A104712 -id:A104712 - 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)

A045501 Third-from-right diagonal of triangle A121207.

Original entry on oeis.org

1, 1, 4, 14, 54, 233, 1101, 5625, 30846, 180474, 1120666, 7352471, 50772653, 367819093, 2787354668, 22039186530, 181408823710, 1551307538185, 13756835638385, 126298933271289, 1198630386463990, 11742905240821910

Offset: 1

Henry Gould

With leading 0 and offset 2: number of permutations beginning with 321 and avoiding 1-23. - Ralf Stephan, Apr 25 2004
Second diagonal in table of binomial recurrence coefficients. Related to A040027. - Vladeta Jovovic, Feb 05 2008
Equals eigensequence of triangle A104712. - Gary W. Adamson, Apr 10 2009
a(n) is the number of set partitions of {1,2,...,n+1} in which the last block has length 2; the blocks are arranged in order of their least element. - Don Knuth, Jun 12 2017

S. Kitaev, Generalized pattern avoidance with additional restrictions, Sem. Lothar. Combinat. B48e (2003).
S. Kitaev and T. Mansour, Simultaneous avoidance of generalized patterns, arXiv:math/0205182 [math.CO], 2002.

Cf. A045499, A045500.
Cf. A104712. - Gary W. Adamson, Apr 10 2009
Column k=2 of A124496.

a[1] = a[2] = 1; a[n_] := a[n] = Sum[Binomial[n, k+1]*a[k], {k, 0, n-1}];
Array[a, 22] (* Jean-François Alcover, Jul 14 2018, after Vladeta Jovovic *)

{a(n)=local(A=x+x^2); for(i=1, n, A=x+x*subst(A, x, x/(1-x+x*O(x^n)))/(1-x)^2); polcoeff(A, n)} /* Paul D. Hanna, Mar 23 2012 */

# The function Gould_diag is defined in A121207.
A045501_list = lambda size: Gould_diag(3, size)
print(A045501_list(24)) # Peter Luschny, Apr 24 2016

a(n+1) = Sum_{k=0..n} binomial(n+2, k+2)*a(k). - Vladeta Jovovic, Nov 10 2003
With offset 2, e.g.f.: x^2 + exp(exp(x))/2 * Integral_{0..x} t^2*exp(-exp(t)+t) dt. - Ralf Stephan, Apr 25 2004
G.f.: A(x) = Sum_{k>=0} x^(k+1)/((1-k*x)^2 * Product_{m=0..k} (1 - m*x)). - Vladeta Jovovic, Feb 05 2008
O.g.f. satisfies: A(x) = x + x*A( x/(1-x) ) / (1-x)^2. - Paul D. Hanna, Mar 23 2012

More terms from Vladeta Jovovic, Nov 10 2003

Entry revised by N. J. A. Sloane, Dec 11 2006

A014831 a(1)=1; for n>1, a(n) = 8*a(n-1) + n.

Original entry on oeis.org

1, 10, 83, 668, 5349, 42798, 342391, 2739136, 21913097, 175304786, 1402438299, 11219506404, 89756051245, 718048409974, 5744387279807, 45955098238472, 367640785907793, 2941126287262362, 23529010298098915, 188232082384791340, 1505856659078330741, 12046853272626645950

Offset: 1

N. J. A. Sloane

			For n=5, a(5) = 1*15 + 7*20 + 7^2*15 + 7^3*6 + 7^4*1 = 5349. [_Bruno Berselli_, Nov 13 2015]

Colin Barker, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 18.
Index entries for linear recurrences with constant coefficients, signature (10,-17,8).

Cf. A000420, A104712.

a:=n->sum((8^(n-j)-1)/7,j=0..n): seq(a(n), n=1..19); # Zerinvary Lajos, Jan 15 2007
a:= n-> (Matrix ([[1, 0, 1], [1, 1, 1], [0, 0, 8]])^n)[2, 3]: seq (a(n), n=1..25); # Alois P. Heinz, Aug 06 2008

Table[(8^(n + 1) - 7 n - 8)/49, {n, 1, 25}] (* Bruno Berselli, Nov 13 2015 *)
nxt[{n_,a_}]:={n+1,8a+n+1}; NestList[nxt,{1,1},30][[;;,2]] (* Harvey P. Dale, Aug 09 2025 *)

Vec(x/((1 - x)^2*(1 - 8*x)) + O(x^25)) \\ Colin Barker, Jun 03 2020

a(n) = (8^(n+1) - 7*n - 8)/49. - Rolf Pleisch, Oct 21 2010
a(n) = Sum_{i=0..n-1} 7^i*binomial(n+1,n-1-i). - Bruno Berselli, Nov 13 2015
From Colin Barker, Jun 03 2020: (Start)
G.f.: x/((1 - x)^2*(1 - 8*x)).
a(n) = 10*a(n-1) - 17*a(n-2) + 8*a(n-3) for n > 3. (End)
E.g.f.: exp(x)*(8*exp(7*x) - 7*x - 8)/49. - Elmo R. Oliveira, Mar 29 2025

A014830 a(1)=1; for n > 1, a(n) = 7*a(n-1) + n.

Original entry on oeis.org

1, 9, 66, 466, 3267, 22875, 160132, 1120932, 7846533, 54925741, 384480198, 2691361398, 18839529799, 131876708607, 923136960264, 6461958721864, 45233711053065, 316635977371473, 2216451841600330, 15515162891202330, 108606140238416331, 760242981668914339, 5321700871682400396

Offset: 1

N. J. A. Sloane

			For n=5, a(5) = 1*15 + 6*20 + 6^2*15 + 6^3*6 + 6^4*1 = 3267. - _Bruno Berselli_, Nov 13 2015

Colin Barker, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 18.
Index entries for linear recurrences with constant coefficients, signature (9,-15,7).

Row n=7 of A126885.

Cf. A000400, A104712.

a:=n->sum((7^(n-j)-1)/6,j=0..n): seq(a(n), n=1..19); # Zerinvary Lajos, Jan 15 2007

a[1] = 1; a[n_] := 7*a[n-1]+n; Table[a[n], {n, 10}] (* Zak Seidov, Feb 06 2011 *)
LinearRecurrence[{9, -15, 7}, {1, 9, 66}, 30] (* Harvey P. Dale, Jul 22 2013 *)

Vec(x/((1 - x)^2*(1 - 7*x)) + O(x^25)) \\ Colin Barker, Jun 03 2020

a(n) = (7^(n+1) - 6*n - 7)/36. - Rolf Pleisch, Oct 19 2010
a(1)=1, a(2)=9, a(3)=66; for n > 3, a(n) = 9*a(n-1) - 15*a(n-2) + 7*a(n-3). - Harvey P. Dale, Jul 22 2013
a(n) = Sum_{i=0..n-1} 6^i*binomial(n+1,n-1-i). - Bruno Berselli, Nov 13 2015
G.f.: x/((1 - x)^2*(1 - 7*x)). - Colin Barker, Jun 03 2020
E.g.f.: exp(x)*(7*exp(6*x) - 6*x - 7)/36. - Elmo R. Oliveira, Mar 29 2025

A014832 a(1)=1; for n>1, a(n) = 9*a(n-1) + n.

Original entry on oeis.org

1, 11, 102, 922, 8303, 74733, 672604, 6053444, 54481005, 490329055, 4412961506, 39716653566, 357449882107, 3217048938977, 28953440450808, 260580964057288, 2345228676515609, 21107058088640499, 189963522797764510, 1709671705179880610, 15387045346618925511, 138483408119570329621

Offset: 1

N. J. A. Sloane

			For n=5, a(5) = 1*15 + 8*20 + 8^2*15 + 8^3*6 + 8^4*1 = 8303. [_Bruno Berselli_, Nov 13 2015]

Seiichi Manyama, Table of n, a(n) for n = 1..1000
Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 18.
Index entries for linear recurrences with constant coefficients, signature (11,-19,9).

Cf. A001018, A104712.

a:=n->sum((9^(n-j)-1)/8,j=0..n): seq(a(n), n=1..18); # Zerinvary Lajos, Jan 15 2007
a:= n-> (Matrix([[1,0,1],[1,1,1],[0,0,9]])^n)[2,3]: seq(a(n), n=1..18); # Alois P. Heinz, Aug 06 2008

RecurrenceTable[{a[1]==1,a[n]==9a[n-1]+n},a,{n,20}] (* or *) LinearRecurrence[ {11,-19,9},{1,11,102},20] (* Harvey P. Dale, May 01 2012 *)

a(n) = (9^(n+1) - 8*n - 9)/64. - Rolf Pleisch, Oct 22 2010
From Harvey P. Dale, May 01 2012: (Start)
a(1)=1, a(2)=11, a(3)=102; for n>3, a(n) = 11*a(n-1) - 19*a(n-2) + 9*a(n-3).
G.f.: -x/((x-1)^2*(9*x-1)). (End)
a(n) = Sum_{i=0..n-1} 8^i*binomial(n+1,n-1-i). - Bruno Berselli, Nov 13 2015
E.g.f.: exp(x)*(9*exp(8*x) - 8*x - 9)/64. - Elmo R. Oliveira, Mar 29 2025

A126277 Triangle generated from Eulerian numbers.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 7, 4, 1, 5, 11, 15, 5, 1, 6, 15, 26, 31, 6, 1, 7, 19, 37, 57, 63, 7, 1, 8, 23, 48, 83, 120, 127, 8, 1, 9, 27, 59, 109, 177, 247, 255, 9, 1, 10, 31, 70, 135, 234, 367, 502, 511, 10

Offset: 1

Gary w. Adamson, Dec 23 2006

N-th diagonal starting from the right = binomial transform of [1, N, q, q, q, ...) where q = 2*N - 2. Given the infinite set of triangles "T" composed of partial column sums of the polygonal numbers, the N-th diagonal starting from the right = row sums of triangle "T": (T=3 = A104712; T=4 = A125165; T=5 = A125232; T=6 = A125233; T=7 = A125234, T=8 = A125235; and so on). For example, 3rd diagonal from the right = the offset Eulerian numbers, (1, 4, 11, 26, 57, 120, ...) = row sums of Triangle A104712 having partial column sums of the triangular numbers: 1; 3, 1; 6, 4, 1; 10, 10, 5, 1; 15, 20, 15, 6, 1; ... Row sums = A124671: (1, 3, 7, 16, 37, 85, 191, ...).

			First few rows of the triangle:
  1;
  1,  2;
  1,  3,  3;
  1,  4,  7,  4;
  1,  5, 11, 15,   5;
  1,  6, 15, 26,  31,   6;
  1,  7, 19, 37,  57,  63,   7;
  1,  8, 23, 48,  83, 120, 127,   8;
  1,  9, 27, 59, 109, 177, 247, 255,   9;
  1, 10, 31, 70, 135, 234, 367, 502, 511, 10;
  ...
T(7,4) = 37 = A000295(4) + T(6,4) = 11 + 26.

G. C. Greubel, Rows n=1..100 of triangle, flattened

Cf. A126268, A000295, A104712, A125165, A104712, A125232 - A125235.

T[n_,1]:=1; T[n_,n_]:=n; T[n_,k_]:= T[n-1,k] + 2^k - k - 1; Table[T[n,k], {n,1,15}, {k,1,n}]//Flatten (* G. C. Greubel, Oct 23 2018 *)

{T(n,k) = if(k==1, 1, if(k==n, n, 2^k - k - 1 + T(n-1,k)))};
for(n=1,10, for(k=1,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Oct 23 2018

Given right border = (1,2,3,...), T(n,k) = A000295(k) + T(n-1,k); where A000295 = the Eulerian numbers starting (0, 1, 4, 11, 26, 57, ...).

A243662 Triangle read by rows: the reversed x = 1+q Narayana triangle at m=2.

Original entry on oeis.org

1, 3, 1, 12, 8, 1, 55, 55, 15, 1, 273, 364, 156, 24, 1, 1428, 2380, 1400, 350, 35, 1, 7752, 15504, 11628, 4080, 680, 48, 1, 43263, 100947, 92169, 41895, 9975, 1197, 63, 1, 246675, 657800, 708400, 396704, 123970, 21560, 1960, 80, 1, 1430715, 4292145, 5328180, 3552120, 1381380, 318780, 42504, 3036, 99, 1

Offset: 1

N. J. A. Sloane, Jun 13 2014

See Novelli-Thibon (2014) for precise definition.
From Tom Copeland, Dec 13 2022: (Start)
The row polynomials are the nonvanishing numerator polynomials generated in the compositional, or Lagrange, inversion in x about the origin of the odd o.g.f. Od1(x,t) = x*(t*(1-x^2)-x^2) / (1-x^2) = t*x - x^3 - x^5 - x^7 - x^9 - ... .
For example, from the Lagrange inversion formula (LIF), the tenth derivative in x of (x/Od1(x,t))^11 / 11! = (1/((t*(1-x^2)-x^2) / (1-x^2)))^11 / 11! at x = 0 is (t^4 + 24*t^3 + 156*t^2 + 364*t + 273) / t^16. These polynomials are also generated by the iterated derivatives ((1/(D Od1(x,t)) D)^n g(x) evaluated at x = 0 where D = d/dx.
An explicit generating function for the polynomials can be obtained by finding the solution of the cubic equation y - t*x - y*x^2 + (1+t)*x^3 = 0 for x in terms of y and t that satisfies y(x=0;t) = 0 = x(y=0;t).
The row polynomials are also the polynomials generated in the compositional inverse of O(x,t) = x / (1+(1+t)x)*(1+x)^2) = x + (-t - 3)*x^2 + (t^2 + 4 t + 6)*x^3 + (-t^3 - 5*t^2 - 10*t - 10)*x^4 + ..., containing the truncated Pascal polynomials of A104712 / A325000.
For example, from the LIF, the third derivative of ((1 + (1+t)*x)*(1+x)^2)^4 / 4! at x = 0 is 55 + 55*t + 15*t^2 + t^3.
A natural refinement of this array was provided in a letter by Isaac Newton in 1676--a set of partition polynomials for generating the o.g.f. of the compositional inverse of the generic odd o.g.f. x + u_1 x^3 + u_2 x^5 + ... in the infinite set of indeterminates u_n. (End)
T(n,k) is the number of noncrossing cacti with n+1 nodes and n+1-k blocks. See A361242. - Andrew Howroyd, Apr 13 2023

			Triangle begins:
     1;
     3,    1;
    12,    8,    1;
    55,   55,   15,   1;
   273,  364,  156,  24,  1;
  1428, 2380, 1400, 350, 35, 1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (rows 1 <= n <= 150, flattened)
Paul Barry, On the inversion of Riordan arrays, arXiv:2101.06713 [math.CO], 2021.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014. See Fig. 10.

Cf. A001764, A001263, A243663 (m=3).
Row sums give A003168.
Row reversed triangle is A102537.

T[m_][n_, k_] := Binomial[(m + 1) n + 1 - k, n - k] Binomial[n, k - 1]/n;
Table[T[2][n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 12 2019 *)

T(n)=[Vecrev(p) | p<-Vec(serreverse(x/((1+x+x*y)*(1+x)^2) + O(x*x^n)))]
{ my(A=T(10)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Apr 13 2023

T(n,k) = (binomial(3*n+1,n) * binomial(n,k-1) * binomial(n-1,k-1)) / (binomial(3*n,k-1) * (3*n+1)) = (A001764(n) * A001263(n,k) * k) / binomial(3*n,k-1) for 1 <= k <= n (conjectured). - Werner Schulte, Nov 22 2018
T(n,k) = binomial(3*n+1-k,n-k) * binomial(n,k-1) / n for 1 <= k <= n, more generally: T_m(n,k) = binomial((m+1)*n+1-k,n-k) * binomial(n,k-1) / n for 1 <= k <= n and some fixed integer m > 1. - Werner Schulte, Nov 22 2018
G.f.: A(x,y) is the series reversion of x/((1 + x + x*y)*(1 + x)^2). - Andrew Howroyd, Apr 13 2023

Data and Example (T(2,2) and T(5,3)) corrected and more terms added by Werner Schulte, Nov 22 2018

A104713 Triangle T(n,k) = binomial(n,k), read by rows, 3 <= k <=n .

Original entry on oeis.org

1, 4, 1, 10, 5, 1, 20, 15, 6, 1, 35, 35, 21, 7, 1, 56, 70, 56, 28, 8, 1, 84, 126, 126, 84, 36, 9, 1, 120, 210, 252, 210, 120, 45, 10, 1, 165, 330, 462, 462, 330, 165, 55, 11, 1, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 286, 715, 1287, 1716

Offset: 3

Gary W. Adamson, Mar 19 2005

			First few rows of the triangle are:
1;
4, 1;
10, 5, 1;
20, 15, 6, 1;
35, 35, 21, 7, 1;
56, 70, 56, 28, 8, 1;
...

Cf. A007318, A104712, A002662 (row sums).

A104713 := proc(n,k)
        binomial(n,k) ;
end proc;
seq(seq( A104713(n,k),k=3..n),n=3..16) ; # R. J. Mathar, Oct 29 2011

T(n,k) = A007318(n,k) for n>=3, 3<=k<=n.
From Peter Bala, Jul 16 2013: (Start)
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^4, x/(1 - x)).
O.g.f.: 1/(1 - t)^3*1/(1 - (1 + x)*t) = 1 + (4 + x)*t + (10 + 5*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^3 (exp(t)*(exp(x*t) - 1 - x*t - (x*t)^2/2!)) = 1 + (4 + x)*t + (10 + 5*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [4,5,6,...] on the main subdiagonal and 0's elsewhere. (End)

A380851 Riordan array ((1-x)^(m-1), x/(1-x)) with factor r^(2*n) on row n, for m = 3/2, r = 2.

Original entry on oeis.org

1, -2, 4, -2, 8, 16, -4, 24, 96, 64, -10, 80, 480, 640, 256, -28, 280, 2240, 4480, 3584, 1024, -84, 1008, 10080, 26880, 32256, 18432, 4096, -264, 3696, 44352, 147840, 236544, 202752, 90112, 16384, -858, 13728, 192192, 768768, 1537536, 1757184, 1171456, 425984, 65536

Offset: 0

Igor Victorovich Statsenko, Feb 06 2025

			Triangle starts:
       k = 0      1       2        3        4        5       6
  n=0:     1;
  n=1:    -2,     4;
  n=2:    -2,     8,     16;
  n=3:    -4,    24,     96,      64;
  n=4:   -10,    80,    480,     640,     256;
  n=5:   -28,   280,   2240,    4480,    3584,    1024;
  n=6:   -84,  1008,  10080,   26880,   32256,   18432,   4096;

Igor Victorovich Statsenko, Identification of Riordan generalizations of binomial coefficients, Innovation science No 02-1, State Ufa, Aeterna Publishing House, 2025, pp. 11-18, see table 10. In Russian.

Columns: A002420 (k=0); A240530 (k=1).

Triangle for m=-3, r=1: A104713; for m=-2, r=1: A104712; for m=-1, r=1: A135278; for m=0, r=1: A007318; for m=1, r=1: A097805; for m=2, r=1: A159854.

T:=(m,r,n,k)->add(binomial(i+m,m)*binomial(n+1,n-k-i)*r^(2*n)*(-1)^(i),i=0..n-k): m:=3/2: r:=2: seq(print(seq(T(m,r,n,k), k=0..n)), n=0..10);

T[n_, k_] := 4^n Binomial[n, k] Hypergeometric2F1[3/2, k - n, k + 1, 1];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten  (* Peter Luschny, Feb 07 2025 *)

# Using function riordan_array from A256893.
RA = riordan_array((1 - x)^(3/2 - 1), x/(1-x), 7)
for n in range(7): print(4^n * RA.row(n)[:n+1])  # Peter Luschny, Feb 28 2025

T(n,k) = Sum_{i=0..n-k} binomial(i+m, m)*binomial(n+1, n-k-i)*r^(2*n)*(-1)^(i), for m = 3/2 and r = 2.
From Peter Luschny, Feb 07 2025: (Start)
T(n,k) = r^(2*n)*JacobiP(n - k, 1 + k, m - 1 - n, -1).
T(n,k) = 4^n*binomial(n, k)*hypergeom([3/2, k - n], [k + 1], 1). (End)

A168577 Pascal's triangle, first two columns and diagonal removed.

Original entry on oeis.org

3, 6, 4, 10, 10, 5, 15, 20, 15, 6, 21, 35, 35, 21, 7, 28, 56, 70, 56, 28, 8, 36, 84, 126, 126, 84, 36, 9, 45, 120, 210, 252, 210, 120, 45, 10, 55, 165, 330, 462, 462, 330, 165, 55, 11, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 78, 286, 715, 1287, 1716, 1716, 1287

Offset: 2

Roger L. Bagula, Nov 30 2009

Row sums are 3, 10, 25, 56, 119, 246, .. (A000247).

			3;
6, 4;
10, 10, 5;
15, 20, 15, 6;
21, 35, 35, 21, 7;
28, 56, 70, 56, 28, 8;
36, 84, 126, 126, 84, 36, 9;
45, 120, 210, 252, 210, 120, 45, 10;
55, 165, 330, 462, 462, 330, 165, 55, 11;
66, 220, 495, 792, 924, 792, 495, 220, 66, 12;
78, 286, 715, 1287, 1716, 1716, 1287, 715, 286, 78, 13;

Cf. A007318, A104712

p[x_, n_] = ((x + 1)^n - x^n - n*x - 1)/x^2;
Table[CoefficientList[p[x, n], x], {n, 3, 13}];
Flatten[%]

T(n,k)= [x^k] ((x + 1)^n - x^n - n*x - 1), 2<=k

T(n,k) = binomial(n,k).

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