A008292 -id:A008292 - cp's OEIS frontend

A066912 Fourth column of the Eulerian triangle A008292 in square array format.

0, 1, 26, 302, 2416, 15619, 88234, 455192, 2203488, 10187685, 45533450, 198410786, 848090912, 3572085255, 14875399450, 61403313100, 251732291184, 1026509354985, 4168403181210, 16871482830550, 68111623139600

Offset: 0

Randall L Rathbun, Jan 22 2002

nonn

Essentially the same as A000498.

a(n)=4^(n+3)-(n+4)*3^(n+3)+1/2*(n+3)*(n+4)*2^(n+3)-1/6*(n+2)*(n+3)*(n+4)

a(n) = 4^(n+3) - (n+4)*3^(n+3) + (1/2)*(n+3)*(n+4)*2^(n+3) - (1/6)*(n+2)*(n+3)*(n+4).

More terms from Larry Reeves (larryr(AT)acm.org), Jun 11 2002

A089249 Triangular array read by rows illustrating the connection between A000522 and A008292.

Original entry on oeis.org

1, 3, 4, 6, 16, 11, 10, 40, 55, 26, 15, 80, 165, 156, 57, 21, 140, 385, 546, 399, 120

Offset: 1

Alford Arnold, Dec 11 2003

The general case corresponding to other diagonals of A046802 can be derived from A046802(m,n) = Sum C(m-1,n-1) * A008292(m-r,n-1).

			The fifth row of the array is 15 80 165 156 57 resulting from A089249 (1 4 11 26 57 ) times ( 15 20 15 6 1)

Row sums = the third diagonal of A046802.

Cf. A000522, A008292, A046802, A066810, A089072.

A125108 Column sums of a Gaussian polynomial-shaped array. Row sums generate the Eulerian array A008292.

Original entry on oeis.org

1, 2, 4, 10, 26, 72, 202, 580

Offset: 1

Alford Arnold, Dec 25 2006

Column sums of the Gaussian polynomial template count numeric partitions. Row sums of the Gaussian polynomial template generate Pascal's triangle. A105552 has the same shape as the template and counts compositions. Row sums of the Eulerian array counts permutations of n object.

			The column sums begin 1 2 4 10 26 72 202 580 ... because the structure of the Array begin as follows:
1..................................................................
......1............................................................
......1............................................................
............1......................................................
............2......2...............................................
............1......................................................
..................1................................................
..................3......5......3..................................
..................3......5......3..................................
..................1................................................
............................1......................................
............................4.......9.......9.......4..............
............................6.......16......22......16.......6.....
............................4.......9.......9.......4..............
............................1......................................
etc.

Cf. A000041, A000079, A000142, A007318, A008292, A047970, A060351, A105552.

A147563 Irregular triangle, T(n, k) = [x^k] p(n, x), where p(n, x) = 4Sum_{j=0..n} A008292(n+1, j) (x/2)^j * (1-x/2)^(n-j), read by rows.

Original entry on oeis.org

4, 4, 4, 4, -2, 4, 16, -8, 4, 44, -6, -16, 4, 4, 104, 84, -136, 34, 4, 228, 606, -584, -24, 102, -17, 4, 480, 2832, -1088, -2208, 1488, -248, 4, 988, 11122, 5536, -20840, 8896, 832, -992, 124, 4, 2008, 39772, 74296, -118190, -2144, 51952, -22112, 2764

Offset: 0

Roger L. Bagula, Nov 07 2008

			Irregular triangle begins as:
  4;
  4;
  4,    4,    -2;
  4,   16,    -8;
  4,   44,    -6,   -16,       4;
  4,  104,    84,  -136,      34;
  4,  228,   606,  -584,     -24,   102,   -17;
  4,  480,  2832, -1088,   -2208,  1488,  -248;
  4,  988, 11122,  5536,  -20840,  8896,   832,   -992,  124;
  4, 2008, 39772, 74296, -118190, -2144, 51952, -22112, 2764;

G. C. Greubel, Row n = 0..50 of the irregular triangle, flattened

Cf. A008292.

A008292:= func< n,k | (&+[(-1)^j*Binomial(n+1, j)*(k-j)^n: j in [0..k]]) >;
T:= func< n,k | (-1/2)^(k-2)*(&+[(-1)^j*Binomial(n-j,k-j)*A008292(n+1,j+1): j in [0..k]]) >;
[Floor(T(n,k)): k in [0..2*Floor(n/2)], n in [0..16]]; // G. C. Greubel, Oct 27 2022; Mar 03 2023

(* First program *)
nmax:= 15;
p[x_, n_]= (1-x)^(n+1)*PolyLog[-n, x]/x;
b= Table[CoefficientList[p[x, n], x], {n, nmax+1}];
F[n_]:= CoefficientList[4*Sum[b[[n+1]][[m+1]]*(x/2)^(n-m)*(1-x/2)^m, {m, 0, n}], x];
T[n_]:= If[IntegerQ[F[n]], F[n], Sign[F[n]]*Abs[Round[F[n] - 1/2]]];
Table[T[n], {n, 0, nmax}]//Flatten
(* Second program *)
A008292[n_, k_]:= Sum[(-1)^j*(k-j)^n*Binomial[n+1,j], {j,0,k}];
F[n_, k_]:= (-1/2)^(k-2)*Sum[(-1)^j*Binomial[n-j, k-j]*A008292[n+1, j+ 1], {j,0,k}];
T[n_, k_]:= If[IntegerQ[F[n,k]], F[n,k], Sign[F[n,k]]*Abs[Round[F[n, k] - 1/2]]];
Table[T[n, k], {n,0,16}, {k, 0, 2*Floor[n/2]}]//Flatten (* G. C. Greubel, Mar 03 2023 *)

def A008292(n,k): return sum( (-1)^j*binomial(n+1, j)*(k-j)^n for j in range(k+1) )
def A147563(n,k): return floor((-1/2)^(k-2)*sum((-1)^j*binomial(n-j, k-j)*A008292(n+1,j+1) for j in range(k+1)))
flatten([[A147563(n,k) for k in range(2*floor(n/2) + 1)] for n in range(16)]) # G. C. Greubel, Oct 27 2022; Mar 03 2023

T(n, k) = coefficients [x^k]( p(n, x) ), where p(n, x) = 4*Sum_{j=0..n} A008292(n+1, j) * (x/2)^j * (1-x/2)^(n-j).

T(n, k) = round( (-1/2)^(k-2) * Sum_{j=0..k} (-1)^j*binomial(n-j, k-j) * A008292(n+1, j+1) ). - G. C. Greubel, Mar 03 2023

Edited by G. C. Greubel, Oct 27 2022

A174303 A symmetrical triangle: T(n,k) = A008292(n+1, k) * f(n,k), where f(n,k) = 2^k when floor(n/2) >= k, otherwise 2^(n-k).

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 22, 22, 1, 1, 52, 264, 52, 1, 1, 114, 1208, 1208, 114, 1, 1, 240, 4764, 19328, 4764, 240, 1, 1, 494, 17172, 124952, 124952, 17172, 494, 1, 1, 1004, 58432, 705872, 2499040, 705872, 58432, 1004, 1, 1, 2026, 191360, 3641536, 20965664, 20965664, 3641536, 191360, 2026, 1

Offset: 0

Roger L. Bagula, Mar 15 2010

Row sums are: {1, 2, 10, 46, 370, 2646, 29338, 285238, 4029658, ...}.

			Triangle begins as:
  1;
  1,    1;
  1,    8,     1;
  1,   22,    22,      1;
  1,   52,   264,     52,       1;
  1,  114,  1208,   1208,     114,      1;
  1,  240,  4764,  19328,    4764,    240,     1;
  1,  494, 17172, 124952,  124952,  17172,   494,    1;
  1, 1004, 58432, 705872, 2499040, 705872, 58432, 1004,   1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Eric Weisstein's World of Mathematics, Eulerian Number

Cf. A008292, A144463, A144470, A174301.

Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Floor(n/2) ge k select 2^k*Eulerian(n+1,k) else 2^(n-k)*Eulerian(n+1,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 15 2019

Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1,j]*(k-j+1)^n, {j,0,k+1}];
Table[Eulerian[n+1,m]*If[Floor[n/2] >= m, 2^m, 2^(n-m)], {n,0,10}, {m,0,n} ]//Flatten (* modified by G. C. Greubel, Apr 15 2019 *)

{eulerian(n,k) = sum(j=0,k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n)};
for(n=0,10, for(k=0,n, print1(eulerian(n+1,k)*if(floor(n/2)>=k, 2^k, 2^(n-k)), ", "))) \\ G. C. Greubel, Apr 15 2019

def Eulerian(n,k): return sum((-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1))
def T(n,k):
   if floor(n/2)>=k: return 2^k*Eulerian(n+1,k)
   else: return 2^(n-k)*Eulerian(n+1,k)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 15 2019

T(n,k) = Eulerian(n+1, k)*if(floor(n/2) greater than or equal to k then 2^m otherwise 2^(n-k)), where the Eulerian numbers are defined as A008292(n,k).

Edited by G. C. Greubel, Apr 15 2019

A176204 Triangle T(n, k) = 4 * A008292(n+1, k) - 3, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 41, 41, 1, 1, 101, 261, 101, 1, 1, 225, 1205, 1205, 225, 1, 1, 477, 4761, 9661, 4761, 477, 1, 1, 985, 17169, 62473, 62473, 17169, 985, 1, 1, 2005, 58429, 352933, 624757, 352933, 58429, 2005, 1, 1, 4049, 191357, 1820765, 5241413, 5241413, 1820765, 191357, 4049, 1

Offset: 0

Roger L. Bagula, Apr 11 2010

 This sequence belongs to the class defined by T(n, m, q) = 2*T(n, m, q-1) - 1. The first few q values gives the sequences: A008292(n+1, k) (q=0), A176200 (q=1), this sequence (q=2).
Row sums are: {1, 2, 15, 84, 465, 2862, 20139, 161256, 1451493, 14515170, 159667167, ...}.
Former title: A recursive symmetrical triangular sequence based on Eulerian numbers: q=2: T(n, m, q) = 2*T(n, m, q-1) - 1.

			Triangle begins as:
  1;
  1,    1;
  1,   13,      1;
  1,   41,     41,       1;
  1,  101,    261,     101,       1;
  1,  225,   1205,    1205,     225,       1;
  1,  477,   4761,    9661,    4761,     477,       1;
  1,  985,  17169,   62473,   62473,   17169,     985,      1;
  1, 2005,  58429,  352933,  624757,  352933,   58429,   2005,    1;
  1, 4049, 191357, 1820765, 5241413, 5241413, 1820765, 191357, 4049, 1;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A007318, A109128, A131061, A168625, A176200.

Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >;
[[4*Eulerian(n+1,k) -3: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Mar 12 2020

A008292:= (n,k) -> add((-1)^j*binomial(n+1,j)*(k-j+1)^n, j=0..k+1);
A176204:= (n,k,q) -> 2^q*( A008292(n+1,k) -1) + 1;
seq(seq( A176204(n,k,2), k=0..n), n=0..12); # G. C. Greubel, Mar 12 2020

Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1, j]*(k-j+1)^n, {j,0,k+1}];
T[n_, m_, q_]:= 2^q*Eulerian[n+1, m] - 2^q +1;
Table[T[n, m, 2], {n,0,12}, {m,0,n}]//Flatten (* modified by G. C. Greubel, Mar 12 2020 *)

Eulerian(n,k) = sum(j=0,k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n);
T(n,k,q) = 2^q*Eulerian(n+1,k) - (2^q - 1); \\ G. C. Greubel, Mar 12 2020

def Eulerian(n,k): return sum((-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1))
def T(n,k,q): return 2^q*Eulerian(n+1,k) - 2^q + 1
[[T(n,k,2) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Mar 12 2020

T(n, m, q) = 2*T(n, m, q-1) - 1, with T(n, m, 0) = A008292(n+1, m).
From G. C. Greubel, Mar 12 2020: (Start)
T(n, k, q) = 2^q * A008292(n+1, k) - (2^q - 1).
Sum_{k=0..n} T(n, k, q) = (n+1)*( 2^q * n! - 2^q + 1) (row sums). (End)

Edited by G. C. Greubel, Mar 12 2020

A176492 Triangle T(n,k) = A176492(n,k) + A008292(n+1,k+1) - 1 read along rows 0<=k<=n.

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 45, 45, 1, 1, 129, 365, 129, 1, 1, 353, 2293, 2293, 353, 1, 1, 965, 12937, 28397, 12937, 965, 1, 1, 2677, 69261, 290993, 290993, 69261, 2677, 1, 1, 7561, 360853, 2661809, 4987461, 2661809, 360853, 7561, 1, 1, 21705, 1852053, 22618437

Offset: 0

Roger L. Bagula, Apr 19 2010

Row sums are 1, 2, 15, 92, 625, 5294, 56203, 725864, 11047909, 193052642, 3795725791,....

			1;
1, 1;
1, 13, 1;
1, 45, 45, 1;
1, 129, 365, 129, 1;
1, 353, 2293, 2293, 353, 1;
1, 965, 12937, 28397, 12937, 965, 1;
1, 2677, 69261, 290993, 290993, 69261, 2677, 1;
1, 7561, 360853, 2661809, 4987461, 2661809, 360853, 7561, 1;
, 21705, 1852053, 22618437, 72034125, 72034125, 22618437, 1852053, 21705, 1;
1, 63117, 9421457, 182707997, 926399717, 1558541213, 926399717, 182707997, 9421457, 63117, 1;

Cf. A007318, A008292, A060187, A176487.

A176492 := proc(n,k)
    A176491(n,k)+A008292(n+1,k+1)-1 ;
end proc: # R. J. Mathar, Jun 16 2015

(*A060187*)
p[x_, n_] = (1 - x)^(n + 1)*Sum[(2*k + 1)^n*x^k, {k, 0, Infinity}];
f[n_, m_] := CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m + 1]];
<< DiscreteMath`Combinatorica`;
t[n_, m_, 0] := Binomial[n, m];
t[n_, m_, 1] := Eulerian[1 + n, m];
t[n_, m_, 2] := f[n, m];
t[n_, m_, q_] := t[n, m, q] = t[n, m, q - 2] + t[n, m, q - 3] - 1;
Table[Flatten[Table[Table[t[n, m, q], {m, 0, n}], {n, 0, 10}]], {q, 0, 10}]

A177823 Triangle of Eulerian numbers squared: A008292(n,m)^2 read by rows.

Original entry on oeis.org

1, 1, 1, 1, 16, 1, 1, 121, 121, 1, 1, 676, 4356, 676, 1, 1, 3249, 91204, 91204, 3249, 1, 1, 14400, 1418481, 5837056, 1418481, 14400, 1, 1, 61009, 18429849, 243953161, 243953161, 18429849, 61009, 1, 1, 252004, 213393664, 7785238756, 24395316100, 7785238756, 213393664, 252004, 1, 1, 1026169

Offset: 1

Roger L. Bagula, Dec 13 2010

Row sums are A168562.

			1;
1, 1;
1, 16, 1;
1, 121, 121, 1;
1, 676, 4356, 676, 1;
1, 3249, 91204, 91204, 3249, 1;
1, 14400, 1418481, 5837056, 1418481, 14400, 1;

Cf. A008292, A014732, A014748, A014765, A168562

<< DiscreteMath`Combinatorica`;
a = Table[Table[Eulerian[n + 1, m]^2, {m, 0, n}], {n, 0, 10}];
Flatten[%]

A178048 Triangle T(n, m) = ( |-A008292(n+1,m+1)^2 + 2binomial(n, m)^2| + A008292(n+1,m+1)binomial(n, m) )/2 read by rows.

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 68, 68, 1, 1, 374, 2340, 374, 1, 1, 1742, 47012, 47012, 1742, 1, 1, 7524, 717948, 2942288, 717948, 7524, 1, 1, 31320, 9259560, 122248688, 122248688, 9259560, 31320, 1, 1, 127946, 106900560, 3895086794, 12203119800, 3895086794, 106900560, 127946, 1

Offset: 0

Roger L. Bagula, May 18 2010

			The triangle starts in row n=0 with columns 0 <= m <= n as
  1;
  1,      1;
  1,      8,         1;
  1,     68,        68,          1;
  1,    374,      2340,        374,           1;
  1,   1742,     47012,      47012,        1742,          1;
  1,   7524,    717948,    2942288,      717948,       7524,         1;
  1,  31320,   9259560,  122248688,   122248688,    9259560,     31320,      1;
  1, 127946, 106900560, 3895086794, 12203119800, 3895086794, 106900560, 127946, 1;

Cf. A141686, A008292.

A178048 := proc(n,m) binomial(n,m)*A008292(n+1,m+1)+abs( -A008292(n+1,m+1)^2+2*binomial(n,m)^2) ; %/2; end proc:
seq(seq(A178048(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Nov 26 2010

<< DiscreteMath`Combinatorica`
t[n_, m_] = (Abs[2*Binomial[n, m]^2 - Eulerian[n + 1, m]^2] + Binomial[n, m]*Eulerian[n + 1, m])/2;
Table[Table[t[n, m], {m, 0, n}], {n, 0, 10}];
Flatten[%]

T(n, m) = T(n,n-m).

Definition corrected by R. J. Mathar, Nov 26 2010

A178232 A triangle sequence derived from setting an Euler numbers A122045 generalization equal to the Eulerian numbers A008292 to get a generating function expansion: p(x,t) = ((-1 + exp(x)) (-1 + x)/(-1 + exp(tx) + t - exp(t) x)).

Original entry on oeis.org

0, 0, 1, 6, 1, 1, 36, 8, 3, 7, 1, 240, 60, -20, 81, 11, 21, 1, 1800, 480, -510, 822, 143, 173, 123, 51, 1, 15120, 4200, -7560, 8526, 2450, 239, 2381, 435, 715, 113, 1, 141120, 40320, -102480, 93744, 43512, -21320, 36991, 2943, 11035, 4035, 3139, 239, 1

Offset: 0

Roger L. Bagula, May 23 2010

The first column gives the Lah numbers A001286: (n - 1)*n!/2;
{0,0,1, 6, 36, 240, 1800, 15120, 141120, 1451520, ...}.
Row sums are {0, 0, 1, 8, 55, 394, 3083, 26620, 253279, 2642390, 30052699, ...}.
The equation solved in the integer q was
  q*exp(x*t)/(q - 1 + exp(t)) - (1 - t)/(1 - t*exp(x*(1 - t))) = 0.
Factors and the n! first term from taken out in Mathematica to give a more simple set of coefficients.
The idea in solving for an integer q here is to get a polynomial that behaves as a generalization of both types.
No q-form value for q=n=0,1 is expected.

			{0},
{0},
{1},
{6, 1, 1},
{36, 8, 3, 7, 1},
{240, 60, -20, 81, 11, 21, 1},
{1800, 480, -510, 822, 143, 173, 123, 51, 1},
{15120, 4200, -7560, 8526, 2450, 239, 2381, 435, 715, 113, 1},
{141120, 40320, -102480, 93744, 43512, -21320, 36991, 2943, 11035, 4035, 3139, 239, 1},
{1451520, 423360, -1391040, 1103760, 763056, -585432, 527544, 71353, 82513, 107377, 39589, 36349, 11947, 493, 1},
{16329600, 4838400, -19504800, 13940640, 13361040, -12088080, 7137270, 2643650, -749001, 2527719, 165459, 900099, 256743, 251073, 41883, 1003, 1}

Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 78-79.
L. Comtet, Advanced Combinatorics, Reidel, Holland, 1978, page 245.

L. Carlitz, q-Bernoulli numbers and polynomials, Duke Math. J. Volume 15, Number 4 (1948), 987-1000.

Cf. A008292, A122045, A156222.

p[t_] = ((-1 + Exp[x]) (-1 + x)/(-1 + Exp[t*x] + t - Exp[t]* x));
a = Table[ CoefficientList[FullSimplify[ExpandAll[(FullSimplify[ExpandAll[ -(1/((-1 + Exp[x])*(-1 + x)))*x^(n + 1)*n!*SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]]] - n!)/(x^2*(-1 + x))]], x], {n, 0, 10}] Flatten[a]

p(x,t) = ((-1 + exp(x)) (-1 + x)/(-1 + exp(t*x) + t - exp(t)* x)).

cp's OEIS Frontend

A066912 Fourth column of the Eulerian triangle A008292 in square array format.

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A089249 Triangular array read by rows illustrating the connection between A000522 and A008292.

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A125108 Column sums of a Gaussian polynomial-shaped array. Row sums generate the Eulerian array A008292.

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A147563 Irregular triangle, T(n, k) = [x^k] p(n, x), where p(n, x) = 4*Sum_{j=0..n} A008292(n+1, j) * (x/2)^j * (1-x/2)^(n-j), read by rows.

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A174303 A symmetrical triangle: T(n,k) = A008292(n+1, k) * f(n,k), where f(n,k) = 2^k when floor(n/2) >= k, otherwise 2^(n-k).

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A176204 Triangle T(n, k) = 4 * A008292(n+1, k) - 3, read by rows.

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A176492 Triangle T(n,k) = A176492(n,k) + A008292(n+1,k+1) - 1 read along rows 0<=k<=n.

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A177823 Triangle of Eulerian numbers squared: A008292(n,m)^2 read by rows.

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A147563 Irregular triangle, T(n, k) = [x^k] p(n, x), where p(n, x) = 4Sum_{j=0..n} A008292(n+1, j) (x/2)^j * (1-x/2)^(n-j), read by rows.

A178048 Triangle T(n, m) = ( |-A008292(n+1,m+1)^2 + 2binomial(n, m)^2| + A008292(n+1,m+1)binomial(n, m) )/2 read by rows.

A178232 A triangle sequence derived from setting an Euler numbers A122045 generalization equal to the Eulerian numbers A008292 to get a generating function expansion: p(x,t) = ((-1 + exp(x)) (-1 + x)/(-1 + exp(tx) + t - exp(t) x)).