oeis.org

1, 12, 119, 1177, 12217, 135302, 1606446, 20450052, 278604252, 4051377504, 62702222112, 1029832270848, 17899305402240, 328353314855040, 6341705227082880, 128655706986282240, 2735782096305749760, 60855815849067648000, 1413487524282196608000
Offset: 4

Author

Alois P. Heinz, Apr 15 2017

Keywords

Comments

Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.

Crossrefs

Column k=4 of A185105.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<5, [0$4, 1][n+1],
          ((n-2)*(3*n^2-13*n+6)*a(n-1)-(3*n^4-26*n^3+76*n^2
          -81*n+16)*a(n-2)+(n-3)^4*n*a(n-3))/((n-1)*(n-4)))
        end:
    seq(a(n), n=4..30);
  • Mathematica
    a[3]=0; a[4]=1; a[5]=12; a[n_] := a[n] = ((n-2)(3n^2 - 13n + 6) a[n-1] - ( 3n^4 - 26n^3 + 76n^2 - 81n + 16)a[n-2]+(n-3)^4 n a[n-3])/((n-1)(n-4));
    Table[a[n], {n, 4, 30}] (* Jean-François Alcover, Jun 01 2018, from Maple *)

Formula

a(n) = A185105(n,4).
a(n) ~ n!*n/16. - Vaclav Kotesovec, Apr 25 2017
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