A003506 -id:A003506 - cp's OEIS frontend

A241269 Denominator of c(n) = (n^2+n+2)/((n+1)(n+2)(n+3)).

3, 6, 15, 60, 105, 21, 126, 360, 495, 330, 429, 1092, 1365, 420, 1020, 2448, 2907, 1710, 1995, 4620, 5313, 759, 3450, 7800, 8775, 4914, 5481, 12180, 13485, 3720, 8184, 17952, 19635, 10710, 11655, 25308, 27417, 3705, 15990, 34440, 37023, 19866, 21285, 45540

Offset: 0

Paul Curtz, Apr 18 2014

All terms are multiples of 3.
Difference table of c(n):
   1/3,     1/6,    2/15,    7/60,    2/21,...
  -1/6,    -1/30,  -1/60,   -1/84,   -1/105,...
   2/15,    1/60,   1/210,   1/420,   1/630,...
  -7/60,   -1/84,  -1/420,  -1/1260, -1/2520,... .
This is an autosequence of the second kind; the inverse binomial transform is the signed sequence. The main diagonal is the first upper diagonal multiplied by 2.
Denominators of the main diagonal: A051133(n+1).
Denominators of the first upper diagonal; A000911(n).
c(n) is a companion to A026741(n)/A045896(n).
Based on the Akiyama-Tanigawa transform applied to 1/(n+1) which yields the Bernoulli numbers A164555(n)/A027642(n).
Are the numerators of the main diagonal (-1)^n? If yes, what is the value of 1/3 - 1/30 + 1/210,... or 1 - 1/10 + 1/70 - 1/420, ... , from A002802(n)?
Is a(n+40) - a(n) divisible by 10?
No: a(5) = 21 but a(45) = 12972. # Robert Israel, Jul 17 2023
Are the common divisors to A014206(n) and A007531(n+3) of period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2?
Reduce c(n) = f(n) = b(n)/a(n) = 1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, ... .
Consider the successively interleaved autosequences (also called eigensequences) of the second kind and of the first kind
  1,    1/2,  1/3,    1/4,     1/5,     1/6,   ...
  0,    1/6,  1/6,    3/20,    2/15,    5/42,  ...
  1/3,  1/6,  2/15,   7/60,   11/105,   2/21,  ...
  0,    1/10, 1/10,  13/140,   3/35,    5/63,  ...
  1/5,  1/10, 3/35,  11/140,  23/315,  43/630, ...
  0,    1/14, 1/14,  17/252,   4/63,   ...
This array is Au1(m,n). Au1(0,0)=1, Au1(0,1)=1/2.
Au1(m+1,n) = 2*Au1(m,n+1) - Au1(m,n).
First row: see A003506, Leibniz's Harmonic Triangle.
Second row: A026741/A045896.
a(n) is the denominator of the third row f(n).
The first column is 1, 0, 1/3, 0, 1/5, 0, 1/7, 0, ... . Numerators: A093178(n+1). This incites, considering tan(1), to introduce before the first row
Ta0(n) = 0, 1/2, 1/2, 5/12, 1/3, 4/15, 13/60, 151/840, ... .

Robert Israel, Table of n, a(n) for n = 0..10000

seq(denom((n^2+n+2)/((n+1)*(n+2)*(n+3))),n=0..1000);

Denominator[Table[(n^2+n+2)/Times@@(n+{1,2,3}),{n,0,50}]] (* Harvey P. Dale, Mar 27 2015 *)

for(n=0, 100, print1(denominator((n^2+n+2)/((n+1)*(n+2)*(n+3))), ", ")) \\ Colin Barker, Apr 18 2014

c(n) = A014206(n)/A007531(n+3).
The sum of the difference table main diagonal is 1/3 - 1/30 + 1/210 - ... = 10*A086466-4 = 4*(sqrt(5)*log(phi)-1) = 0.3040894... - Jean-François Alcover, Apr 22 2014
a(n) = (n+1)*(n+2)*(n+3)/gcd(4*n - 4, n^2 + n + 2), where gcd(4*n - 4, n^2 + n + 2) is periodic with period 16. - Robert Israel, Jul 17 2023

More terms from Colin Barker, Apr 18 2014

A174341 a(n) = Numerator of Bernoulli(n, 1) + 1/(n+1).

Original entry on oeis.org

2, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, -37, 1, 37, 1, -211, 1, 2311, 1, -407389, 1, 37153, 1, -1181819909, 1, 76977929, 1, -818946931, 1, 277930363757, 1, -84802531453217, 1, 90219075042851, 1, -711223555487930419, 1, 12696640293313423, 1, -6367871182840222481, 1, 35351107998094669831, 1, -83499808737903072705023, 1, 12690449182849194963361, 1

Offset: 0

Paul Curtz, Mar 16 2010

a(n) is numerator of (A164555(n)/A027642(n) + 1/(n+1)).
1/(n+1) and Bernoulli(n,1) are autosequences in the sense that they remain the same (up to sign) under inverse binomial transform. This feature is kept for their sum, a(n)/A174342(n) = 2, 1, 1/2, 1/4, 1/6, 1/6, 1/6, 1/8, 7/90, 1/10, ...
Similar autosequences are also A000045, A001045, A113405, A000975 preceded by two zeros, and A140096.
Conjecture: the numerator of (A164555(n)/(n+1) + A027642(n)/(n+1)^2) is a(n) and the denominator of this fraction is equal to 1 if and only if n+1 is prime or 1. Cf. A309132. - Thomas Ordowski, Jul 09 2019
The "if" part of the conjecture is true: see the theorems in A309132 and A326690. The values of the numerator when n+1 is prime are A327033. - Jonathan Sondow, Aug 15 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..300
OEIS Wiki, Autosequence

Cf. A164555, A027642, A174342 (denominators), A025529, A003506, A309132, A326690, A327033.

[2,1] cat [Numerator(Bernoulli(n)+1/(n+1)): n in [2..40]]; // Vincenzo Librandi, Jul 18 2019

A174341 := proc(n) bernoulli(n,1)+1/(n+1); numer(%) end proc: # R. J. Mathar, Nov 19 2010

a[n_] := Numerator[BernoulliB[n, 1] + 1/(n + 1)];
Table[a[n], {n, 0, 47}] (* Peter Luschny, Jul 13 2019 *)

B(n)=if(n!=1, bernfrac(n), -bernfrac(n));
a(n)=numerator(B(n) + 1/(n + 1));
for(n=0, 50, print1(a(n),", ")) \\ Indranil Ghosh, Jun 19 2017

a(n)=numerator(bernpol(n, 1) + 1/(n + 1)); \\ Michel Marcus, Jun 26 2025

from sympy import bernoulli, Integer
def a(n): return (bernoulli(n) + 1/Integer(n + 1)).numerator # Indranil Ghosh, Jun 19 2017

Reformulation of the name by Peter Luschny, Jul 13 2019

A253946 a(n) = 6*binomial(n+1, 6).

Original entry on oeis.org

6, 42, 168, 504, 1260, 2772, 5544, 10296, 18018, 30030, 48048, 74256, 111384, 162792, 232560, 325584, 447678, 605682, 807576, 1062600, 1381380, 1776060, 2260440, 2850120, 3562650, 4417686, 5437152, 6645408, 8069424, 9738960, 11686752, 13948704, 16564086

Offset: 5

Serhat Bulut, Jan 20 2015

For a set of integers {1, 2, ..., n}, a(n) is the sum of the 3 smallest elements of each subset with 5 elements, which is 6*C(n+1, 6) (for n >= 5), hence a(n) = 6*C(n+1, 6) = 6 * A000579(n+1).

			For A = {1, 2, 3, 4, 5, 6} the subsets with 5 elements are {1, 2, 3, 4, 5}, {1, 2, 3, 4, 6}, {1, 2, 3, 5, 6}, {1, 2, 4, 5, 6}, {1, 3, 4, 5, 6}, {2, 3, 4, 5, 6}.
The sum of 3 smallest elements of each subset: a(6) = (1 + 2 + 3) + (1 + 2 + 3) + (1 + 2 + 3) + (1 + 2 + 4) + (1 + 3 + 4) + (2 + 3 + 4) = 42 = 6*C(6 + 1, 6) = 6*A000579(6+1).

Colin Barker, Table of n, a(n) for n = 5..1000
Serhat Bulut and Oktay Erkan Temizkan, Subset Sum Problem, 2015.
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000579.

Sixth column of A003506.

[6*Binomial(n+1, 6): n in [5..40]]; // Vincenzo Librandi, Feb 13 2015

A253946:=n->6*binomial(n+1,6): seq(A253946(n), n=5..50); # Wesley Ivan Hurt, Feb 13 2015

Drop[Plus @@ Flatten[Part[#, 1 ;; 3] & /@ Subsets[Range@ #, {5}]] & /@
  Range@ 30, 4] (* Michael De Vlieger, Jan 20 2015 *)
6Binomial[Range[6, 29], 6] (* Alonso del Arte, Feb 05 2015 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1},{6,42,168,504,1260,2772,5544},40] (* Harvey P. Dale, May 14 2019 *)

Vec(6*x^5/(1-x)^7 + O(x^100)) \\ Colin Barker, Apr 03 2015

a(n) = 6*C(n+1,6) = 6*A000579(n+1).
G.f.: 6*x^5 / (1-x)^7. - Colin Barker, Apr 03 2015
From Amiram Eldar, Jan 09 2022: (Start)
Sum_{n>=5} 1/a(n) = 1/5.
Sum_{n>=5} (-1)^(n+1)/a(n) = 32*log(2) - 661/30. (End)

More terms from Vincenzo Librandi, Feb 13 2015

A061928 Array T(n,m) = 1/beta(n+1,m+1) read by antidiagonals.

Original entry on oeis.org

6, 12, 12, 20, 30, 20, 30, 60, 60, 30, 42, 105, 140, 105, 42, 56, 168, 280, 280, 168, 56, 72, 252, 504, 630, 504, 252, 72, 90, 360, 840, 1260, 1260, 840, 360, 90, 110, 495, 1320, 2310, 2772, 2310, 1320, 495, 110, 132, 660, 1980, 3960, 5544, 5544, 3960

Offset: 1

Frank Ellermann, May 22 2001

beta(n+1,m+1) = Integral_{x=0..1} x^n * (1-x)^m dx for real n, m.

			Antidiagonals:
   6,
  12, 12,
  20, 30, 20,
  30, 60, 60, 30,
  ...
Array:
   6  12  20   30   42
  12  30  60  105  168
  20  60 140  280  504
  30 105 280  630 1260
  42 168 504 1260 2772

G. Boole, A Treatise On The Calculus of Finite Differences, Dover, 1960, p. 26.

Rows: 1/b(n, 2): A002378, 1/b(n, 3): A027480, 1/b(n, 4): A033488. Diagonals: 1/b(n, n): A002457, 1/b(n, n+1) A005430, 1/b(n, n+2): A000917.

T(i, j)=A003506(i+1, j+1).

t[n_, m_] := 1/Beta[n+1, m+1]; Take[ Flatten[ Table[ t[n+1-m, m], {n, 1, 10}, {m, 1, n}]], 52] (* Jean-François Alcover, Oct 11 2011 *)

A(i,j)=if(i<1||j<1,0,1/subst(intformal(x^i*(1-x)^j),x,1)) /* Michael Somos, Feb 05 2004 */

A(i,j)=if(i<1||j<1,0,1/sum(k=0,i,(-1)^k*binomial(i,k)/(j+1+k))) /* Michael Somos, Feb 05 2004 */

from sympy import factorial as f
def T(n, m): return f(n + m + 1)/(f(n)*f(m))
for n in range(1, 11): print([T(m, n - m + 1) for m in range(1, n + 1)]) # Indranil Ghosh, Apr 29 2017

beta(n+1, m+1) = gamma(n+1)*gamma(m+1)/gamma(n+m+2) = n!*m!/(n+m+1)!.

A141611 Triangle read by rows: T(n, k) = (n-k+1)(k+1)binomial(n, k).

Original entry on oeis.org

1, 2, 2, 3, 8, 3, 4, 18, 18, 4, 5, 32, 54, 32, 5, 6, 50, 120, 120, 50, 6, 7, 72, 225, 320, 225, 72, 7, 8, 98, 378, 700, 700, 378, 98, 8, 9, 128, 588, 1344, 1750, 1344, 588, 128, 9, 10, 162, 864, 2352, 3780, 3780, 2352, 864, 162, 10, 11, 200, 1215, 3840, 7350, 9072, 7350, 3840, 1215, 200, 11

Offset: 0

Roger L. Bagula and Gary W. Adamson, Aug 22 2008

Read as a square array, this array factorizes as M*transpose(M), where M = ( k*binomial(n, k) )A003506(n,k).%20-%20_Peter%20Bala">{n,k>=1} = A003506(n,k). - _Peter Bala, Mar 06 2017

			Triangle begins as:
   1;
   2,   2;
   3,   8,    3;
   4,  18,   18,    4;
   5,  32,   54,   32,    5;
   6,  50,  120,  120,   50,    6;
   7,  72,  225,  320,  225,   72,    7;
   8,  98,  378,  700,  700,  378,   98,    8;
   9, 128,  588, 1344, 1750, 1344,  588,  128,    9;
  10, 162,  864, 2352, 3780, 3780, 2352,  864,  162,  10;
  11, 200, 1215, 3840, 7350, 9072, 7350, 3840, 1215, 200, 11;
  ...
From _Peter Bala_, Mar 06 2017: (Start)
Factorization as a square array
  /1         \ /1  2  3  4...\ /1  2   3   4...\
  |2  2      | |   2  6 12...| |2  8  12  32...|
  |3  6  3   |*|      3 12...|=|3 18  54 120...|
  |4 12 12 4 | |         4...| |4 32 120 320...|
  |...       | |             | |...            |
(End)

Indranil Ghosh, Table of n, a(n) for n = 1..5151 (Rows 0..100 of triangle, flattened)

Cf. A003506, A007466 (row sums), A037966, A085373.

A141611:= func< n,k | (k+1)*(n-k+1)*Binomial(n,k) >;
[A141611(n,k): k in [0..n], n in [0..14]]; // G. C. Greubel, Sep 22 2024

T[n_, m_]:= (n-m+1)*(m+1)*Binomial[n,m];
Table[T[n, m], {n,0,12}, {m,0,n}]//Flatten

T(n,m)=(n - m + 1)*(m + 1)*binomial(n, m) \\ Charles R Greathouse IV, Feb 15 2017

def A141611(n,k): return (k+1)*(n-k+1)*binomial(n,k)
flatten([[A141611(n,k) for k in range(n+1)] for n in range(15)]) # G. C. Greubel, Sep 22 2024

T(n, k) = (k+1)*(n-k+1)*binomial(n,k).
Sum_{k=0..n} T(n, k) = A007466(n+1) (row sums).
O.g.f.: (1 - (1 + t)*x + 2*t*x^2)/(1 - (1 + t)*x)^3 = 1 + (2 + 2*t)*x + (3 + 8*t + 3*t^2)*x^2 + (4 + 18*t + 18*t^2 + 4*t^3)*x^3 + .... - Peter Bala, Mar 06 2017
From G. C. Greubel, Sep 22 2024: (Start)
T(2*n, n) = A037966(n+1).
T(2*n-1, n) = 2*A085373(n-1), for n >= 1.
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n) - 2*[n=2]. (End)

Offset corrected by G. C. Greubel, Sep 22 2024

A215652 Exponential Riordan array [exp(x*exp(-x)),x].

Original entry on oeis.org

1, 1, 1, -1, 2, 1, -2, -3, 3, 1, 9, -8, -6, 4, 1, -4, 45, -20, -10, 5, 1, -95, -24, 135, -40, -15, 6, 1, 414, -665, -84, 315, -70, -21, 7, 1, 49, 3312, -2660, -224, 630, -112, -28, 8, 1, -10088, 441, 14904, -7980, -504, 1134, -168, -36, 9, 1

Offset: 0

Peter Bala, Sep 11 2012

For commuting lower unitriangular matrices A and B we define A raised to the matrix power B, denoted A^^B, to be the matrix Exp(B*Log(A)). Here Exp denotes the matrix exponential and the matrix logarithm Log(A) is defined as sum {n >= 1} (-1)^(n+1)*(A-1)^n/n. Call the present triangle X and let P denote Pascal's triangle A007318. Then X solves the matrix equation X^^P = P. Equivalently, the infinite tower of matrix powers X^^(X^^(X^^(....))) equals P. Note that the infinite tower of powers P^^(P^^(P^^(...))) of the Pascal triangle equals the hyperbinomial array A088956. Thus we might view the present array as the hypobinomial triangle.

			Triangle begins
.n\k.|....0.....1.....2.....3.....4.....5.....6.....7
= = = = = = = = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|....1.....1
..2..|...-1.....2.....1
..3..|...-2....-3.....3.....1
..4..|....9....-8....-6.....4.....1
..5..|...-4....45...-20...-10.....5.....1
..6..|..-95...-24...135...-40...-15.....6.....1
..7..|..414..-665...-84...315...-70...-21.....7.....1
...

G. Helms, Pascalmatrix tetrated

Cf. A003506, A003725 (column 0), A007318, A088956.

max = 9; MapIndexed[ Take[#1, #2[[1]]]&, CoefficientList[ Series[ Exp[x*t]*Exp[x*Exp[-x]], {x, 0, max}, {t, 0, max}], {x, t}]*Range[0, max]!, 1] // Flatten (* Jean-François Alcover, Jan 08 2013 *)

T(n,k) = binomial(n,k)*A003725(n-k).
The triangle equals P^^Q, where P is Pascal's triangle and Q is the inverse of P. Column 0 equals A003725.
E.g.f.: exp(x*t)*exp(x*exp(-x)) = 1 + (1 + t)*x + (-1 + 2*t + t^2)*x^2/2! + (-2 - 3*t + 3*t^2 + t^3)*x^3/3! + ....
The infinitesimal generator for this triangle is the generalized exponential Riordan array [x*exp(-x),x], which factors as [x,x]*[exp(-x),x] = A132440*A007318^(-1). The infinitesimal generator begins
..0
..1....0
.-2....2....0
..3...-6....3....0
.-4...12..-12....4....0
This is a signed version of the triangle of denominators from Leibniz's harmonic triangle - see A003506.

A134400 M * A007318, where M = triangle with (1, 1, 2, 3, ...) in the main diagonal and the rest zeros.

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 3, 9, 9, 3, 4, 16, 24, 16, 4, 5, 25, 50, 50, 25, 5, 6, 36, 90, 120, 90, 36, 6, 7, 49, 147, 245, 245, 147, 49, 7, 8, 64, 224, 448, 560, 448, 224, 64, 8, 9, 81, 324, 756, 1134, 1134, 756, 324, 81, 9, 10, 100, 450, 1200, 2100, 2520, 2100, 1200, 450, 100, 10

Offset: 0

Gary W. Adamson, Oct 23 2007

Row sums = A134401: (1, 2, 8, 24, 64, 160, 384, ...).
Triangle T(n,k), read by rows, given by [1,1,-1,1,0,0,0,0,0,...] DELTA [1,1,-1,1,0,0,0,0,0,...] where DELTA is the operator defined in A084938. A134402*A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 26 2007
For n > 0, from n athletes, select a team of k players and then choose a coach who is allowed to be on the team or not. - Geoffrey Critzer, Mar 13 2010
Row sums are A036289 if first term changed to zero. Diagonal sums are A023610, starting with the 2nd diagonal. Partial sums of diagonals are A002940 if first term changed to zero. - John Molokach, Jul 06 2013
For n > 0, T(n,k) is the number of states in Sokoban puzzle with n non-obstacles cells and k boxes (see Russell and Norvig at page 157). - Stefano Spezia, Dec 03 2023

			First few rows of the triangle:
  1;
  1,  1;
  2,  4,   2;
  3,  9,   9,   3;
  4, 16,  24,  16,   4;
  5, 25,  50,  50,  25,   5;
  6, 36,  90, 120,  90,  36,  6;
  7, 49, 147, 245, 245, 147, 49, 7;
  ...

Stuart Russell and Peter Norvig, Artificial Intelligence: A Modern Approach, Fourth Edition, Hoboken: Pearson, 2021.

Wikipedia, Sokoban

Cf. A002940, A003506, A007318, A036289, A134401, A134402.

T(2n,n) give A002011(n-1) for n>=1.

with(combstruct): for n from 0 to 10 do seq(`if`(n=0, 1, n)* count( Combination(n), size=m), m=0..n) od; # Zerinvary Lajos, Apr 09 2008

Join[{1},Table[Table[n*Binomial[n, k], {k,0, n}], {n, 10}]] //Flatten (* Geoffrey Critzer, Mar 13 2010 adapted by Stefano Spezia, Dec 03 2023 *)

From Geoffrey Critzer, Mar 13 2010: (Start)
T(0,0) = 1 and T(n,k) = n * binomial(n,k) for n > 0.
E.g.f. for column k is: (x^k/k!)*exp(x)*(x+k). (End)
T(n,k) = A003506(n,k) + A003506(n,k-1). - Geoffrey Critzer, Mar 13 2010
G.f.: (1-x-x*y+x^2+x^2*y+x^2*y^2)/(1-2*x-2*x*y+x^2+2*x^2*y+x^2*y^2). - Philippe Deléham, Nov 14 2013
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-2,k-2), T(0,0)=T(1,0)=T(1,1)=1, T(2,0)=T(2,2)=2, T(2,1)=4, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 14 2013
E.g.f.: 1 + exp(y*x)*exp(x)*(y*x + x). - Geoffrey Critzer, Mar 15 2015

a(55)-a(65) from Stefano Spezia, Dec 03 2023

A216973 Exponential Riordan array [x*exp(x),x].

Original entry on oeis.org

0, 1, 0, 2, 2, 0, 3, 6, 3, 0, 4, 12, 12, 4, 0, 5, 20, 30, 20, 5, 0, 6, 30, 60, 60, 30, 6, 0, 7, 42, 105, 140, 105, 42, 7, 0, 8, 56, 168, 280, 280, 168, 56, 8, 0, 9, 72, 252, 504, 630, 504, 252, 72, 9, 0, 10, 90, 360, 840, 1260, 1260, 840, 360, 90, 10, 0

Offset: 0

Peter Bala, Sep 21 2012

This is the triangle of denominators from Leibniz's harmonic triangle, A003506, augmented with a main diagonal of 0's.

Note, the usual definition of the exponential Riordan array [f(x), x*g(x)] associated with a pair of power series f(x) and g(x) requires f(0) to be nonzero. Here we don't make this assumption. - Peter Bala, Feb 13 2017

			Triangle begins
.n\k.|..0.....1.....2.....3.....4.....5.....6
= = = = = = = = = = = = = = = = = = = = = = =
..0..|..0
..1..|..1.....0
..2..|..2.....2.....0
..3..|..3.....6.....3.....0
..4..|..4....12....12.....4.....0
..5..|..5....20....30....20.....5.....0
..6..|..6....30....60....60....30.....6.....0
...

A003506, A007318, A059297, A116071, A132440, A215652.

A216973_row := proc(n) x*exp(x)*exp(x*t): series(%,x,n+1): n!*coeff(%,x,n):
seq(coeff(%,t,k), k=0..n) end:
for n from 0 to 10 do A216973_row(n) od; # Peter Luschny, Feb 03 2017

(* The function RiordanArray is defined in A256893. *)
RiordanArray[# Exp[#]&, Identity, 11, True] // Flatten (* Jean-François Alcover, Jul 16 2019 *)

T(n,k) = (n-k)*binomial(n,k) for 0 <= k <= n.
E.g.f.: x*exp(x)*exp(x*t) = 1 + x + (2 + 2*t)*x^2/2! + (3 + 6*t + 3*t^2)*x^3/3! + ....
The exponential Riordan array [x*exp(x),x] factors as [x,x]*[exp(x),x] = A132440*A007318.
This array is the infinitesimal generator for A116071; that is, Exp(A216973) = A116071, where Exp denotes the matrix exponential. A signed version of the array is the infinitesimal generator for A215652.
The first column of the array Exp(t*A216973) is the sequence of idempotent polynomials, the row polynomials of A059297.

A174002 a(n) = n*binomial(n+4, 4).

Original entry on oeis.org

0, 5, 30, 105, 280, 630, 1260, 2310, 3960, 6435, 10010, 15015, 21840, 30940, 42840, 58140, 77520, 101745, 131670, 168245, 212520, 265650, 328900, 403650, 491400, 593775, 712530, 849555, 1006880, 1186680, 1391280, 1623160, 1884960, 2179485

Offset: 0

Reinhard Zumkeller, Mar 05 2010, Mar 17 2010

This sequence can be computed from Pascal's triangle. Find the fifth number in a row and multiply it by the second number of the next row. - Alonso del Arte, Jan 21 2018

Index entries for linear recurrences with constant coefficients, signature (6, -15, 20, -15, 6, -1).

Cf. A033488, A027480, A003506.

[ (n^5+10*n^4+35*n^3+50*n^2+24*n)/24: n in [0..40] ]; // Vincenzo Librandi, Dec 28 2010

Table[n Binomial[n + 4, 4], {n, 0, 40}] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 5, 30, 105, 280, 630}, 40] (* Harvey P. Dale, Dec 03 2011 *)

a(n) = (n^5 + 10*n^4 + 35*n^3 + 50*n^2 + 24*n) / 24.
For n > 0: a(n) = A003506(n+4, 5).
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), with a(0)=0, a(1)=5, a(2)=30, a(3)=105, a(4)=280, a(5)=630. - Harvey P. Dale, Dec 03 2011
G.f.: 5*x/(1-x)^6. - Colin Barker, Mar 18 2012

Title switched with first Formula section entry, at the suggestion of Alonso del Arte, by Jon E. Schoenfield, Jan 28 2018

A178819 Pascal's prism (3-dimensional array) read by folded antidiagonal cross-sections: (h+i; h, i-j, j), h >= 0, i >= 0, 0 <= j <= i.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 3, 3, 1, 3, 6, 3, 3, 3, 1, 1, 4, 4, 6, 12, 6, 4, 12, 12, 4, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 5, 20, 30, 20, 5, 10, 30, 30, 10, 10, 20, 10, 5, 5, 1, 1, 6, 6, 15, 30, 15, 20, 60, 60, 20, 15, 60, 90, 60, 15, 6, 30, 60, 60, 30, 6, 1, 6, 15, 20, 15, 6, 1

Offset: 0

Harlan J. Brothers, Jun 16 2010

P_h = level h of Pascal's prism where P_1 = Pascal's triangle (A007318) and P_2 = denominators of Leibniz harmonic triangle (A003506). A sequence of length k through P is defined by P for n = {1, 2, 3, ..., k}.

			Prism begins (levels 1-4):
1
1 1
1 2 1
1 3 3 1
1
2 2
3 6 3
4 12 12 4
1
3 3
6 12 6
10 30 30 10
1
4 4
10 20 10
20 60 60 20

H. J. Brothers, Pascal's prism, The Mathematical Gazette, 96 (July 2012), 213-220.
H. J. Brothers, Pascal's Prism: Supplementary Material

Level 1 = A007318.
Level 2 = A003506.
Level 3 = A094305.
Level 4 = A178820.
Level 5 = A178821.
Level 6 = A178822.
Sums of shallow diagonals for each level correspond to rows of square A037027.
Contains A109649 and A046816.
P = A000984.
P = A006480.
P = A000897.
P<3n-2, 3n-2, n> = A113424.

end = 5; Column/@Table[Multinomial[h, i-j, j], {h, 0, end}, {i, 0, end}, {j, 0, i}]

a_(h, i, j) = (h+i-2; h-1, i-j, j-1), h >= 1, i >= 1, 1 <= j <= i.
Recurrence:
For P_h, element a is given by: a_(1, 1) = 1; a_(i, j) = ((i+h-2)/(i-1)) (a_(i-1, j) + a_(i-1, j-1)).

Keyword tabf by Michel Marcus, Oct 22 2017

cp's OEIS Frontend

A241269 Denominator of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)).

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A174341 a(n) = Numerator of Bernoulli(n, 1) + 1/(n+1).

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A253946 a(n) = 6*binomial(n+1, 6).

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A061928 Array T(n,m) = 1/beta(n+1,m+1) read by antidiagonals.

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A141611 Triangle read by rows: T(n, k) = (n-k+1)*(k+1)*binomial(n, k).

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A215652 Exponential Riordan array [exp(x*exp(-x)),x].

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A241269 Denominator of c(n) = (n^2+n+2)/((n+1)(n+2)(n+3)).

A141611 Triangle read by rows: T(n, k) = (n-k+1)(k+1)binomial(n, k).