A033487 -id:A033487 - cp's OEIS frontend

A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).

1, 2, 1, 3, 3, 2, 4, 6, 8, 6, 5, 10, 20, 30, 24, 6, 15, 40, 90, 144, 120, 7, 21, 70, 210, 504, 840, 720, 8, 28, 112, 420, 1344, 3360, 5760, 5040, 9, 36, 168, 756, 3024, 10080, 25920, 45360, 40320, 10, 45, 240, 1260, 6048, 25200, 86400, 226800, 403200, 362880

Offset: 1

Ross La Haye, Nov 15 2005

For k > 1, a(n,k) = the number of permutations of the symmetric group S_n that are pure k-cycles.
Reverse signed array is A238363. For a relation to (Cauchy-Euler) derivatives of the Vandermonde determinant, see Chervov link. - Tom Copeland, Apr 10 2014
Dividing the k-th column of T by (k-1)! for each column generates A135278 (the f-vectors, or face-vectors for the n-simplices). Then ignoring the first column gives A104712, so T acting on the column vector (-0,d,-d^2/2!,d^3/3!,...) gives the Euler classes for hypersurfaces of degree d in CP^n. Cf. A104712 and Dugger link therein. - Tom Copeland, Apr 11 2014
With initial i,j,n=1, given the n X n Vandermonde matrix V_n(x_1,...,x_n) with elements a(i=row,j=column)=(x_j)^(i-1), its determinant |V_n|, and the column vector of n ones C=(1,1,...,1), the n-th row of the lower triangular matrix T is given by the column vector determined by (1/|V_n|) * V_n(:x_1*d/dx_1:,...,:x_n*d/dx_n:)|V_n| * C, where :x_j*d/dx_j:^n = (x_j)^n*(d/dx_j)^n. - Tom Copeland, May 20 2014
For some other combinatorial interpretations of the first three columns of T, see A208535 and the link to necklace polynomials therein. Because of the simple relation of the array to the Pascal triangle, it can easily be related to many other arrays, e.g., T(p,k)/(p*(k-1)!) with p prime gives the prime rows of A185158 and A051168 when the non-integers are rounded to 0. - Tom Copeland, Oct 23 2014

			a(3,3) = 2 because (3-1)!C(3,3) = 2.
1;
2 1;
3 3 2;
4 6 8 6;
5 10 20 30 24;
6 15 40 90 144 120;
7 21 70 210 504 840 720;
8 28 112 420 1344 3360 5760 5040;
9 36 168 756 3024 10080 25920 45360 40320;

A. Chervov, A sum involving derivatives of Vandermonde

/* As triangle: */ [[Factorial(k-1)*Binomial(n,k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 21 2014

Flatten[Table[(k - 1)!Binomial[n, k], {n, 10}, {k, n}]]

a(n, k) = (k-1)!C(n, k) = P(n, k)/k.
E.g.f. (by columns) = exp(x)((x^k)/k).
a(n, 1) = A000027(n);
a(n, 2) = A000217(n-1);
a(n, 3) = A007290(n);
a(n, 4) = A033487(n-3).
a(n, n) = A000142(n-1);
a(n, n-1) = A001048(n-1) for n > 1.
Sum[a(n, k), {k, 1, n}] = A002104(n);
Sum[a(n, k), {k, 2, n}] = A006231(n).
a(n,k) = sum(j=k..n-1, j!/(j-k)!) (cf. Chervov link). - Tom Copeland, Apr 10 2014
From Tom Copeland, Apr 28 2014: (Start)
E.g.f. by row: [(1+t)^n-1]/t.
E.g.f. of row e.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
O.g.f. of row e.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
E.g.f. of row o.g.f.s: -exp(x) * log(1-t*x). (End)

A052149 Number of nonsquare rectangles on an n X n board.

Original entry on oeis.org

0, 4, 22, 70, 170, 350, 644, 1092, 1740, 2640, 3850, 5434, 7462, 10010, 13160, 17000, 21624, 27132, 33630, 41230, 50050, 60214, 71852, 85100, 100100, 117000, 135954, 157122, 180670, 206770, 235600, 267344, 302192, 340340, 381990, 427350, 476634

Offset: 1

Ronald Arms (ron.arms(AT)stanfordalumni.org), Jan 23 2000

Partial sums of A045991 (n^3-n^2). - Jeremy Gardiner, Jun 30 2013

			a(10) = 10 * 9 * 11 * 32 / 12 = 2640.
a(5) = 170 and the sum from 1 to 5 is 15, giving 1*(15-1)=14, 2*(15-2)=26, 2*(15-3)=36, 4*(15-4)=44 and 5*(15-5)=50; adding 14+26+36+44+50=170. Do the same for each n and get a(n). - _J. M. Bergot_, Oct 31 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Project Euler, Sum square difference: Problem 6.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A035291, A045991, A033487, A007290, A000537, A000330, A000914.

I:=[0, 4, 22, 70, 170]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..45]]; // Vincenzo Librandi, Apr 28 2012

a:=n->sum(j^3-j^2, j=0..n): seq(a(n), n=1..37); # Zerinvary Lajos, May 08 2008

CoefficientList[Series[2*x*(2+x)/(1-5*x+10*x^2-10*x^3+ 5*x^4-x^5), {x,0,50}], x] (* Vincenzo Librandi, Apr 28 2012 *)
LinearRecurrence[{5,-10,10,-5,1},{0,4,22,70,170},40] (* Harvey P. Dale, Jul 30 2019 *)

a(n) = sum(k=1,n,(k-1)*k^2) \\ Michel Marcus, Nov 09 2012

a(n) = n*(n-1)*(n+1)*(3*n+2)/12.
G.f.: 2*x^2*(2+x)/(1-5*x+10*x^2-10*x^3+5*x^4-x^5). - Colin Barker, Jan 04 2012
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Vincenzo Librandi, Apr 28 2012
a(n) = A033487(n-1) - A007290(n+1) starting at n=1. - J. M. Bergot, Jun 04 2012
a(n) = Sum_{k=1..n} (k-1)*k^2. - Michel Marcus, Nov 09 2012
a(n) = A000537(n) - A000330(n) = 2*A000914(n-1). - Luciano Ancora, Mar 16 2015
From Amiram Eldar, Jan 10 2022: (Start)
Sum_{n>=2} 1/a(n) = 81*log(3)/5 - 9*sqrt(3)*Pi/5 - 192/25.
Sum_{n>=2} (-1)^n/a(n) = 18*sqrt(3)*Pi/5 - 48*log(2)/5 - 318/25. (End)

A109649 Entries in 3-dimensional version of Pascal triangle: trinomial coefficients.

Original entry on oeis.org

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

Offset: 0

Philippe Deléham, Aug 03 2005

Greatest numbers in each 2D triangle form A022916 (multinomial coefficient n!/([n/3]![(n+1)/3]![(n+2)/3]!)).
2D triangle sums are powers of 3.
See A046816 for another version.

			.1 3 3 1 ... Here is the third slice of the pyramid
. 3 6 3
.. 3 3
... 1 .....

Cf. A022916, A046816.

Cf. A007318, A002378, A027480, A033487, A033488.

Coefficients of x, y, z in (x+y+z)^n.

A033486 a(n) = n(n + 1)(n + 2)*(n + 3)/2.

Original entry on oeis.org

0, 12, 60, 180, 420, 840, 1512, 2520, 3960, 5940, 8580, 12012, 16380, 21840, 28560, 36720, 46512, 58140, 71820, 87780, 106260, 127512, 151800, 179400, 210600, 245700, 285012, 328860, 377580, 431520, 491040, 556512, 628320, 706860, 792540, 885780, 987012

Offset: 0

N. J. A. Sloane

a(n) is the area of an irregular quadrilateral with vertices at (1,1), (n+1, n+2), ((n+1)^2, (n+2)^2) and ((n+1)^3, (n+2)^3). - Art Baker, Dec 08 2018

G. C. Greubel, Table of n, a(n) for n = 0..5000 (terms 0..680 from Vincenzo Librandi)
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000332, A050534, A033488, A033487, A060008.

List([0..40],n->n*(n+1)*(n+2)*(n+3)/2); # Muniru A Asiru, Dec 08 2018

[n*(n+1)*(n+2)*(n+3)/2: n in [0..40]]; // Vincenzo Librandi, Apr 28 2011

[seq(12*binomial(n+3,4),n=0..32)]; # Zerinvary Lajos, Nov 24 2006

Table[n*(n + 1)*(n + 2)*(n + 3)/2, {n, 0, 50}] (* David Nacin, Mar 01 2012 *)
LinearRecurrence[{5,-10,10,-5,1},{0,12,60,180,420},40] (* Harvey P. Dale, Feb 04 2015 *)

a(n)=n*(n+1)*(n+2)*(n+3)/2 \\ Charles R Greathouse IV, Oct 07 2015

[12*binomial(n+3,4) for n in range(40)] # G. C. Greubel, Dec 08 2018

a(n) = 6*A034827(n+3) = 12*A000332(n+3).
G.f.: 12*x/(1 - x)^5. - Colin Barker, Mar 01 2012
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) with a(0) = 0, a(1) = 12, a(2) = 60, a(3) = 180, a(4) = 420. - Harvey P. Dale, Feb 04 2015
E.g.f.: (24*x + 36*x^2 + 12*x^3 + x^4)*exp(x)/2. - Franck Maminirina Ramaharo, Dec 08 2018
From Amiram Eldar, Sep 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*(3*log(2)-2)/9. (End)

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)binomial(n,k)binomial(n+k,k) (n >= k >= 0).

Original entry on oeis.org

-1, -1, 2, -1, 6, -6, -1, 12, -30, 20, -1, 20, -90, 140, -70, -1, 30, -210, 560, -630, 252, -1, 42, -420, 1680, -3150, 2772, -924, -1, 56, -756, 4200, -11550, 16632, -12012, 3432, -1, 72, -1260, 9240, -34650, 72072, -84084, 51480, -12870, -1, 90, -1980, 18480, -90090, 252252, -420420, 411840, -218790, 48620, -1, 110, -2970, 34320, -210210, 756756, -1681680, 2333760, -1969110, 923780, -184756

Offset: 0

N. J. A. Sloane, Jan 17 2020

This is Table I of Ser (1933), page 92.
From Petros Hadjicostas, Jul 09 2020: (Start)
Essentially Ser (1933) in his book (and in particular for Tables I-IV) finds triangular arrays that allow him to express the coefficients of various kinds of series in terms of the coefficients of other series.
He uses Newton's series (or some variation of it), factorial series, and inverse factorial series. Unfortunately, he uses unusual notation, and as a result it is difficult to understand what he is actually doing.
Rivoal (2008, 2009) essentially uses factorial series and transformations to other kinds of series to provide new proofs of the irrationality of log(2), zeta(2), and zeta(3). As a result, the triangular array T(n,k) appears in various parts of his papers.
We believe Table I (p. 92) in Ser (1933), regarding the numbers T(n,k), corresponds to four different formulas. We have deciphered the first two of them. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k=0..n) begins:
  -1;
  -1,  2;
  -1,  6,   -6;
  -1, 12,  -30,   20;
  -1, 20,  -90,  140,    -70;
  -1, 30, -210,  560,   -630,   252;
  -1, 42, -420, 1680,  -3150,  2772,   -924;
  -1, 56, -756, 4200, -11550, 16632, -12012, 3432;
  ...
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92) is
  -1;
  -1/2, 1/2;
  -1/3, 1/2,   -1/6;
  -1/4, 9/20,  -1/4,  1/20;
  -1/5, 2/5,   -2/7,  1/10, -1/70;
  -1/6, 5/14, -25/84, 5/36, -1/28,  1/252;
  -1/7, 9/28, -25/84, 1/6,  -9/154,  1/84, -1/924;
   ... (End)

J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, pp. 92-93.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
A. Buhl, Book review: J. Ser - Les calculs formels des séries de factorielles, L'Enseignement Mathématique, 32 (1933), p. 275.
L. A. MacColl, Review: J. Ser, Les calculs formels des séries de factorielles, Bull. Amer. Math. Soc., 41(3) (1935), p. 174.
L. M. Milne-Thomson, Review of Les calculs formels des séries de factorielles. By J. Ser. Pp. vii, 98. 20 fr. 1933. (Gauthier-Villars), The Mathematical Gazette, Vol. 18, No. 228 (May, 1934), pp. 136-137.
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages)
Tanguy Rivoal, Applications arithmétiques de l'interpolation lagrangienne, preprint (2008); see pp. 1 and 15.
Tanguy Rivoal, Applications arithmétiques de l'interpolation lagrangienne, Int. J. Number Theory 5.2 (2009), 185-208; see pp. 185 and 199.

A063007 is the same triangle without the minus signs, and has much more information.
Columns 1 and 2 are A002378 and A033487; the last three diagonals are A002544, A002457, A000984.
Cf. A331431, A331432.

/* As triangle: */ [[(-1)^(k+1) * Factorial(n+k) / (Factorial(k) * Factorial(k) * Factorial(n-k)): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jan 19 2020

Table[CoefficientList[-Hypergeometric2F1[-n, n + 1, 1, x], x], {n, 0, 9}] // Flatten (* Georg Fischer, Jan 18 2020 after Peter Luschny in A063007 *)

def T(n,k): return (-1)^(k+1)*falling_factorial(n+k,2*k)/factorial(k)^2
flatten([[T(n,k) for k in (0..n)] for n in (0..10)]) # Peter Luschny, Jul 09 2020

T(n,k) can also be written as (-1)^(k+1)*(n+k)!/(k!*k!*(n-k)!).
From Petros Hadjicostas, Jul 09 2020: (Start)
Ser's first formula from his Table I (p. 92) is the following:
Sum_{k=0..n} T(n,k)*k!/(x*(x+1)*...*(x+k)) = -(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, Sum_{k=0..n} T(n,k)/binomial(m+k, k) = 0 for m = 1..n.
Ser's second formula from his Table I appears also in Rivoal (2008, 2009) in a slightly different form:
Sum_{k=0..n} T(n,k)/(x + k) = (-1)^(n+1)*(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, for m = 1..n, Sum_{k=0..n} T(n,k)/(m + k) = 0. (End)
T(n,k) = (-1)^(k+1)*FallingFactorial(n+k,2*k)/(k!)^2. - Peter Luschny, Jul 09 2020
From Petros Hadjicostas, Jul 10 2020: (Start)
Peter Luschny's formula above is essentially the way the numbers T(n,k) appear in Eq. (7) on p. 86 of Ser's (1933) book. Eq. (7) is essentially equivalent to the first formula above (related to Table I on p. 92).
By inverting that formula (in some way), he gets
n!/(x*(x+1)*...*(x+n)) = Sum_{p=0..n} (-1)^p*(2*p+1)*f_p(n+1)*f_p(x), where f_p(x) = (x-1)*...*(x-p)/(x*(x+1)*...*(x+p)). This is equivalent to Eq. (8) on p. 86 of Ser's book.
The rational coefficients A(n,p) = (2*p+1)*f_p(n+1) = (2*p+1)*(n*(n-1)*...*(n+1-p))/((n+1)*...*(n+1+p)) appear in Table II on p. 92 of Ser's book.
If we consider the coefficients T(n,k) and (-1)^(p+1)*A(n,p) as infinite lower triangular matrices, then they are inverses of one another (see the example below). This means that, for m >= s,
Sum_{k=s..m} T(m,k)*(-1)^(s+1)*A(k,s) = I(s=m) = Sum_{k=s..m} (-1)^(k+1)*A(m,k)*T(k,s), where I(s=m) = 1, if s = m, and = 0, otherwise.
Without the (-1)^p, we get the formula
1/(x+n) = Sum_{p=0..n} (2*p+1)*f_p(n+1)*f_p(x),
which apparently is the inversion of the second of Ser's formulas (related to Table I on p. 92).
In all of the above formulas, an empty product is by definition 1, so f_0(x) = 1/x. (End)

Thanks to Bob Selcoe, who noticed a typo in one of the entries, which, when corrected, led to an explicit formula for the whole of Ser's Table I.

A129533 Array read by antidiagonals: T(n,k) = binomial(n+1,2)*binomial(n+k,n+1) for 0 <= k <= n.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 3, 3, 0, 0, 6, 12, 6, 0, 0, 10, 30, 30, 10, 0, 0, 15, 60, 90, 60, 15, 0, 0, 21, 105, 210, 210, 105, 21, 0, 0, 28, 168, 420, 560, 420, 168, 28, 0, 0, 36, 252, 756, 1260, 1260, 756, 252, 36, 0, 0, 45, 360, 1260, 2520, 3150, 2520, 1260, 360, 45, 0, 0, 55, 495

Offset: 0

Emeric Deutsch, Apr 22 2007

Previous name was: Triangle read by rows: T(n,k)=derivative of the q-binomial coefficient [n,k] evaluated at q=1 (0<=k<=n). - N. J. A. Sloane, Jan 06 2016
For example, T(5,2)=30 because [5,2] = q^6 + q^5 + 2*q^4 + 2*q^3 + 2*q^2 + q + 1 with derivative 6q^5 + 5q^4 + 8q^3 + 6q^2 + 4q + 1, having value 30 at q=1. - Emeric Deutsch, Apr 22 2007
Sum of entries in n-th antidiagonal = n(n-1)2^(n-3) = A001788(n-1).
T(n,k) = A094305(n-2, k-1) for n >= 2, k >= 1.
T(n,k) is total number of pips on a set of generalized linear dominoes with n cells (rather than two) and with the number of pips in each cell running from 0 to k (rather than 6). T(2,6) = 168 gives the total number of pips on a standard set of dominoes. We regard a generalized linear domino with n cells and up to k pips per cell as an ordered n-tuple [i_1, i_2, ..., i_n] with 0 <= i_1 <= i_2 <=  ... <= i_n <= k. - Alan Shore and N. J. A. Sloane, Jan 06 2016
T(n,k) can also be written more symmetrically as the trinomial coefficient (n+k; n-1, k-1, 2). - N. J. A. Sloane, Jan 06 2016
As a triangle read by rows, T(n,k) is the total number of inversions over all length n binary words having exactly k 1's. T(n,k) is also the total area above all North East lattice paths from the origin to the point (k,n-k). - Geoffrey Critzer, Mar 22 2018

			Array begins:
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ... (A000004)
0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, ... (A000217)
0, 3, 12, 30, 60, 105, 168, 252, 360, 495, 660, 858, ... (A027480)
0, 6, 30, 90, 210, 420, 756, 1260, 1980, 2970, 4290, ... (A033487)
0, 10, 60, 210, 560, 1260, 2520, 4620, 7920, 12870, ... (A266732)
0, 15, 105, 420, 1260, 3150, 6930, 13860, 25740, ... (A240440)
0, 21, 168, 756, 2520, 6930, 16632, 36036, ... (A266733)
...
If regarded as a triangle, this begins:
  0;
  0,  0;
  0,  1,  0;
  0,  3,  3,  0;
  0,  6, 12,  6,  0;
  0, 10, 30, 30, 10,  0;
  0, 15, 60, 90, 60, 15, 0;
  ...

G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976.

Cemil Karaçam and Alper Vural, Enumerating 2D and 3D lattice paths with arbitrary steps, Mathematica Bohemica, pp. 1-13 (2025). See p. 5.
Index entries for sequences related to dominoes.

Cf. A001788.
Rows give A000004, A000217, A027480, A033487, A266732, A240440, A266733.
A128503 and A094305 are very similar sequences.

dd:=proc(n,m) if m=0 or n=0 then 0 else (m+n)!/(2*(m-1)!*(n-1)!); fi; end;
f:=n->[seq(dd(n,m),m=0..30)];
for n from 0 to 10 do lprint(f(n)); od: # produces sequence as square array
T:=(n,k)->k*(k+1)*binomial(n,k+1)/2: for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

Table[Table[D[Expand[FunctionExpand[QBinomial[n, k, q]]], q] /. q -> 1, {k, 0, n}], {n, 0, 15}] // Grid (* Geoffrey Critzer, Mar 22 2018 *)

T(n,k) = (1/2)*k*(k+1)*binomial(n,k+1).

G.f.: G(q,z) = qz^2/(1-z-qz)^3.

Entry revised by N. J. A. Sloane, Jan 06 2016

A065420 Triangle T(n,k) = binomial(n+2,k+1)*(binomial(n+2,k+1)-1), n >=0, 0 <= k <= n.

Original entry on oeis.org

2, 6, 6, 12, 30, 12, 20, 90, 90, 20, 30, 210, 380, 210, 30, 42, 420, 1190, 1190, 420, 42, 56, 756, 3080, 4830, 3080, 756, 56, 72, 1260, 6972, 15750, 15750, 6972, 1260, 72, 90, 1980, 14280, 43890, 63252, 43890, 14280, 1980, 90, 110, 2970, 27060, 108570, 212982

Offset: 0

Gary W. Adamson, Nov 15 2001

T(n,k) = T(n,n-k). - Robert Israel, Jan 08 2017

			2; 6,6; 12,30,12; 20,90,90,20; ...

Robert Israel, Table of n, a(n) for n = 0..10010 (rows 0 to 140, flattened)

Cf. A002378, A033487.

T:= (n,k) -> binomial(n+2,k+1)*(binomial(n+2,k+1)-1):
seq(seq(T(n,k),k=0..n),n=0..10); # Robert Israel, Jan 08 2017

#(#-1)&/@Table[Binomial[n+2,k+1],{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Sep 02 2018 *)

From Robert Israel, Jan 08 2017:  (Start)
T(n,0) = (n+1)*(n+2) = A002378(n+1).
T(n,1) = n*(n+1)*(n+2)*(n+3)/4 = A033487(n). (End)

More terms from Naohiro Nomoto, Nov 22 2001

A091478 Table of graphs with n (>=0) nodes and k (>=0) edges. Each type of object labeled from its own label set.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 6, 6, 1, 6, 30, 120, 360, 720, 720, 1, 10, 90, 720, 5040, 30240, 151200, 604800, 1814400, 3628800, 3628800, 1, 15, 210, 2730, 32760, 360360, 3603600, 32432400, 259459200, 1816214400, 10897286400, 54486432000, 217945728000, 653837184000, 1307674368000, 1307674368000

Offset: 0

Christian G. Bower, Jan 13 2004

			  1;
  1;
  1, 1;
  1, 3,  6,   6;
  1, 6, 30, 120, 360, 720, 720;
  ...

F. Bergeron, G. Labelle and P. Leroux, Combinatorial Species and Tree-Like Structures, Cambridge, 1998, p. 114 (2.4.44).

Row sums: A091479.
Row lengths: A000124(n-1) for n>=1.
Columns 0-2: A000012, A000217(n-1), A033487(n-2).
a(n,A000217(n-1)) = A052295(n-1).
Cf. A006125, A008406.

a(n, k) = k!*binomial(binomial(n, 2), k).

T(0,0)=1 prepended by Alois P. Heinz, Feb 14 2023

A107394 C(n+2,2)*C(n+4,2).

Original entry on oeis.org

6, 30, 90, 210, 420, 756, 1260, 1980, 2970, 4290

Offset: 0

Zerinvary Lajos, May 25 2005

dead

Duplicates A033487.

			If n=0 then C(0+2,2)*C(0+4,2)= C(2,2)*C(4,2)=1*6=6
If n=4 then C(4+2,2)*C(4+4,2)= C(6,2)*C(8,2)=15*28=420

Cf. A062196.

a(n)=A033487(n+1). [From R. J. Mathar, Aug 24 2008]

A124051 Quasi-mirror of A062196 formatted as a triangular array.

Original entry on oeis.org

3, 6, 8, 10, 30, 15, 15, 80, 90, 24, 21, 175, 350, 210, 35, 28, 336, 1050, 1120, 420, 48, 36, 588, 2646, 4410, 2940, 756, 63, 45, 960, 5880, 14112, 14700, 6720, 1260, 80, 55, 1485, 11880, 38808, 58212, 41580, 13860, 1980, 99, 66, 2200, 22275, 95040, 194040, 199584, 103950, 26400, 2970, 120

Offset: 0

Zerinvary Lajos, Nov 03 2006

			Triangle begins as:
   3;
   6,    8;
  10,   30,    15;
  15,   80,    90,    24;
  21,  175,   350,   210,     35;
  28,  336,  1050,  1120,    420,     48;
  36,  588,  2646,  4410,   2940,    756,     63;
  45,  960,  5880, 14112,  14700,   6720,   1260,    80;
  55, 1485, 11880, 38808,  58212,  41580,  13860,  1980,   99;
  66, 2200, 22275, 95040, 194040, 199584, 103950, 26400, 2970, 120;

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

Cf. A000108, A000894, A062196, A286033.
Columns k: A000217(n+2) (k=0), A002417(n+1) (k=1), A001297(n) (k=2), A105946(n-2) (k=3), A105947(n-3) (k=4), A105948(n-4) (k=5), A107319(n-5) (k=6).
Diagonals: A005563(n+1) (k=n), A033487(n) (k=n-1), A027790(n) (k=n-2), A107395(n-3) (k=n-3), A107396(n-4) (k=n-4), A107397(n-5) (k=n-5), A107398(n-6) (k=n-6), A107399(n-7) (k=n-7).
Sums: A322938(n+1) (row).

A124051:= func< n,k | Binomial(n+1,n-k+1)*Binomial(n+3,n-k+1) >;
[A124051(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 07 2025

for n from 0 to 10 do seq(binomial(n,i-1)*binomial(n+2,n+1-i), i=1..n ) od;

A124051[n_, k_]:= Binomial[n+1,n-k+1]*Binomial[n+3,n-k+1];
Table[A124051[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 07 2025 *)

def A124051(n,k): return binomial(n+1,n-k+1)*binomial(n+3,n-k+1)
print(flatten([[A124051(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Feb 07 2025

From G. C. Greubel, Feb 07 2025: (Start)
T(n, k) = binomial(n+1, n-k+1)*binomial(n+3, n-k+1).
T(2*n, n) = (1/2)*A000894(n) + (5/2)*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*( (1+(-1)^n)*(-1)^(n/2)*A286033((n+4)/2) + (1-(-1)^n)*((-1)^((n+1)/2)*A000108((n+1)/2) - 1) ). (End)

cp's OEIS Frontend

A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).

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A052149 Number of nonsquare rectangles on an n X n board.

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A109649 Entries in 3-dimensional version of Pascal triangle: trinomial coefficients.

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A033486 a(n) = n*(n + 1)*(n + 2)*(n + 3)/2.

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A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)*binomial(n,k)*binomial(n+k,k) (n >= k >= 0).

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A129533 Array read by antidiagonals: T(n,k) = binomial(n+1,2)*binomial(n+k,n+1) for 0 <= k <= n.

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A065420 Triangle T(n,k) = binomial(n+2,k+1)*(binomial(n+2,k+1)-1), n >=0, 0 <= k <= n.

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A033486 a(n) = n(n + 1)(n + 2)*(n + 3)/2.

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)binomial(n,k)binomial(n+k,k) (n >= k >= 0).