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.

Showing 1-4 of 4 results.

A094587 Triangle of permutation coefficients arranged with 1's on the diagonal. Also, triangle of permutations on n letters with exactly k+1 cycles and with the first k+1 letters in separate cycles.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 6, 3, 1, 24, 24, 12, 4, 1, 120, 120, 60, 20, 5, 1, 720, 720, 360, 120, 30, 6, 1, 5040, 5040, 2520, 840, 210, 42, 7, 1, 40320, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 3628800
Offset: 0

Views

Author

Paul Barry, May 13 2004

Keywords

Comments

Also, table of Pochhammer sequences read by antidiagonals (see Rudolph-Lilith, 2015). - N. J. A. Sloane, Mar 31 2016
Reverse of A008279. Row sums are A000522. Diagonal sums are A003470. Rows of inverse matrix begin {1}, {-1,1}, {0,-2,1}, {0,0,-3,1}, {0,0,0,-4,1} ... The signed lower triangular matrix (-1)^(n+k)n!/k! has as row sums the signed rencontres numbers Sum_{k=0..n} (-1)^(n+k)n!/k!. (See A000166). It has matrix inverse 1 1,1 0,2,1 0,0,3,1 0,0,0,4,1,...
Exponential Riordan array [1/(1-x),x]; column k has e.g.f. x^k/(1-x). - Paul Barry, Mar 27 2007
From Tom Copeland, Nov 01 2007: (Start)
T is the umbral extension of n!*Lag[n,(.)!*Lag[.,x,-1],0] = (1-D)^(-1) x^n = (-1)^n * n! * Lag(n,x,-1-n) = Sum_{j=0..n} binomial(n,j) * j! * x^(n-j) = Sum_{j=0..n} (n!/j!) x^j. The inverse operator is A132013 with generalizations discussed in A132014.
b = T*a can be characterized several ways in terms of a(n) and b(n) or their o.g.f.'s A(x) and B(x).
1) b(n) = n! Lag[n,(.)!*Lag[.,a(.),-1],0], umbrally,
2) b(n) = (-1)^n n! Lag(n,a(.),-1-n)
3) b(n) = Sum_{j=0..n} (n!/j!) a(j)
4) B(x) = (1-xDx)^(-1) A(x), formally
5) B(x) = Sum_{j=0,1,...} (xDx)^j A(x)
6) B(x) = Sum_{j=0,1,...} x^j * D^j * x^j A(x)
7) B(x) = Sum_{j=0,1,...} j! * x^j * L(j,-:xD:,0) A(x) where Lag(n,x,m) are the Laguerre polynomials of order m, D the derivative w.r.t. x and (:xD:)^j = x^j * D^j. Truncating the operator series at the j = n term gives an o.g.f. for b(0) through b(n).
c = (0!,1!,2!,3!,4!,...) is the sequence associated to T under the list partition transform and the associated operations described in A133314 so T(n,k) = binomial(n,k)*c(n-k). The reciprocal sequence is d = (1,-1,0,0,0,...). (End)
From Peter Bala, Jul 10 2008: (Start)
This array is the particular case P(1,1) of the generalized Pascal triangle P(a,b), a lower unit triangular matrix, shown below:
n\k|0.....................1...............2.......3......4
----------------------------------------------------------
0..|1.....................................................
1..|a....................1................................
2..|a(a+b)...............2a..............1................
3..|a(a+b)(a+2b).........3a(a+b).........3a........1......
4..|a(a+b)(a+2b)(a+3b)...4a(a+b)(a+2b)...6a(a+b)...4a....1
...
The entries A(n,k) of this array satisfy the recursion A(n,k) = (a+b*(n-k-1))*A(n-1,k) + A(n-1,k-1), which reduces to the Pascal formula when a = 1, b = 0.
Various cases are recorded in the database, including: P(1,0) = Pascal's triangle A007318, P(2,0) = A038207, P(3,0) = A027465, P(2,1) = A132159, P(1,3) = A136215 and P(2,3) = A136216.
When b <> 0 the array P(a,b) has e.g.f. exp(x*y)/(1-b*y)^(a/b) = 1 + (a+x)*y + (a*(a+b)+2a*x+x^2)*y^2/2! + (a*(a+b)*(a+2b) + 3a*(a+b)*x + 3a*x^2+x^3)*y^3/3! + ...; the array P(a,0) has e.g.f. exp((x+a)*y).
We have the matrix identities P(a,b)*P(a',b) = P(a+a',b); P(a,b)^-1 = P(-a,b).
An analog of the binomial expansion for the row entries of P(a,b) has been proved by [Echi]. Introduce a (generally noncommutative and nonassociative) product ** on the ring of polynomials in two variables by defining F(x,y)**G(x,y) = F(x,y)G(x,y) + by^2*d/dy(G(x,y)).
Define the iterated product F^(n)(x,y) of a polynomial F(x,y) by setting F^(1) = F(x,y) and F^(n)(x,y) = F(x,y)**F^(n-1)(x,y) for n >= 2. Then (x+a*y)^(n) = x^n + C(n,1)*a*x^(n-1)*y + C(n,2)*a*(a+b)*x^(n-2)*y^2 + ... + C(n,n)*a*(a+b)*(a+2b)*...*(a+(n-1)b)*y^n. (End)
(n+1) * n-th row = reversal of triangle A068424: (1; 2,2; 6,6,3; ...) - Gary W. Adamson, May 03 2009
Let G(m, k, p) = (-p)^k*Product_{j=0..k-1}(j - m - 1/p) and T(n,k,p) = G(n-1,n-k,p) then T(n, k, 1) is this sequence, T(n, k, 2) = A112292(n, k) and T(n, k, 3) = A136214. - Peter Luschny, Jun 01 2009, revised Jun 18 2019
The higher order exponential integrals E(x,m,n) are defined in A163931. For a discussion of the asymptotic expansions of the 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 - ...) see A130534. The asymptotic expansion of E(x,m=1,n) leads for n >= 1 to the left hand columns of the triangle given above. Triangle A165674 is generated by the asymptotic expansions of E(x,m=2,n). - Johannes W. Meijer, Oct 07 2009
T(n,k) = n!/k! = number of permutations of [n+1] with exactly k+1 cycles and with elements 1,2,...,k+1 in separate cycles. See link and example below. - Dennis P. Walsh, Jan 24 2011
T(n,k) is the number of n permutations that leave some size k subset of {1,2,...,n} fixed. Sum_{k=0..n}(-1)^k*T(n,k) = A000166(n) (the derangements). - Geoffrey Critzer, Dec 11 2011
T(n,k) = A162995(n-1,k-1), 2 <= k <= n; T(n,k) = A173333(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012
The row polynomials form an Appell sequence. The matrix is a special case of a group of general matrices sketched in A132382. - Tom Copeland, Dec 03 2013
For interpretations in terms of colored necklaces, see A213936 and A173333. - Tom Copeland, Aug 18 2016
See A008279 for a relation of this entry to the e.g.f.s enumerating the faces of permutahedra and stellahedra. - Tom Copeland, Nov 14 2016
Also, T(n,k) is the number of ways to arrange n-k nonattacking rooks on the n X (n-k) chessboard. - Andrey Zabolotskiy, Dec 16 2016
The infinitesimal generator of this triangle is the generalized exponential Riordan array [-log(1-x), x] and equals the unsigned version of A238363. - Peter Bala, Feb 13 2017
Formulas for exponential and power series infinitesimal generators for this triangle T are given in Copeland's 2012 and 2014 formulas as T = unsigned exp[(I-A238385)] = 1/(I - A132440), where I is the identity matrix. - Tom Copeland, Jul 03 2017
If A(0) = 1/(1-x), and A(n) = d/dx(A(n-1)), then A(n) = n!/(1-x)^(n+1) = Sum_{k>=0} (n+k)!/k!*x^k = Sum_{k>=0} T(n+k, k)*x^k. - Michael Somos, Sep 19 2021

Examples

			Rows begin {1}, {1,1}, {2,2,1}, {6,6,3,1}, ...
For n=3 and k=1, T(3,1)=6 since there are exactly 6 permutations of {1,2,3,4} with exactly 2 cycles and with 1 and 2 in separate cycles. The permutations are (1)(2 3 4), (1)(2 4 3), (1 3)(2 4), (1 4)(2 3), (1 3 4)(2), and (1 4 3)(2). - _Dennis P. Walsh_, Jan 24 2011
Triangle begins:
     1,
     1,    1,
     2,    2,    1,
     6,    6,    3,    1,
    24,   24,   12,    4,    1,
   120,  120,   60,   20,    5,    1,
   720,  720,  360,  120,   30,    6,    1,
  5040, 5040, 2520,  840,  210,   42,    7,    1
The production matrix is:
      1,     1,
      1,     1,     1,
      2,     2,     1,    1,
      6,     6,     3,    1,    1,
     24,    24,    12,    4,    1,   1,
    120,   120,    60,   20,    5,   1,   1,
    720,   720,   360,  120,   30,   6,   1,   1,
   5040,  5040,  2520,  840,  210,  42,   7,   1,   1,
  40320, 40320, 20160, 6720, 1680, 336,  56,   8,   1,   1
which is the exponential Riordan array A094587, or [1/(1-x),x], with an extra superdiagonal of 1's.
Inverse begins:
   1,
  -1,  1,
   0, -2,  1,
   0,  0, -3,  1,
   0,  0,  0, -4,  1,
   0,  0,  0,  0, -5,  1,
   0,  0,  0,  0,  0, -6,  1,
   0,  0,  0,  0,  0,  0, -7,  1
		

Crossrefs

Programs

  • Haskell
    a094587 n k = a094587_tabl !! n !! k
    a094587_row n = a094587_tabl !! n
    a094587_tabl = map fst $ iterate f ([1], 1)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
    
  • Maple
    T := proc(n, m): n!/m! end: seq(seq(T(n, m), m=0..n), n=0..9);  # Johannes W. Meijer, Oct 07 2009, revised Nov 25 2012
    # Alternative: Note that if you leave out 'abs' you get A021009.
    T := proc(n, k) option remember; if n = 0 and k = 0 then 1 elif k < 0 or k > n then 0 else abs((n + k)*T(n-1, k) - T(n-1, k-1)) fi end: #  Peter Luschny, Dec 30 2021
  • Mathematica
    Flatten[Table[Table[n!/k!, {k,0,n}], {n,0,10}]] (* Geoffrey Critzer, Dec 11 2011 *)
  • Sage
    def A094587_row(n): return (factorial(n)*exp(x).taylor(x,0,n)).list()
    for n in (0..7): print(A094587_row(n)) # Peter Luschny, Sep 28 2017

Formula

T(n, k) = n!/k! if n >= k >= 0, otherwise 0.
T(n, k) = Sum_{i=k..n} |S1(n+1, i+1)*S2(i, k)| * (-1)^i, with S1, S2 the Stirling numbers.
T(n,k) = (n-k)*T(n-1,k) + T(n-1,k-1). E.g.f.: exp(x*y)/(1-y) = 1 + (1+x)*y + (2+2*x+x^2)*y^2/2! + (6+6*x+3*x^2+x^3)*y^3/3!+ ... . - Peter Bala, Jul 10 2008
A094587 = 1 / ((-1)*A129184 * A127648 + I), I = Identity matrix. - Gary W. Adamson, May 03 2009
From Johannes W. Meijer, Oct 07 2009: (Start)
The o.g.f. of right hand column k is Gf(z;k) = (k-1)!/(1-z)^k, k => 1.
The recurrence relations of the right hand columns lead to Pascal's triangle A007318. (End)
Let f(x) = (1/x)*exp(-x). The n-th row polynomial is R(n,x) = (-x)^n/f(x)*(d/dx)^n(f(x)), and satisfies the recurrence equation R(n+1,x) = (x+n+1)*R(n,x)-x*R'(n,x). Cf. A132159. - Peter Bala, Oct 28 2011
A padded shifted version of this lower triangular matrix with zeros in the first column and row except for a one in the diagonal position is given by integral(t=0 to t=infinity) exp[-t(I-P)] = 1/(I-P) = I + P^2 + P^3 + ... where P is the infinitesimal generator matrix A218234 and I the identity matrix. The non-padded version is given by P replaced by A132440. - Tom Copeland, Oct 25 2012
From Peter Bala, Aug 28 2013: (Start)
The row polynomials R(n,x) form a Sheffer sequence of polynomials with associated delta operator equal to d/dx. Thus d/dx(R(n,x)) = n*R(n-1,x). The Sheffer identity is R(n,x + y) = Sum_{k=0..n} binomial(n,k)*y^(n-k)*R(k,x).
Let P(n,x) = Product_{k=0..n-1} (x + k) denote the rising factorial polynomial sequence with the convention that P(0,x) = 1. Then this is triangle of connection constants when expressing the basis polynomials P(n,x + 1) in terms of the basis P(n,x). For example, row 3 is (6, 6, 3, 1) so P(3,x + 1) = (x + 1)*(x + 2)*(x + 3) = 6 + 6*x + 3*x*(x + 1) + x*(x + 1)*(x + 2). (End)
From Tom Copeland, Apr 21 & 26, and Aug 13 2014: (Start)
T-I = M = -A021009*A132440*A021009 with e.g.f. y*exp(x*y)/(1-y). Cf. A132440. Dividing the n-th row of M by n generates the (n-1)th row of T.
T = 1/(I - A132440) = {2*I - exp[(A238385-I)]}^(-1) = unsigned exp[(I-A238385)] = exp[A000670(.)*(A238385-I)] = , umbrally, where I = identity matrix.
The e.g.f. is exp(x*y)/(1-y), so the row polynomials form an Appell sequence with lowering operator d/dx and raising operator x + 1/(1-D).
With L(n,m,x)= Laguerre polynomials of order m, the row polynomials are (-1)^n*n!*L(n,-1-n,x) = (-1)^n*(-1!/(-1-n)!)*K(-n,-1-n+1,x) = n!* K(-n,-n,x) where K is Kummer's confluent hypergeometric function (as a limit of n+s as s tends to zero).
Operationally, (-1)^n*n!*L(n,-1-n,-:xD:) = (-1)^n*x^(n+1)*:Dx:^n*x^(-1-n) = (-1)^n*x*:xD:^n*x^(-1) = (-1)^n*n!*binomial(xD-1,n) = n!*K(-n,-n,-:xD:) where :AB:^n = A^n*B^n for any two operators. Cf. A235706 and A132159.
The n-th row of signed M has the coefficients of d[(-:xD:)^n]/d(:Dx:)= f[d/d(-:xD:)](-:xD:)^n with f(y)=y/(y-1), :Dx:^n= n!L(n,0,-:xD:), and (-:xD:)^n = n!L(n,0,:Dx:). M has the coefficients of [D/(1-D)]x^n. (End)
From Tom Copeland, Nov 18 2015: (Start)
Coefficients of the row polynomials of the e.g.f. Sum_{n>=0} P_n(b1,b2,..,bn;t) x^n/n! = e^(P.(..;t) x) = e^(xt) / (1-b.x) = (1 + b1 x + b2 x^2 + b3 x^3 + ...) e^(xt) = 1 + (b1 + t) x + (2 b2 + 2 b1 t + t^2) x^2/2! + (6 b3 + 6 b2 t + 3 b1 t^2 + t^3) x^3/3! + ... , with lowering operator L = d/dt, i.e., L P_n(..;t) = n * P_(n-1)(..;t), and raising operator R = t + d[log(1 + b1 D + b2 D^2 + ...)]/dD = t - Sum_{n>=1} F(n,b1,..,bn) D^(n-1), i.e., R P_n(..,;t) = P_(n+1)(..;t), where D = d/dt and F(n,b1,..,bn) are the Faber polynomials of A263916.
Also P_n(b1,..,bn;t) = CIP_n(t-F(1,b1),-F(2,b1,b2),..,-F(n,b1,..,bn)), the cycle index polynomials A036039.
(End)
The raising operator R = x + 1/(1-D) = x + 1 + D + D^2 + ... in matrix form acting on an o.g.f. (formal power series) is the transpose of the production matrix M below. The linear term x is the diagonal of ones after transposition. The other transposed diagonals come from D^m x^n = n! / (n-m)! x^(n-m). Then P(n,x) = (1,x,x^2,..) M^n (1,0,0,..)^T is a matrix representation of R P(n-1,x) = P(n,x). - Tom Copeland, Aug 17 2016
The row polynomials have e.g.f. e^(xt)/(1-t) = exp(t*q.(x)), umbrally. With p_n(x) the row polynomials of A132013, q_n(x) = v_n(p.(u.(x))), umbrally, where u_n(x) = (-1)^n v_n(-x) = (-1)^n Lah_n(x), the Lah polynomials with e.g.f. exp[x*t/(t-1)]. This has the matrix form [T] = [q] = [v]*[p]*[u]. Conversely, p_n(x) = u_n (q.(v.(x))). - Tom Copeland, Nov 10 2016
From the Appell sequence formalism, 1/(1-b.D) t^n = P_n(b1,b2,..,bn;t), the generalized row polynomials noted in the Nov 18 2015 formulas, consistent with the 2007 comments. - Tom Copeland, Nov 22 2016
From Peter Bala, Feb 18 2017: (Start)
G.f.: Sum_{n >= 1} (n*x)^(n-1)/(1 + (n - t)*x)^n = 1 + (1 + t)*x + (2 + 2*t + t^2)*x^2 + ....
n-th row polynomial R(n,t) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^k*(x + k - t)^(n-k) = Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*(x + k)^(n-k)*(x + k + t)^k, for arbitrary x. The particular case of the latter sum when x = 0 and t = 1 is identity 10.35 in Gould, Vol.4. (End)
Rodrigues-type formula for the row polynomials: R(n, x) = -exp(x)*Int(exp(-x)* x^n, x), for n >= 0. Recurrence: R(n, x) = x^n + n*R(n-1, x), for n >= 1, and R(0, x) = 1. d/dx(R(n, x)) = R(n, x) - x^n, for n >= 0 (compare with the formula from Peter Bala, Aug 28 2013). - Wolfdieter Lang, Dec 23 2019
T(n, k) = Sum_{i=0..n-k} A048994(n-k, i) * n^i for 0 <= k <= n. - Werner Schulte, Jul 26 2022

Extensions

Edited by Johannes W. Meijer, Oct 07 2009
New description from Dennis P. Walsh, Jan 24 2011

A173333 Triangle read by rows: T(n, k) = n! / k!, 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 3, 1, 24, 12, 4, 1, 120, 60, 20, 5, 1, 720, 360, 120, 30, 6, 1, 5040, 2520, 840, 210, 42, 7, 1, 40320, 20160, 6720, 1680, 336, 56, 8, 1, 362880, 181440, 60480, 15120, 3024, 504, 72, 9, 1, 3628800, 1814400, 604800, 151200, 30240, 5040, 720, 90, 10, 1
Offset: 1

Views

Author

Reinhard Zumkeller, Feb 19 2010

Keywords

Comments

From Wolfdieter Lang, Jun 27 2012: (Start)
T(n-1,k), k=1,...,n-1, gives the number of representative necklaces with n beads (C_N symmetry) of n+1-k distinct colors, say c[1],c[2],...,c[n-k+1], corresponding to the color signature determined by the partition k,1^(n-k) of n. The representative necklaces have k beads of color c[1]. E.g., n=4, k=2: partition 2,1,1, color signature (parts as exponents) c[1]c[1]c[2]c[3], 3=T(3,2) necklaces (write j for color c[j]): cyclic(1123), cyclic(1132) and cyclic(1213). See A212359 for the numbers for general partitions or color signatures. (End)

Examples

			Triangle starts:
n\k      1       2      3      4     5    6   7  8  9 10 ...
1        1
2        2       1
3        6       3      1
4       24      12      4      1
5      120      60     20      5     1
6      720     360    120     30     6    1
7     5040    2520    840    210    42    7   1
8    40320   20160   6720   1680   336   56   8  1
9   362880  181440  60480  15120  3024  504  72  9  1
10 3628800 1814400 604800 151200 30240 5040 720 90 10  1
... - _Wolfdieter Lang_, Jun 27 2012
		

Crossrefs

Row sums give A002627.
Central terms give A006963:
T(2*n-1,n) = A006963(n+1).
T(2*n,n) = A001813(n).
T(2*n,n+1) = A001761(n).
1 < k <= n: T(n,k) = T(n,k-1) / k.
1 <= k <= n: T(n+1,k) = A119741(n,n-k+1).
1 <= k <= n: T(n+1,k+1) = A162995(n,k).
T(n,1) = A000142(n).
T(n,2) = A001710(n) for n>1.
T(n,3) = A001715(n) for n>2.
T(n,4) = A001720(n) for n>3.
T(n,5) = A001725(n) for n>4.
T(n,6) = A001730(n) for n>5.
T(n,7) = A049388(n-7) for n>6.
T(n,8) = A049389(n-8) for n>7.
T(n,9) = A049398(n-9) for n>8.
T(n,10) = A051431(n) for n>9.
T(n,n-7) = A159083(n+1) for n>7.
T(n,n-6) = A053625(n+1) for n>6.
T(n,n-5) = A052787(n) for n>5.
T(n,n-4) = A052762(n) for n>4.
T(n,n-3) = A007531(n) for n>3.
T(n,n-2) = A002378(n-1) for n>2.
T(n,n-1) = A000027(n) for n>1.
T(n,n) = A000012(n).

Programs

  • Haskell
    a173333 n k = a173333_tabl !! (n-1) !! (k-1)
    a173333_row n = a173333_tabl !! (n-1)
    a173333_tabl = map fst $ iterate f ([1], 2)
       where f (row, i) = (map (* i) row ++ [1], i + 1)
    -- Reinhard Zumkeller, Jul 04 2012
  • Mathematica
    Table[n!/k!, {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 01 2019 *)

Formula

E.g.f.: (exp(x*y) - 1)/(x*(1 - y)). - Olivier Gérard, Jul 07 2011
T(n,k) = A094587(n,k), 1 <= k <= n. - Reinhard Zumkeller, Jul 05 2012

A056542 a(n) = n*a(n-1) + 1, a(1) = 0.

Original entry on oeis.org

0, 1, 4, 17, 86, 517, 3620, 28961, 260650, 2606501, 28671512, 344058145, 4472755886, 62618582405, 939278736076, 15028459777217, 255483816212690, 4598708691828421, 87375465144740000, 1747509302894800001, 36697695360790800022, 807349297937397600485
Offset: 1

Views

Author

Henry Bottomley, Jun 20 2000

Keywords

Comments

For n >= 2 also operation count to create all permutations of n distinct elements using Algorithm L (lexicographic permutation generation) from Knuth's The Art of Computer Programming, Vol. 4, chapter 7.2.1.2. Sequence gives number of loop repetitions of the j search loop in step L2. - Hugo Pfoertner, Feb 06 2003
More directly: sum over all permutations of length n-1 of the product of the length of the first increasing run by the value of the first position. The recurrence follows from this definition. - Olivier Gérard, Jul 07 2011
This sequence shares divisibility properties with A000522; each of the primes in A072456 divide only a finite number of terms of this sequence. - T. D. Noe, Jul 07 2005
This sequence also represents the number of subdeterminant evaluations when calculation a determinant by Laplace recursive method. - Reinhard Muehlfeld, Sep 14 2010
Also, a(n) equals the number of non-isomorphic directed graphs of n+1 vertices with 1 component, where each vertex has exactly one outgoing edge, excluding loops and cycle graphs. - Stephen Dunn, Nov 30 2019

Examples

			a(4) = 4*a(3) + 1 = 4*4 + 1 = 17.
Permutations of order 3 .. Length of first run * First position
123..3*1
132..2*1
213..1*2
231..2*2
312..1*3
321..1*3
a(4) = 3+2+2+4+3+3 = 17. - _Olivier Gérard_, Jul 07 2011
		

References

  • D. E. Knuth: The Art of Computer Programming, Volume 4, Combinatorial Algorithms, Volume 4A, Enumeration and Backtracking. Pre-fascicle 2B, A draft of section 7.2.1.2: Generating all permutations. Available online; see link.

Crossrefs

Cf. A079751 (same recursion formula, but starting from a(3)=0), A038155, A038156, A080047, A080048, A080049.
Equals the row sums of A162995 triangle (n>=2). - Johannes W. Meijer, Jul 21 2009
Cf. A070213 (indices of primes).

Programs

  • Haskell
    a056542 n = a056542_list !! (n-1)
    a056542_list = 0 : map (+ 1) (zipWith (*) [2..] a056542_list)
    -- Reinhard Zumkeller, Mar 24 2013
    
  • Magma
    [n le 2 select n-1 else n*Self(n-1)+1: n in [1..20]]; // Bruno Berselli, Dec 13 2013
  • Mathematica
    tmp=0; Join[{tmp}, Table[tmp=n*tmp+1, {n, 2, 100}]] (* T. D. Noe, Jul 12 2005 *)
    FoldList[ #1*#2 + 1 &, 0, Range[2, 21]] (* Robert G. Wilson v, Oct 11 2005 *)

Formula

a(n) = floor((e-2)*n!).
a(n) = A002627(n) - n!.
a(n) = A000522(n) - 2*n!.
a(n) = n! - A056543(n).
a(n) = (n-1)*(a(n-1) + a(n-2)) + 2, n > 2. - Gary Detlefs, Jun 22 2010
1/(e - 2) = 2! - 2!/(1*4) - 3!/(4*17) - 4!/(17*86) - 5!/(86*517) - ... (see A002627 and A185108). - Peter Bala, Oct 09 2013
E.g.f.: (exp(x) - 1 - x) / (1 - x). - Ilya Gutkovskiy, Jun 26 2022

Extensions

More terms from James Sellers, Jul 04 2000

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

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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.
Showing 1-4 of 4 results.