A163626 -id:A163626 - cp's OEIS frontend

A048163 a(n) = Sum_{k=1..n} ((k-1)!)^2*Stirling2(n,k)^2.

1, 2, 14, 230, 6902, 329462, 22934774, 2193664790, 276054834902, 44222780245622, 8787513806478134, 2121181056663291350, 611373265185174628502, 207391326125004608457782, 81791647413265571604175094, 37109390748309009878392597910, 19192672725746588045912535407702

Offset: 1

N. J. A. Sloane, R. K. Guy

nonn

a(n) is also the number of max-closed relations on an ordered n-element domain (see the paper by Jeavons and Cooper, 1995). - Don Knuth, Feb 12 2024

			1
1 + 1 = 2
1 + 9 + 4 = 14
1 + 49 + 144 + 36 = 230
1 + 225 + 2500 + 3600 + 576 = 6902
... - _Philippe Deléham_, May 30 2015

Lovasz, L. and Vesztergombi, K.; Restricted permutations and Stirling numbers. Combinatorics (Proc. Fifth Hungarian Colloq., Keszthely, 1976), Vol. II, pp. 731-738, Colloq. Math. Soc. Janos Bolyai, 18, North-Holland, Amsterdam-New York, 1978.
K. Vesztergombi, Permutations with restriction of middle strength, Stud. Sci. Math. Hungar., 9 (1974), 181-185.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Chad Brewbaker, A combinatorial interpretation of the poly-Bernoulli numbers and two Fermat analogues, INTEGERS Vol. 8 (2008), #A02.
Peter G. Jeavons and Martin C. Cooper, Tractable constraints on ordered domains, Artificial Intelligence 79 (1995), 327-339.
Hyeong-Kwan Ju and Seunghyun Seo, Enumeration of (0,1)-matrices avoiding some 2 X 2 matrices, Discrete Math., 312 (2012), 2473-2481.
Ken Kamano, Lonesum decomposable matrices, arXiv:1701.07157 [math.CO], 2017.
H.-K. Kim et al., Poly-Bernoulli numbers and lonesum matrices, arXiv:1103.4884 [math.CO], 2011.
Anatol N. Kirillov, On some quadratic algebras. I 1/2: Combinatorics of Dunkl and Gaudin elements, Schubert, Grothendieck, Fuss-Catalan, universal Tutte and reduced polynomials, SIGMA, Symmetry Integrability Geom. Methods Appl. 12, Paper 002, 172 p. (2016).

Main diagonal of array A099594.

Cf. A220181, A266695, A306209.

Table[Sum[((k-1)!)^2*StirlingS2[n,k]^2,{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Jun 21 2013 *)

a(n)=if(n<1, 0, polcoeff(sum(m=1, n, m^(m-1)*(m-1)!*x^m/prod(k=1, m-1, 1+m*k*x+x*O(x^n))), n)) \\ Paul D. Hanna, Jan 05 2013
for(n=1,20,print1(a(n),", "))

Stirling2(n, k)=n!*polcoeff(((exp(x+x*O(x^n))-1)^k)/k!, n)
a(n)=sum(k=1,n,(-1)^(n-k)*k^(n-1)*(k-1)!*Stirling2(n-1, k-1))
for(n=1, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Jan 06 2013

a(n) = sum(k=1, n, (k-1)!^2*stirling(n,k,2)^2); \\ Michel Marcus, Jun 22 2018

E.g.f. (with offset 0): Sum((1-exp(-(m+1)*z))^m, m=0..oo)
O.g.f.: Sum_{n>=1} n^(n-1) * (n-1)! * x^n / Product_{k=1..n-1} (1 - n*k*x). - Paul D. Hanna, Jan 05 2013
Limit n->infinity (a(n)/n!)^(1/n)/n = 1/(exp(1)*(log(2))^2) = 0.7656928576... . - Vaclav Kotesovec, Jun 21 2013
a(n) ~ 2*sqrt(Pi) * n^(2*n-3/2) / (sqrt(1-log(2)) * exp(2*n) * (log(2))^(2*n-1)). - Vaclav Kotesovec, May 13 2014
a(n+1) = Sum_{k = 0..n} A163626(n,k)^2. - Philippe Deléham, May 30 2015
a(n) = A306209(2n-2,n-1). - Alois P. Heinz, Feb 01 2019
a(n) = A266695(2n-2). - Alois P. Heinz, Apr 17 2024

Entry revised by N. J. A. Sloane, Jul 05 2012

A371761 Array read by antidiagonals: The number of parades with n girls and k boys that begin with a girl and end with a boy.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 5, 1, 0, 0, 1, 13, 13, 1, 0, 0, 1, 29, 73, 29, 1, 0, 0, 1, 61, 301, 301, 61, 1, 0, 0, 1, 125, 1081, 2069, 1081, 125, 1, 0, 0, 1, 253, 3613, 11581, 11581, 3613, 253, 1, 0, 0, 1, 509, 11593, 57749, 95401, 57749, 11593, 509, 1, 0

Offset: 0

Peter Luschny, Apr 05 2024

The name derives from a proposition by Donald Knuth, who describes the setup of a "girls and boys parade" as follows: "There are m girls {g_1, ..., g_m} and n boys {b_1, ..., b_n}, where g_i is younger than g_{i+1} and b_j is younger than b_{j+1}, but we know nothing about the relative ages of g_i and b_j. In how many ways can they all line up in a sequence such that no girl is directly preceded by an older girl and no boy is directly preceded by an older boy?" [Our notation: A <- D, n <- m, k <- n].
In A344920, the Worpitzky transform is defined as a sequence-to-sequence transformation WT := A -> B, where B(n) = Sum_{k=0..n} A163626(n, k)*A(k). (If A(n) = 1/(n + 1) then B(n) are the Bernoulli numbers (with B(1) = 1/2.)) The rows of the array are the Worpitzky transforms of the powers up to the sign (-1)^k.
The array rows are recursively generated by applying the Akiyama-Tanigawa algorithm to the powers (see the Python implementation below). In this way the array becomes the image of A004248 under the AT-transformation when applied to the rows of A004248. This makes the array closely linked to A344499, which is generated in the same way, but applied to the columns of A004248.
Conjecture: Row n + 1 is row 2^n in table A136301, where a probabilistic interpretation is given (see the link to Parsonnet's paper below).

			Array starts:
[0] 1, 0,   0,     0,      0,       0,        0,         0,          0, ...
[1] 0, 1,   1,     1,      1,       1,        1,         1,          1, ...
[2] 0, 1,   5,    13,     29,      61,      125,       253,        509, ...
[3] 0, 1,  13,    73,    301,    1081,     3613,     11593,      36301, ...
[4] 0, 1,  29,   301,   2069,   11581,    57749,    268381,    1191989, ...
[5] 0, 1,  61,  1081,  11581,   95401,   673261,   4306681,   25794781, ...
[6] 0, 1, 125,  3613,  57749,  673261,  6487445,  55213453,  431525429, ...
[7] 0, 1, 253, 11593, 268381, 4306681, 55213453, 610093513, 6077248381, ...
.
Seen as triangle T(n, k) = A(n - k, k):
  [0] 1;
  [1] 0, 0;
  [2] 0, 1,  0;
  [3] 0, 1,  1,   0;
  [4] 0, 1,  5,   1,   0;
  [5] 0, 1, 13,  13,   1,  0;
  [6] 0, 1, 29,  73,  29,  1, 0;
  [7] 0, 1, 61, 301, 301, 61, 1, 0;
.
A(n, k) as sum of powers:
  A(2, k) =  -3+   2*2^k;
  A(3, k) =   7-  12*2^k+    6*3^k;
  A(4, k) = -15+  50*2^k-   60*3^k+   24*4^k;
  A(5, k) =  31- 180*2^k+  390*3^k-  360*4^k+  120*5^k;
  A(6, k) = -63+ 602*2^k- 2100*3^k+ 3360*4^k- 2520*5^k+  720*6^k;
  A(7, k) = 127-1932*2^k+10206*3^k-25200*4^k+31920*5^k-20160*6^k+5040*7^k;

Peter Luschny, Table of n, a(n) for antidiagonals 0..140.
Beáta Bényi, A Bijection for the Boolean Numbers of Ferrers Graphs, Graphs and Combinatorics (2022) Vol. 38, No. 10.
Beáta Bényi and Péter Hajnal, Combinatorial properties of poly-Bernoulli relatives, arXiv preprint arXiv:1602.08684 [math.CO], 2016. See D_{n, k}.
Don Knuth, Parades and poly-Bernoulli bijections, Mar 31 2024. See (16.2).
Brian Parsonnet, Probability of Derangements.

Variant: A272644.
Rows include: A344920 (row 2, signed), A006230 (row 3).
Row sums of triangle (n>=2): A297195, alternating row sums: A226158.
Diagonal of array: A048144.
Cf. A004248, A075263, A028246, A173018, A136301, A371762, A136126 (similar), A099594, A344499.

egf := 1/(exp(w) + exp(z) - exp(w + z)): serw := n -> series(egf, w, n + 1):
# Returns row n (>= 0) with length len (> 0):
R := n -> len -> local k;
seq(k!*coeff(series(n!*coeff(serw(n), w, n), z, len), z, k), k = 0..len - 1):
seq(lprint(R(n)(9)), n = 0..7);
# Explicit with Stirling2 :
A := (n, k) -> local j; add(j!^2*Stirling2(n, j)*Stirling2(k, j), j = 0..min(n, k)): seq(lprint(seq(A(n, k), k = 0..8)), n = 0..7);
# Using the unsigned Worpitzky transform.
WT := (a, len) -> local n, k;
seq(add((-1)^(n - k)*k!*Stirling2(n + 1, k + 1)*a(k), k=0..n), n = 0..len-1):
Arow := n -> WT(x -> x^n, 8): seq(lprint(Arow(n)), n = 0..8);
# Two recurrences:
A := proc(n, k) option remember; local j; if k = 0 then return k^n fi;
add(binomial(n, j)*(A(n-j, k-1) + A(n-j+1, k-1)), j = 1..n) end:
A := proc(n, k) option remember; local j; if n = 0 then 0^k else
add(binomial(k + `if`(j=0,0,1), j+1)*A(n-1, k-j), j = 0..k) fi end:

(* Using the unsigned Worpitzky transform. *)
Unprotect[Power]; Power[0, 0] = 1;
W[n_, k_] := (-1)^(n - k) k! StirlingS2[n + 1, k + 1];
WT[a_, len_] := Table[Sum[W[n, k] a[k], {k, 0, n}], {n, 0, len-1}];
A371761row[n_, len_] := WT[#^n &, len];
Table[A371761row[n, 9], {n, 0, 8}] // MatrixForm
(* Row n >= 1 by linear recurrence: *)
RowByLRec[n_, len_] := LinearRecurrence[Table[-StirlingS1[n+1, n+1-k], {k, 1, n}],
A371761row[n, n+1], len]; Table[RowByLRec[n, 9], {n, 1, 8}] // MatrixForm

from functools import cache
from math import comb as binomial
@cache
def A(n, k):
    if n == 0: return int(k == 0)
    return sum(binomial(k + int(j > 0), j + 1) * A(n - 1, k - j)
           for j in range(k + 1))
for n in range(8): print([A(n, k) for k in range(8)])

# The Akiyama-Tanigawa algorithm for powers generates the rows.
def ATPowList(n, len):
    A = [0] * len
    R = [0] * len
    for k in range(len):
        R[k] = k**n   # Changing this to R[k] = (n + 1)**k generates A344499.
        for j in range(k, 0, -1):
            R[j - 1] = j * (R[j] - R[j - 1])
        A[k] = R[0]
    return A
for n in range(8): print([n], ATPowList(n, 9))

def A371761(n, k): return sum((-1)^(j - k) * factorial(j) * stirling_number2(k + 1, j + 1) * j^n for j in range(k + 1))
for n in range(9): print([A371761(n, k) for k in range(8)])

A(n, k) = k! * [z^k] (n! * [w^n] 1/(exp(w) + exp(z) - exp(w + z))).
A(n, k) = k! * [w^k] (Sum_{j=0..n} A075263(n, n - j) * exp(j*w)).
A(n, k) = Sum_{j=0..k} (-1)^(j-k) * Stirling2(k + 1, j + 1) * j! * j^n.
A(n, k) = Sum_{j=0..min(n,k)} (j!)^2 * Stirling2(n, j) * Stirling2(k, j).
A(n, k) = Sum_{j=0..n} (-1)^(n-j)*A028246(n, j) * j^k; this is explicit:
A(n, k) = Sum_{j=0..n} Sum_{m=0..n} binomial(n-m, n-j) * Eulerian1(n, m) * j^k *(-1)^(n-j), where Eulerian1 = A173018.
A(n, k) = Sum_{j=0..k} binomial(k + [j>0], j+1)*A(n-1, k-j) for n > 0.
A(n, k) = Sum_{j=1..n} binomial(n, j)*(A(n-j, k-1) + A(n-j+1, k-1)) for n,k >= 1.
Row n (>=1) satisfies a linear recurrence:
A(n, k) = -Sum_{j=1..n} Stirling1(n + 1, n + 1 - j)*A(n, k - j) if k > n.
A(n, k) = [x^k] (Sum_{j=0..n} A371762(n, j)*x^j) / (Sum_{j=0..n} Stirling1(n + 1, n + 1 - j)*x^j).
A(n, k) = A(k, n). (From the symmetry of the bivariate exponential g.f.)
Let T(n, k) = A(n - k, k) and G(n) = Sum_{k=0..n} (-1)^k*T(n, k) the alternating row sums of the triangle. Then G(n) = (n + 2)*Euler(n + 1, 1) and as shifted Genocchi numbers G(n) = -2*(n + 2)*PolyLog(-n - 1, -1) = -A226158(n + 2).

A130850 Triangle read by rows, 0 <= k <= n, T(n,k) = Sum_{j=0..n} A(n,j)*binomial(n-j,k) where A(n,j) are the Eulerian numbers A173018.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 12, 7, 1, 24, 60, 50, 15, 1, 120, 360, 390, 180, 31, 1, 720, 2520, 3360, 2100, 602, 63, 1, 5040, 20160, 31920, 25200, 10206, 1932, 127, 1, 40320, 181440, 332640, 317520, 166824, 46620, 6050, 255, 1, 362880, 1814400, 3780000, 4233600, 2739240, 1020600, 204630, 18660, 511, 1

Offset: 0

Philippe Deléham, Aug 20 2007

Old name was: Triangle T(n,k), 0<=k<=n, read by rows given by [1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,2,0,3,0,4,0,5,0,6,0,...] where DELTA is the operator defined in A084938.
Vandervelde (2018) refers to this as the Worpitzky number triangle - N. J. A. Sloane, Mar 27 2018 [Named after the German mathematician Julius Daniel Theodor Worpitzky (1835-1895). - Amiram Eldar, Jun 24 2021]
Triangle given by A123125*A007318 (as infinite lower triangular matrices), A123125 = Euler's triangle, A007318 = Pascal's triangle; A007318*A123125 gives A046802.
Taylor coefficients of Eulerian polynomials centered at 1. - Louis Zulli, Nov 28 2015
A signed refinement is A263634. - Tom Copeland, Nov 14 2016
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of this entry (the Worpitsky triangle, A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020

			Triangle begins:
1
1      1
2      3       1
6      12      7       1
24     60      50      15      1
120    360     390     180     31      1
720    2520    3360    2100    602     63      1
5040   20160   31920   25200   10206   1932    127    1
40320  181440  332640  317520  166824  46620   6050   255   1
362880 1814400 3780000 4233600 2739240 1020600 204630 18660 511 1
...

G. C. Greubel, Table of n, a(n) for n = 0..1034 [a(438) and a(901) corrected by _Georg Fischer_, Nov 10 2021]
Nguyen-Huu-Bong, Some Combinatorial Properties of Summation Operators, J. Comb. Theory, Ser. A, Vo. 11, No. 3 (1971), pp. 213-221. See Table on page 214.
John K. Sikora, On Calculating the Coefficients of a Polynomial Generated Sequence Using the Worpitzky Number Triangles, arXiv:1806.00887 [math.NT], 2018.
Sam Vandervelde, The Worpitzky Numbers Revisited, Amer. Math. Monthly, Vol. 125, No. 3 (2018), pp. 198-206.

Essentially reverse of A028246.
Variants are A130850, A075263, A163626.
Cf. A000142, A001710, A005460, A005461, A005462, A005463, A005464, A263634.
Cf. A019538, A028246, A046802, A090582, A119879, A123125, A248727.

Table[(n-k)!*StirlingS2[n+1, n-k+1], {n, 0, 10}, {k, 0, n}] (* G. C. Greubel, Nov 15 2015 *)

t(n, k) = (n-k)!*stirling(n+1, n-k+1, 2);
tabl(nn) = for (n=0, 10, for (k=0, n, print1(t(n,k),", ")); print()); \\ Michel Marcus, Nov 16 2015

from sage.combinat.combinat import eulerian_number
def A130850(n, k):
    return add(eulerian_number(n, j)*binomial(n-j, k) for j in (0..n))
for n in (0..7): [A130850(n, k) for k in (0..n)] # Peter Luschny, May 21 2013

T(n,k) = (-1)^k*A075263(n,k).
T(n,k) = (n-k)!*A008278(n+1,k+1).
T(n,n-1) = 2^n - 1 for n > 0. - Derek Orr, Dec 31 2015
E.g.f.: x/(e^(-x*t)*(1+x)-1). - Tom Copeland, Nov 14 2016
Sum_{k=1..floor(n/2)} T(n,2k) = Sum_{k=0..floor(n/2)} T(n,2k+1) = A000670(n). - Jacob Sprittulla, Oct 03 2021

New name from Peter Luschny, May 21 2013

A210672 a(0)=1; thereafter a(n) = 2Sum_{k=1..n} binomial(2n,2k)a(n-k).

Original entry on oeis.org

1, 2, 26, 842, 50906, 4946282, 704888186, 138502957322, 35887046307866, 11855682722913962, 4863821092813045946, 2425978759725443056202, 1445750991051368583278426, 1014551931766896667943384042, 828063237870027116855857421306, 777768202388460616924079724057482

Offset: 0

N. J. A. Sloane, Mar 28 2012

Consider the sequence defined by a(0) = 1; thereafter a(n) = c*Sum_{k = 1..n} binomial(2n,2k)*a(n-k). For c = -3, -2, -1, 1, 2, 3, 4 this is A210676, A210657, A028296, A094088, A210672, A210674, A249939.
Exp( Sum_{n >= 1} a(n)*x^n/n) is the o.g.f. for A255929. - Peter Bala, Mar 13 2015
The Stirling-Bernoulli transform of Fibonacci(n+1) = 1, 1, 2, 3, 5, 8, 13, ... is 1, 0, 2, 0, 26, 0, 842, 0, 50906, 0, ... - Philippe Deléham, May 25 2015

G. C. Greubel, Table of n, a(n) for n = 0..220
Hacène Belbachir, Yahia Djemmada, On central Fubini-like numbers and polynomials, arXiv:1811.06734 [math.CO], 2018. See p. 4.

Cf. A028296, A094088, A210657, A210674, A210676, A255929, A241171.

f:=proc(n,k) option remember;  local i;
if n=0 then 1
else k*add(binomial(2*n,2*i)*f(n-i,k),i=1..floor(n)); fi; end;
g:=k->[seq(f(n,k),n=0..40)];
g(2);

nmax=20; Table[(CoefficientList[Series[1/(3-2*Cosh[x]), {x, 0, 2*nmax}], x] * Range[0, 2*nmax]!)[[2*n+1]], {n,0,nmax}] (* Vaclav Kotesovec, Mar 14 2015 *)

a(n) ~ 2*sqrt(Pi/5) * n^(2*n+1/2) / (exp(2*n) * (log((1+sqrt(5))/2))^(2*n+1)). - Vaclav Kotesovec, Mar 13 2015
E.g.f.: 1/(3-2*cosh(x)) (even coefficients). - Vaclav Kotesovec, Mar 14 2015
a(n) = Sum_{k = 0..2*n} A163626(2*n,k)*A000045(n+1). - Philippe Deléham, May 25 2015
a(n) = Sum_{k=0..n} A241171(n, k)*2^k. - Peter Luschny, Sep 03 2022

A228910 a(n) = 8^n - 77^n + 216^n - 355^n + 354^n - 213^n + 72^n - 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 5040, 181440, 3780000, 59875200, 801496080, 9574044480, 105398092800, 1091804313600, 10794490827120, 102896614941120, 952741767650400, 8617145057539200, 76461500619902160, 667855517349303360, 5757691363157764800, 49099453300298016000, 414884142077935345200

Offset: 0

Richard V. Scholtz, III, Sep 07 2013

Calculates the eighth column of coefficients with respect to the derivatives, d^n/dx^n(y), of the logistic equation when written as y=1/[1+exp(-x)].

Seiichi Manyama, Table of n, a(n) for n = 0..1107
Index entries for linear recurrences with constant coefficients, signature (36,-546,4536,-22449,67284,-118124,109584,-40320).

Cf. A008277, A049434.
The eighth column of results of A163626.
Cf. A000225, A028243, A028244, A028245, A032180, A228909.

[8^(n)-7*7^(n)+21*6^(n)-35*5^(n)+35*4^(n)-21*3^(n)+7*2^(n)-1: n in [0..30]]; // G. C. Greubel, Nov 19 2017

Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1 - y[x]); Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n - 1][y][x], x]; row[n_] := CoefficientList[ Derivative[n][y][x], y[x]] // Rest; Join[{0, 0, 0, 0, 0, 0, 0}, Table[ -row[n], {n, 7, 23}] [[All, 8]]] (* Jean-François Alcover, Dec 16 2014 *)
Table[7!*StirlingS2[n + 1, 8], {n, 0, 20}] (* Vaclav Kotesovec, Dec 16 2014 *)
Table[8^n - 7*7^n + 21*6^n - 35*5^n + 35*4^n - 21*3^n + 7*2^n - 1, {n, 0, 20}] (* Vaclav Kotesovec, Dec 16 2014 *)
CoefficientList[Series[5040*x^7 / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Dec 16 2014 after Colin Barker *)

a(n)=8^(n)-7*7^(n)+21*6^(n)-35*5^(n)+35*4^(n)-21*3^(n)+7*2^(n)-1.

for(n=0,30, print1(5040*stirling(n+1,8,2), ", ")) \\ G. C. Greubel, Nov 19 2017

a(n) = 5040 * S2(n+1,8), n>=0.
G.f.: 5040*x^7 / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)). - Colin Barker, Sep 20 2013
E.g.f.: Sum_{k=1..8} (-1)^(8-k)*binomial(8-1,k-1)*exp(k*x). - Wolfdieter Lang, May 03 2017

Offset corrected by Vaclav Kotesovec, Dec 16 2014

A028245 a(n) = 5^(n-1) - 44^(n-1) + 63^(n-1) - 4*2^(n-1) + 1 (essentially Stirling numbers of second kind).

Original entry on oeis.org

0, 0, 0, 0, 24, 360, 3360, 25200, 166824, 1020600, 5921520, 33105600, 180204024, 961800840, 5058406080, 26308573200, 135666039624, 694994293080, 3542142833040, 17980946172000, 90990301641624

Offset: 1

N. J. A. Sloane, Doug McKenzie mckfam4(AT)aol.com

For n>=2, a(n) is equal to the number of functions f: {1,2,...,n-1}->{1,2,3,4,5} such that Im(f) contains 4 fixed elements. - Aleksandar M. Janjic and Milan Janjic, Mar 08 2007

Seiichi Manyama, Table of n, a(n) for n = 1..1431
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (15,-85,225,-274,120).

Cf. A000481, A008277, A163626, A000225, A028243, A028244.

[5^(n-1) - 4*4^(n-1) + 6*3^(n-1) - 4*2^(n-1) + 1: n in [1..30]]; // G. C. Greubel, Nov 19 2017

24StirlingS2[Range[30],5] (* Harvey P. Dale, Jun 18 2013 *)
Table[5^(n - 1) - 4*4^(n - 1) + 6*3^(n - 1) - 4*2^(n - 1) + 1, {n, 21}] (* or *)
Rest@ CoefficientList[Series[-24 x^5/((x - 1) (4 x - 1) (3 x - 1) (2 x - 1) (5 x - 1)), {x, 0, 21}], x] (* Michael De Vlieger, Sep 24 2016 *)

for(n=1,30, print1(24*stirling(n,5,2), ", ")) \\ G. C. Greubel, Nov 19 2017

a(n) = 24*S(n, 5) = 24*A000481(n). - Emeric Deutsch, May 02 2004
G.f.: -24*x^5/((x-1)*(4*x-1)*(3*x-1)*(2*x-1)*(5*x-1)). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009; checked and corrected by R. J. Mathar, Sep 16 2009
E.g.f.: (Sum_{k=0..5} (-1)^(5-k)*binomial(5,k)*exp(k*x))/5. with a(0) = 0. - Wolfdieter Lang, May 03 2017

A278075 Coefficients of the signed Fubini polynomials in ascending order, F_n(x) = Sum_{k=0..n} (-1)^nStirling2(n,k)k!*(-x)^k.

Original entry on oeis.org

1, 0, 1, 0, -1, 2, 0, 1, -6, 6, 0, -1, 14, -36, 24, 0, 1, -30, 150, -240, 120, 0, -1, 62, -540, 1560, -1800, 720, 0, 1, -126, 1806, -8400, 16800, -15120, 5040, 0, -1, 254, -5796, 40824, -126000, 191520, -141120, 40320, 0, 1, -510, 18150, -186480, 834120, -1905120, 2328480, -1451520, 362880

Offset: 0

Peter Luschny, Jan 09 2017

sign

Signed version of A131689.

Integral_{x=0..1} F_n(x) = B_n(1) where B_n(x) are the Bernoulli polynomials.

			Triangle of coefficients starts:
[1]
[0,  1]
[0, -1,    2]
[0,  1,   -6,    6]
[0, -1,   14,  -36,    24]
[0,  1,  -30,  150,  -240,   120]
[0, -1,   62, -540,  1560, -1800,    720]
[0,  1, -126, 1806, -8400, 16800, -15120, 5040]

Peter Luschny, Illustration of the polynomials.
Peter Luschny, The Bernoulli Manifesto.
Grzegorz Rządkowski, Bernoulli numbers and solitons - revisited, Journal of Nonlinear Mathematical Physics, (2010) 17:1, 121-126.
J. Worpitzky, Studien über die Bernoullischen und Eulerschen Zahlen, Journal für die reine und angewandte Mathematik, 94 (1883), 203-232.

Row sums are A000012, diagonal is A000142.
Cf. A131689 (unsigned), A019538 (n>0, k>0), A090582.
Let F(n, x) = Sum_{k=0..n} T(n,k)*x^k then, apart from possible differences in the sign or the offset, we have: F(n, -5) = A094418(n), F(n, -4) = A094417(n), F(n, -3) = A032033(n), F(n, -2) = A004123(n), F(n, -1) = A000670(n), F(n, 0) = A000007(n), F(n, 1) = A000012(n), F(n, 2) = A000629(n), F(n, 3) = A201339(n), F(n, 4) = A201354(n), F(n, 5) = A201365(n).
Cf. A163626, A173018.

function T(n, k)
    if k < 0 || k > n return 0 end
    if n == 0 && k == 0 return 1 end
    k*(T(n-1, k-1) - T(n-1, k))
end
for n in 0:7
    println([T(n,k) for k in 0:n])
end
# Peter Luschny, Mar 26 2020

F := (n,x) -> add((-1)^n*Stirling2(n,k)*k!*(-x)^k, k=0..n):
for n from 0 to 10 do PolynomialTools:-CoefficientList(F(n,x), x) od;

T[ n_, k_] := If[ n < 0 || k < 0, 0, (-1)^(n - k) k! StirlingS2[n, k]]; (* Michael Somos, Jul 08 2018 *)

{T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(n + i) * binomial(k, i) * i^n))};
/* Michael Somos, Jul 08 2018 */

T(n, k) = (-1)^(n-k) * Stirling2(n, k) * k!.
E.g.f.: 1/(1-x*(1-exp(-t))) = Sum_{n>=0} F_n(x) t^n/n!.
T(n, k) = k*(T(n-1, k-1) - T(n-1, k)) for 0 <= k <= n, T(0, 0) = 1, otherwise 0.
Bernoulli numbers are given by B(n) = Sum_{k = 0..n} T(n, k) / (k+1) with B(1) = 1/2. - Michael Somos, Jul 08 2018
Let F_n(x) be the row polynomials of this sequence and W_n(x) the row polynomials of A163626. Then F_n(1 - x) = W_n(x) and Integral_{x=0..1} U(n, x) = Bernoulli(n, 1) for U in {W, F}. - Peter Luschny, Aug 10 2021
T(n, k) = [z^k] Sum_{k=0..n} Eulerian(n, k)*z^(k+1)*(z-1)^(n-k-1) for n >= 1, where Eulerian(n, k) = A173018(n, k). - Peter Luschny, Aug 15 2022

A050946 "Stirling-Bernoulli transform" of Fibonacci numbers.

Original entry on oeis.org

0, 1, 1, 7, 13, 151, 421, 6847, 25453, 532231, 2473141, 63206287, 352444093, 10645162711, 69251478661, 2413453999327, 17943523153933, 708721089607591, 5927841361456981, 261679010699505967, 2431910546406522973, 118654880542567722871, 1212989379862721528101

Offset: 0

N. J. A. Sloane, Jan 02 2000

nonn

From Paul Curtz, Oct 11 2013: (Start)
Differences table:
     0,    1,    1,    7,   13,  151,  421, 6847, ...
     1,    0,    6,    6,  138,  270, 6426, ...
    -1,    6,    0,  132,  132, 6156, ...
     7,   -6,  132,    0, 6024, ...
   -13,  138, -132, 6024, ...
   151, -270, 6156, ...
  -421, 6426, ...
  6847, ... .
a(n) is an autosequence of first kind: the inverse binomial transform is the sequence signed, the main diagonal is A000004=0's.
The "Stirling-Bernoulli transform" applied to an autosequence of first kind is an autosequence of first kind.
Now consider the Akiyama-Tanigawa transform or algorithm applied to A000045(n):
     0,   1,   1,   2,   3,   5, 8, ...
    -1,   0,  -3,  -4, -10, -18, ...    = -A006490
    -1,   6,   3,  24,  40, ...
    -7,   6, -63, -64, ...
   -13, 138,   3, ...
  -151, 270, ...
  -421, ... .
Hence -a(n). The Akiyama-Tanigawa algorithm applied to an autosequence of first kind is an autosequence of first kind.
a(n+5) - a(n+1) = 150, 420, 6840, ... is divisible by 30.
For an autosequence of the second kind, the inverse binomial transform is the sequence signed with the main diagonal double of the first upper diagonal.
The Akiyama-Tanigawa algorithm applied to an autosequence leads to an autosequence of the same kind. Example: the A-T algorithm applied to the autosequence of second kind 1/n leads to the autosequence of the second kind A164555(n)/A027642(n).
Note that a2(n) = 2*a1(n+1) - a1(n) applied to the autosequence of the first kind a1(n) is a corresponding autosequence of the second kind. (End)

Alois P. Heinz, Table of n, a(n) for n = 0..447
C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7

Cf. A000045, A051782, A105796, A163626.

with(combinat):
a:= n-> add((-1)^(k+1) *k! *stirling2(n+1, k+1)*fibonacci(k), k=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, May 17 2013

CoefficientList[Series[E^x*(1-E^x)/(1-3*E^x+E^(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Aug 13 2013 *)
t[0, k_] := Fibonacci[k]; t[n_, k_] := t[n, k] = (k+1)*(t[n-1, k] - t[n-1, k+1]); a[n_] := t[n, 0] // Abs; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Oct 22 2013, after Paul Curtz *)

{a(n)=polcoeff(sum(m=0, n, fibonacci(m)*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 20 2011 */

O.g.f.: Sum_{n>=1} Fibonacci(n) * n! * x^n / Product_{k=1..n} (1+k*x). - Paul D. Hanna, Jul 20 2011
A100872(n)=a(2*n) and A100868(n)=a(2*n-1).
From Paul Barry, Apr 20 2005: (Start)
E.g.f.: exp(x)*(1-exp(x))/(1-3*exp(x)+exp((2*x))).
a(n) = Sum_{k=0..n} (-1)^(n-k)*S2(n, k)*k!*Fibonacci(k). [corrected by Ilya Gutkovskiy, Apr 04 2019] (End)
a(n) ~ c * n! / (log((3+sqrt(5))/2))^(n+1), where c = 1/sqrt(5) if n is even and c = 1 if n is odd. - Vaclav Kotesovec, Aug 13 2013
a(n) = -1 * Sum_{k = 0..n} A163626(n,k)*A000045(k). - Philippe Deléham, May 29 2015

A228909 a(n) = 7^n - 66^n + 155^n - 204^n + 153^n - 6*2^n + 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 720, 20160, 332640, 4233600, 46070640, 451725120, 4115105280, 35517081600, 294293759760, 2362955474880, 18509835445920, 142172988048000, 1074905737084080, 8023358912869440, 59263889194762560, 433988913576556800, 3155502239364459600, 22807773973299268800

Offset: 0

Richard V. Scholtz, III, Sep 07 2013

Essentially Stirling Numbers of the Second Kind, with an offset index, and multiplied by 720.

Calculates the seventh column of coefficients with respect to the derivatives, d^n/dx^n(y), of the logistic equation when written as y=1/[1+exp(-x)].

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (28,-322,1960,-6769,13132,-13068,5040).

Cf. A000771, A008277.

Represents the seventh column of results of A163626.

[7^n - 6*6^n + 15*5^n - 20*4^n + 15*3^n - 6*2^n + 1: n in [0..30]]; // G. C. Greubel, Nov 19 2017

Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1 - y[x]); Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n - 1][y][x], x]; row[n_] := CoefficientList[ Derivative[n][y][x], y[x]] // Rest; Join[{0, 0, 0, 0, 0, 0}, Table[row[n], {n, 6, 23}] [[All, 7]]] (* Jean-François Alcover, Dec 16 2014 *)
Table[7^n - 6*6^n + 15*5^n - 20*4^n + 15*3^n - 6*2^n + 1, {n, 0, 20}] (* Vaclav Kotesovec, Dec 16 2014 *)
Table[6!*StirlingS2[n + 1, 7], {n, 0, 20}] (* Vaclav Kotesovec, Dec 16 2014 *)

a(n)=7^(n)-6*6^(n)+15*5^(n)-20*4^(n)+15*3^(n)-6*2^(n)+1

concat([0,0,0,0,0,0], Vec(-720*x^6/((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)) + O(x^100))) \\ Colin Barker, Dec 16 2014

a(n) = 720 * S(n+1,7), n>=0.
G.f.: -720*x^6 / ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)). - Colin Barker, Dec 16 2014
E.g.f.: Sum_{k=1..7} (-1)^(7-k)*binomial(7-1,k-1)*exp(k*x). - Wolfdieter Lang, May 03 2017

Offset corrected by Jean-François Alcover, Dec 16 2014
a(20) corrected by Jean-François Alcover, Dec 16 2014
Formula adapted for new offset by Vaclav Kotesovec, Dec 16 2014

A228911 a(n) = 9^n - 88^n + 287^n - 566^n + 705^n - 564^n + 283^n - 8*2^n + 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 40320, 1814400, 46569600, 898128000, 14495120640, 207048441600, 2706620716800, 33094020960000, 384202115256960, 4280991956841600, 46150861752777600, 484294916235312000, 4970346251077025280, 50075960398487654400, 496745174491651008000

Offset: 0

Richard V. Scholtz, III, Sep 07 2013

Calculates the ninth column of coefficients with respect to the derivatives, d^n/dx^n(y), of the logistic equation when written as y=1/[1+exp(-x)].

Essentially 40320 * A049447. - Joerg Arndt, Sep 24 2016

Seiichi Manyama, Table of n, a(n) for n = 0..1047
Index entries for linear recurrences with constant coefficients, signature (45,-870,9450,-63273,269325,-723680,1172700,-1026576,362880).

The ninth column of results of A163626.

Cf. A228910 (also for more crossrefs).

[9^n-8*8^n+28*7^n-56*6^n+70*5^n-56*4^n+28*3^n-8*2^n+1: n in [0..32]]; // Vincenzo Librandi, Oct 11 2017

Derivative[0][y][x] = y[x]; Derivative[1][y][x] = y[x]*(1 - y[x]); Derivative[n_][y][x] := Derivative[n][y][x] = D[Derivative[n - 1][y][x], x]; row[n_] := CoefficientList[ Derivative[n][y][x], y[x]] // Rest; Join[{0, 0, 0, 0, 0, 0, 0, 0}, Table[row[n], {n, 8, 22}] [[All, 9]]] (* Jean-François Alcover, Dec 16 2014 *)
Table[8!*StirlingS2[n + 1, 9], {n, 0, 22}] (* Vaclav Kotesovec, Dec 16 2014 *)
Table[9^n-8*8^n+28*7^n-56*6^n+70*5^n-56*4^n+28*3^n-8*2^n+1, {n, 0, 22}] (* Vaclav Kotesovec, Dec 16 2014 *)
CoefficientList[Series[-40320*x^8/ ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)*(9*x-1)), {x, 0, 20}], x] (* Vaclav Kotesovec, Dec 16 2014 after Colin Barker *)
lst={}; Do[f= 40320 StirlingS2[n, 9];  AppendTo[lst, f], {n, 1, 5!}]; lst (* Vincenzo Librandi, Oct 11 2017 *)

a(n)=9^n-8*8^n+28*7^n-56*6^n+70*5^n-56*4^n+28*3^n-8*2^n+1

G.f.: -40320*x^8/ ((x-1)*(2*x-1)*(3*x-1)*(4*x-1)*(5*x-1)*(6*x-1)*(7*x-1)*(8*x-1)*(9*x-1)). - Colin Barker, Sep 20 2013

E.g.f.: Sum_{k=1..9} (-1)^(9-k)*binomial(9-1,k-1)*exp(k*x). - Wolfdieter Lang, May 03 2017

Offset corrected by Vaclav Kotesovec, Dec 16 2014

cp's OEIS Frontend

A048163 a(n) = Sum_{k=1..n} ((k-1)!)^2*Stirling2(n,k)^2.

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A371761 Array read by antidiagonals: The number of parades with n girls and k boys that begin with a girl and end with a boy.

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A130850 Triangle read by rows, 0 <= k <= n, T(n,k) = Sum_{j=0..n} A(n,j)*binomial(n-j,k) where A(n,j) are the Eulerian numbers A173018.

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A210672 a(0)=1; thereafter a(n) = 2*Sum_{k=1..n} binomial(2n,2k)*a(n-k).

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A228910 a(n) = 8^n - 7*7^n + 21*6^n - 35*5^n + 35*4^n - 21*3^n + 7*2^n - 1.

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A028245 a(n) = 5^(n-1) - 4*4^(n-1) + 6*3^(n-1) - 4*2^(n-1) + 1 (essentially Stirling numbers of second kind).

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A210672 a(0)=1; thereafter a(n) = 2Sum_{k=1..n} binomial(2n,2k)a(n-k).

A228910 a(n) = 8^n - 77^n + 216^n - 355^n + 354^n - 213^n + 72^n - 1.

A028245 a(n) = 5^(n-1) - 44^(n-1) + 63^(n-1) - 4*2^(n-1) + 1 (essentially Stirling numbers of second kind).

A278075 Coefficients of the signed Fubini polynomials in ascending order, F_n(x) = Sum_{k=0..n} (-1)^nStirling2(n,k)k!*(-x)^k.

A228909 a(n) = 7^n - 66^n + 155^n - 204^n + 153^n - 6*2^n + 1.

A228911 a(n) = 9^n - 88^n + 287^n - 566^n + 705^n - 564^n + 283^n - 8*2^n + 1.