A007318 -id:A007318 - cp's OEIS frontend

A168292 T(n,k) = 24A046802(n+1,k+1) - 9A008518(n,k) - 8*A007318(n,k), triangle read by rows (0 <= k <= n).

7, 7, 7, 7, 38, 7, 7, 99, 99, 7, 7, 220, 546, 220, 7, 7, 461, 2236, 2236, 461, 7, 7, 942, 8001, 15596, 8001, 942, 7, 7, 1903, 26697, 89921, 89921, 26697, 1903, 7, 7, 3824, 85660, 463520, 796594, 463520, 85660, 3824, 7, 7, 7665, 268530, 2224350, 6068400

Offset: 0

Roger L. Bagula, Nov 22 2009

			Triangle begins:
     7;
     7,    7;
     7,   38,     7;
     7,   99,    99,      7;
     7,  220,   546,    220,      7;
     7,  461,  2236,   2236,    461,      7;
     7,  942,  8001,  15596,   8001,    942,     7;
     7, 1903, 26697,  89921,  89921,  26697,  1903,    7;
     7, 3824, 85660, 463520, 796594, 463520, 85660, 3824, 7;
      ... reformatted. - _Franck Maminirina Ramaharo_, Oct 21 2018

Triangles related to Eulerian numbers: A008292, A046802, A060187, A123125.

Cf. A142147, A142175, A168287, A168288, A168289, A168290, A168291, A168293.

A123125(n, k) := sum((-1)^(k - j)*(binomial(n - j, k - j))*stirling2(n, j)*j!, j, 0, k)$
A046802(n, k) := sum(binomial(n - 1, r)*A123125(r, k - 1), r, k - 1, n - 1)$
A008518(n, k) := A123125(n, k) + A123125(n, k + 1)$
T(n, k) := 24*A046802(n + 1, k + 1) - 9*A008518(n, k) - 8*binomial(n, k)$
create_list(T(n, k), n, 0, 10, k, 0, n);
/* Franck Maminirina Ramaharo, Oct 21 2018 */

E.g.f.: 24*(1 - x)*exp(t)/(1 - x*exp(t*(1 - x))) - 9*(exp(t) - x*exp(t*x))/(exp(t*x) - x*exp(t)) - 8*exp(t*(1 + x)).

Edited, new name from Franck Maminirina Ramaharo, Oct 21 2018

A168293 T(n,k) = 12A046802(n+1,k+1) - 9A008518(n,k) - 2*A007318(n,k), triangle read by rows (0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 14, 1, 1, 33, 33, 1, 1, 64, 186, 64, 1, 1, 119, 724, 724, 119, 1, 1, 222, 2415, 5120, 2415, 222, 1, 1, 421, 7491, 28799, 28799, 7491, 421, 1, 1, 812, 22456, 142268, 257866, 142268, 22456, 812, 1, 1, 1587, 66342, 649554, 1934544, 1934544, 649554

Offset: 0

Roger L. Bagula, Nov 22 2009

			Triangle begins:
    1;
    1,   1;
    1,  14,     1;
    1,  33,    33,      1;
    1,  64,   186,     64,      1;
    1, 119,   724,    724,    119,      1;
    1, 222,  2415,   5120,   2415,    222,     1;
    1, 421,  7491,  28799,  28799,   7491,   421,   1;
    1, 812, 22456, 142268, 257866, 142268, 22456, 812, 1:
     ... reformatted. - _Franck Maminirina Ramaharo_, Oct 21 2018

Triangles related to Eulerian numbers: A008292, A046802, A060187, A123125.

Cf. A142147, A142175, A168287, A168288, A168289, A168290, A168291, A168292.

A123125(n, k) := sum((-1)^(k - j)*(binomial(n - j, k - j))*stirling2(n, j)*j!, j, 0, k)$
A046802(n, k) := sum(binomial(n - 1, r)*A123125(r, k - 1), r, k - 1, n - 1)$
A008518(n, k) := A123125(n, k) + A123125(n, k + 1)$
T(n, k) := 12*A046802(n + 1, k + 1) - 9*A008518(n, k) - 2*binomial(n, k)$
create_list(T(n, k), n, 0, 10, k, 0, n);
/* Franck Maminirina Ramaharo, Oct 21 2018 */

E.g.f.: 12*(1 - x)*exp(t)/(1 - x*exp(t*(1 - x))) - 9*(exp(t) - x*exp(t*x))/(exp(t*x) - x*exp(t)) - 2*exp(t*(1 + x)).

Edited, and new name by Franck Maminirina Ramaharo, Oct 21 2018

A288876 a(n) = binomial(n+4, n)^2. Square of the fifth diagonal sequence of A007318 (Pascal). Fifth diagonal sequence of A008459.

Original entry on oeis.org

1, 25, 225, 1225, 4900, 15876, 44100, 108900, 245025, 511225, 1002001, 1863225, 3312400, 5664400, 9363600, 15023376, 23474025, 35820225, 53509225, 78411025, 112911876, 160022500, 223502500, 308002500, 419225625, 564110001, 751034025, 990046225, 1293121600, 1674446400, 2150733376

Offset: 0

Wolfdieter Lang, Jul 27 2017

This is also the square of the fifth (k = 4) column sequence (without leading zeros) of the Pascal triangle A007318. For the triangle with the squares of the entries of Pascal's triangle see A008459.
For the square of the (d+1)-th diagonal sequence of A007318, PD2(d,n) = binomial(d + n, n)^2, d >= 0, one finds the o.g.f. GPD2(d, x) = Sum_{n>=0} PD2(d,n)*x^n in the following way. Compute the compositional inverse (Lagrange inversion formula) of y(t,x) = x*(1 - t/(1-x)) w.r.t. x, that is x = x(t,y). Then -log(1 - x(t,y)) = Sum_{d=0} y^(d+1)/(d+1)*GPD2(d, x). The r.h.s. can be called the logarithmic generating function (l.g.f.) of the o.g.f.s of the square of the diagonals of Pascal's triangle.
This computation was inspired by an article by P. Bala (see a link in A112007) on the diagonal sequences of special Sheffer triangles (1, f(t)) (Sheffer triangles are there called exponential Riordan triangles, and f is called F). This can be generalized to Sheffer (g, f). For general Riordan triangles R = (G(x), F(x)) a similar analysis can be done. The present entry is then obtained for example of the Pascal triangle P = (1/(1-x), x/(1-x)).
The o.g.f.s for the square of the diagonals of Pascal's triangle turn out to be GPD2(d, x) = P(d,x)/(1 - x)^(2*d+1), with the numerator polynomials given by row n of triangle A008459 (squares of the entries of Pascal's triangle): P(d, x) = Sum_{k=0..d} A008459(d, k)*x^k.

Cf. A007318, A008459.

The squares of the first diagonals are in A000012, A000290(n+1), A000537, A001249 (for d = 0..3).

[Binomial(n+4, n)^2: n in [0..30]]; // Vincenzo Librandi, Aug 02 2017

Table[Binomial[n + 4, n]^2, {n, 0, 30}] (* Michael De Vlieger, Jul 30 2017 *)

a(n) = binomial(n+4, n)^2 \\ Felix Fröhlich, Jul 31 2017

a(n) = binomial(n+4, n)^2, n >= 0.
O.g.f.: (1 + 16*x + 36*x^2 + 16*x^3 + x^4)/(1 - x)^9. (See a comment above and row n=4 of A008459.)
E.g.f: exp(x)*(1 + 24*x + 176*x^2/2! + 624*x^3/3! + 1251*x^4/4!+ 1500*x^5/5!+ 1070*x^6/6! + 420*x^7/7! + 70*x^8/8!), computed from the o.g.f with the formulas (23) - (25) of the W. Lang link given in A060187.
From Amiram Eldar, Sep 20 2022: (Start)
Sum_{n>=0} 1/a(n) = 160*Pi^2/3 - 1576/3.
Sum_{n>=0} (-1)^n/a(n) = 512*log(2)/3 - 352/3. (End)

A102547 Triangle read by rows, formed from antidiagonals of the antidiagonals (A011973) of Pascal's triangle (A007318).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 4, 1, 1, 5, 3, 1, 6, 6, 1, 7, 10, 1, 1, 8, 15, 4, 1, 9, 21, 10, 1, 10, 28, 20, 1, 1, 11, 36, 35, 5, 1, 12, 45, 56, 15, 1, 13, 55, 84, 35, 1, 1, 14, 66, 120, 70, 6, 1, 15, 78, 165, 126, 21, 1, 16, 91, 220, 210, 56, 1, 1, 17, 105, 286, 330, 126, 7, 1, 18, 120

Offset: 0

Gerald McGarvey, Feb 24 2005

Row sums are A000930, antidiagonal sums are A003269.
Row n contains 1+floor(n/3) terms.
T(n,k) is the number of compositions of n+3 with k+1 parts, all at least 3. Example: T(9,2) = binomial(5,2) = 10 because we have 336, 363, 633, 345, 354, 435, 453, 534, 543, and 444. - Emeric Deutsch, Aug 15 2010
T(n+2,k) is the number of k-subsets of {1..n} with values at least 3 apart. For example, T(7,2) = 3 corresponds to the subsets {1,4},{1,5},{2,5} of {1..5}. - Enrique Navarrete, Dec 19 2021

			Triangle begins:
  1;
  1;
  1;
  1, 1;
  1, 2;
  1, 3;
  1, 4, 1;
  1, 5, 3;

Alois P. Heinz, Rows n = 0..250, flattened
Michael A. Allen, Connections between Combinations Without Specified Separations and Strongly Restricted Permutations, Compositions, and Bit Strings, arXiv:2409.00624 [math.CO], 2024. See p. 13.
Richard J. Mathar, Tiling n x m rectangles with 1 X 1 and s X s squares, arXiv:1609.03964 [math.CO], 2016, Section 4.2.
Richard J. Mathar, Bivariate Generating Functions Enumerating Non-Bonding Dominoes on Rectangular Boards, arXiv:2404.18806 [math.CO], 2024. See p. 7.
Michel Rigo, Manon Stipulanti, and Markus A. Whiteland, Gapped Binomial Complexities in Sequences, Univ. Liège (Belgium 2023).

Cf. A007318, A011973, A003269, A000930 (row sums), A349862 (max row values).

/* As triangle */ [[Binomial(n-2*k,k): k in [0..n div 3]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 23 2019

for n from 0 to 20 do seq(binomial(n-2*k, k), k = 0 .. floor((1/3)*n)) end do; # yields sequence in triangular form. - Emeric Deutsch, Aug 15 2010

nn=20;Map[Select[#,#>0&]&,CoefficientList[Series[1/(1-x)/(1-y x^3/(1-x)),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Jun 25 2014 *)

T(n,k) = binomial(n-2k,k) (0 <= k <= n/3). - Emeric Deutsch, Aug 15 2010

G.f.: 1/(1 - x)/(1 - y*x^3/(1 - x)) = 1/(1 - x - y*x^3). - Geoffrey Critzer, Jun 25 2014

A136104 A007318 * A002110; a(n) = Sum_{k=0..n} binomial(n,k)*A002110(k).

Original entry on oeis.org

1, 3, 11, 55, 375, 3731, 47743, 777771, 14770535, 331611235, 9205305591, 285781156343, 10308779559631, 418386835375575, 18097509979840775, 846748292083023991, 44182142790019823943, 2570069981187508600331, 157428743473326543397855, 10449715795107936675445215, 739751959772798881608189731

Offset: 0

Gary W. Adamson, Dec 14 2007

nonn

			a(3) = 55 = (1, 3, 3, 1) dot (1, 2, 6, 30) = (1 + 6 + 18 + 30), where A002110 = (1, 2, 6, 30, 210, 2310, ...).

Cf. A002110, A007188, A007318, A276085.
Leftmost column of A276586.
Cf. also A001339, A121572.

b:= proc(n) option remember; `if`(n=0, 1, ithprime(n)*b(n-1)) end:
a:= n-> add(binomial(n, k)*b(k), k=0..n):
seq(a(n), n=0..20);  # Alois P. Heinz, Sep 20 2016

b[n_] := b[n] = If[n==0, 1, Prime[n]*b[n-1]]; a[n_] := Sum[Binomial[n, k]*b[k], {k, 0, n}]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Feb 22 2017, translated from Maple *)

Binomial transform of primorial numbers, A002110.

a(n) = A276085(A007188(n)). - Antti Karttunen, Sep 18 2016

A few more terms from L. Edson Jeffery, Apr 11 2011

Explicit binomial sum formula added to the name by Antti Karttunen, Sep 19 2016

A152844 Triangle read by rows, A007318 rows repeated six times .

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 6, 4, 1, 1, 4, 6, 4, 1, 1, 4, 6, 4, 1, 1, 4, 6, 4, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 5, 10, 10, 5, 1, 1

Offset: 0

Philippe Deléham, Dec 14 2008

Diagonal sums : A103374 .

			Triangle begins : 1 ; 1 ; 1 ; 1 ; 1 ; 1 ; 1,1 ; 1,1 ; 1,1 ; 1,1 ; 1,1 ; 1,1 ; 1,2,1 ; 1,2,1 ; 1,2,1 ; 1,2,1 ; 1,2,1 ; 1,2,1 ; 1,3,3,1 ; ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A007318, A152198, A152815, A152828, A152830, A152831.

Flatten[Table[#, {6}] & /@ Table[Binomial[n, k], {n,0,10}, {k, 0, n}]] (* G. C. Greubel, May 03 2017 *)

A230206 Trapezoid of dot products of row 3 (signs alternating) with sequential 4-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 4-tuples (C(3,0), -C(3,1), C(3,2), -C(3,3)) and (C(n-1,k-3), C(n-1,k-2), C(n-1,k-1), C(n-1,k)), n >= 1, 0 <= k <= n+2.

Original entry on oeis.org

-1, 3, -3, 1, -1, 2, 0, -2, 1, -1, 1, 2, -2, -1, 1, -1, 0, 3, 0, -3, 0, 1, -1, -1, 3, 3, -3, -3, 1, 1, -1, -2, 2, 6, 0, -6, -2, 2, 1, -1, -3, 0, 8, 6, -6, -8, 0, 3, 1, -1, -4, -3, 8, 14, 0, -14, -8, 3, 4, 1, -1, -5, -7, 5, 22, 14, -14, -22, -5, 7

Offset: 1

Dixon J. Jones, Oct 11 2013

The array is trapezoidal rather than triangular because C(n,k) is not uniquely defined for all negative n and negative k.
Row sums are 0.
Coefficients of (x-1)^3 (x+1)^(n-1) for n > 0.

			Trapezoid begins
  -1,  3, -3,  1;
  -1,  2,  0, -2,  1;
  -1,  1,  2, -2, -1,  1;
  -1,  0,  3,  0, -3,  0,  1;
  -1, -1,  3,  3, -3, -3,  1, 1;
  -1, -2,  2,  6,  0, -6, -2, 2, 1;
  -1, -3,  0,  8,  6, -6, -8, 0, 3, 1;

Dixon J. Jones, Rows n = 1..100 for irregular triangle, flattened
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.

Using row j of the alternating Pascal triangle as generator: A007318 (j=0), A008482 and A112467 (j=1 after the first term in each), A182533 (j=2 after the first two rows), A230207-A230212 (j=4 to j=9).

m:=3; [[k le 0 select (-1 )^m else (&+[(-1)^(j+m)* Binomial(m,j) *Binomial(n-1,k-j): j in [0..(n+m-1)]]): k in [0..(n+m-1)]]: n in [1..10]]; // G. C. Greubel, Nov 29 2018

Flatten[Table[CoefficientList[(x - 1)^3 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=3; Table[If[k == 0, (-1)^m, Sum[(-1)^(j+m)*Binomial[m, j]*Binomial[n-1, k-j], {j, 0, n+m-1}]], {n, 1, 10}, {k, 0, n+m-1}]//Flatten (* G. C. Greubel, Nov 29 2018 *)

m=3; for(n=1, 10, for(k=0, n+m-1, print1(if(k==0, (-1)^m, sum(j=0, n+m-1, (-1)^(j+m)*binomial(m,j)*binomial(n-1,k-j))), ", "))) \\ G. C. Greubel, Nov 29 2018

m=3; [[sum((-1)^(j+m)*binomial(m,j)*binomial(n-1,k-j) for j in range(n+m)) for k in range(n+m)] for n in (1..10)] # G. C. Greubel, Nov 29 2018

T(n,k) = Sum_{i=0..n+m-1} (-1)^(i+m)*C(m,i)*C(n-1,k-i), n >= 1, with T(n,0) = (-1)^m and m=3.

A230212 Trapezoid of dot products of row 9 (signs alternating) with sequential 10-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 10-tuples (C(9,0), -C(9,1), ..., C(9,8), -C(9,9)) and (C(n-1,k-9), C(n-1,k-8), ..., C(n-1,k)), n >= 1, 0 <= k <= n+8.

Original entry on oeis.org

-1, 9, -36, 84, -126, 126, -84, 36, -9, 1, -1, 8, -27, 48, -42, 0, 42, -48, 27, -8, 1, -1, 7, -19, 21, 6, -42, 42, -6, -21, 19, -7, 1, -1, 6, -12, 2, 27, -36, 0, 36, -27, -2, 12, -6, 1, -1, 5, -6, -10, 29, -9, -36, 36, 9, -29, 10, 6, -5, 1, -1, 4, -1, -16

Offset: 1

Dixon J. Jones, Oct 12 2013

The array is trapezoidal rather than triangular because C(n,k) is not uniquely defined for all negative n and negative k.
Row sums are 0.
Coefficients of (x-1)^9 (x+1)^(n-1), n > 0.

			Trapezoid begins:
  -1, 9, -36,  84, -126, 126, -84,  36,  -9,   1;
  -1, 8, -27,  48,  -42,   0,  42, -48,  27,  -8,   1;
  -1, 7, -19,  21,    6, -42,  42,  -6, -21,  19,  -7,  1;
  -1, 6, -12,   2,   27, -36,   0,  36, -27,  -2,  12, -6,  1;
  -1, 5,  -6, -10,   29,  -9, -36,  36,   9, -29,  10,  6, -5,  1;
  -1, 4,  -1, -16,   19,  20, -45,   0,  45, -20, -19, 16,  1, -4,  1;
  -1, 3,   3, -17,    3,  39, -25, -45,  45,  25, -39, -3, 17, -3, -3, 1;
  etc.

G. C. Greubel, Rows n=1..50 of trapezoid, flattened
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, Graça Tomaz, Combinatorial Identities Associated with a Multidimensional Polynomial Sequence, J. Int. Seq., Vol. 21 (2018), Article 18.7.4.

Using row j of the alternating Pascal triangle as generator: A007318 (j=0), A008482 and A112467 (j=1 after the first term in each), A182533 (j=2 after the first two rows), A230206-A230211 (j=3 to j=8).

m:=9; [[k le 0 select (-1 )^m else (&+[(-1)^(j+m)* Binomial(m,j) *Binomial(n-1,k-j): j in [0..(n+m-1)]]): k in [0..(n+m-1)]]: n in [1..10]]; // G. C. Greubel, Nov 28 2018

Flatten[Table[CoefficientList[(x - 1)^9 (x + 1)^n, x], {n, 0, 7}]] (* T. D. Noe, Oct 25 2013 *)
m=9; Table[If[k == 0, (-1)^m, Sum[(-1)^(j+m)*Binomial[m, j]*Binomial[n-1, k-j], {j, 0, n+m-1}]], {n, 1, 10}, {k, 0, n+m-1}]//Flatten (* G. C. Greubel, Nov 28 2018 *)

m=9; for(n=1, 10, for(k=0, n+m-1, print1(if(k==0, (-1)^m, sum(j=0, n+m-1, (-1)^(j+m)*binomial(m,j)*binomial(n-1,k-j))), ", "))) \\ G. C. Greubel, Nov 28 2018

m=9; [[sum((-1)^(j+m)*binomial(m,j)*binomial(n-1,k-j) for j in range(n+m)) for k in range(n+m)] for n in (1..10)] # G. C. Greubel, Nov 28 2018

T(n,k) = Sum_{i=0..n+m-1} (-1)^(i+m)*C(m,i)*C(n-1,k-i), n>=1, with T(n,0) = (-1)^m and m=9.

A064405 Number of even entries (A048967) minus the number of odd entries (A001316) in row n of Pascal's triangle (A007318).

Original entry on oeis.org

-1, -2, -1, -4, 1, -2, -1, -8, 5, 2, 3, -4, 5, -2, -1, -16, 13, 10, 11, 4, 13, 6, 7, -8, 17, 10, 11, -4, 13, -2, -1, -32, 29, 26, 27, 20, 29, 22, 23, 8, 33, 26, 27, 12, 29, 14, 15, -16, 41, 34, 35, 20, 37, 22, 23, -8, 41, 26, 27, -4, 29, -2, -1, -64, 61, 58, 59, 52, 61, 54, 55, 40, 65, 58, 59, 44, 61, 46, 47, 16, 73, 66, 67, 52, 69, 54

Offset: 0

Robert G. Wilson v, Sep 29 2001

Seiichi Manyama, Table of n, a(n) for n = 0..8191

Cf. A000079, A001316, A007318, A048967.

Table[ n + 1 - 2Sum[ Mod[ Binomial[ n, k ], 2 ], {k, 0, n} ], {n, 0, 100} ]

PARI
```
a(n)=sum(i=0,n,(-1)^binomial(n,i))
```

a(n)=if(n<1,-1,if(n%2==0,a(n/2)+n/2,2*a((n-1)/2)))

a(n) = Sum_{k=0..n} (-1)^binomial(n, k); a(2^n) = 2^n-3; a(2^n+1)=2^n-6; more generally there's a sequence z(k) such that for any k>=0 and for 2^n >k, a(2^n+k) = 2^n+z(k); for k=0, 1, 2, 3, 4, 5, 6, 7, 8... z(k) = -3, -6, -5, -12, -3, -10, -9, -24, 1, ... - Benoit Cloitre, Oct 18 2002
a(2n) = a(n) + n, a(2n+1) = 2a(n). - Ralf Stephan, Mar 05 2004
a(n) = -Sum_{k=0..n} moebius(binomial(n, k) mod 2). - Paul Barry, Apr 29 2005
a(2^n-1) = -2^n. - Seiichi Manyama, Aug 24 2022

A128315 Inverse Moebius transform of signed A007318.

Original entry on oeis.org

1, 0, 1, 2, -2, 1, -1, 4, -3, 1, 2, -4, 6, -4, 1, 0, 4, -9, 10, -5, 1, 2, -6, 15, -20, 15, -6, 1, -2, 11, -24, 36, -35, 21, -7, 1, 3, -10, 29, -56, 70, -56, 28, -8, 1, 0, 6, -30, 80, -125, 126, -84, 36, -9, 1, 2, -10, 45, -120, 210, -252, 210, -120, 45, -10, 1, -2, 18, -67, 176, -335, 463, -462, 330, -165, 55, -11, 1

Offset: 1

Gary W. Adamson, Feb 25 2007

A128316 = A000012 * A128315.

			First few rows of the triangle:
   1;
   0,  1;
   2, -2,  1;
  -1,  4, -3,  1;
   2, -4,  6, -4,  1;
   0,  4, -9, 10, -5, 1;
  ...

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

Cf. A000034, A000984, A007318, A010693, A048272, A051731, A128316.
Cf. A130595, A325940, A363615, A363616.
Cf. A000012 (row sums).

A128315:= func< n,k | (&+[0^(n mod j)*(-1)^(k+j)*Binomial(j-1, k-1): j in [k..n]]) >;
[A128315(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jun 22 2024

A128315[n_, k_]:= (-1)^k*DivisorSum[n, (-1)^#*Binomial[#-1, k-1] &];
Table[A128315[n,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jun 22 2024 *)

def A128315(n,k): return sum( 0^(n%j)*(-1)^(k+j)*binomial(j-1,k-1) for j in range(k,n+1))
flatten([[A128315(n,k) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Jun 22 2024

T(n, k) = A051731(n, k) * A130595(n-1, k-1) as infinite lower triangular matrices.
T(n, 1) = A048272(n).
Sum_{k=1..n} T(n, k) = A000012(n) = 1 (row sums).
From G. C. Greubel, Jun 22 2024: (Start)
T(n, k) = (-1)^k * Sum_{d|n} (-1)^d * binomial(d-1, k-1).
T(n, 2) = A325940(n), n >= 2.
T(n, 3) = A363615(n), n >= 3.
T(n, 4) = A363616(n), n >= 4.
T(2*n-1, n) = (-1)^(n-1)*A000984(n-1), n >= 1.
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (-1)^(n-1)*A081295(n).
Sum_{k=1..n} k*T(n, k) = A000034(n-1), n >= 1.
Sum_{k=1..n} (k+1)*T(n, k) = A010693(n-1), n >= 1. (End)

a(43) = 28 inserted and more terms from Georg Fischer, Jun 05 2023

cp's OEIS Frontend

A168292 T(n,k) = 24*A046802(n+1,k+1) - 9*A008518(n,k) - 8*A007318(n,k), triangle read by rows (0 <= k <= n).

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A168293 T(n,k) = 12*A046802(n+1,k+1) - 9*A008518(n,k) - 2*A007318(n,k), triangle read by rows (0 <= k <= n).

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A288876 a(n) = binomial(n+4, n)^2. Square of the fifth diagonal sequence of A007318 (Pascal). Fifth diagonal sequence of A008459.

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A102547 Triangle read by rows, formed from antidiagonals of the antidiagonals (A011973) of Pascal's triangle (A007318).

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A136104 A007318 * A002110; a(n) = Sum_{k=0..n} binomial(n,k)*A002110(k).

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A152844 Triangle read by rows, A007318 rows repeated six times .

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A230206 Trapezoid of dot products of row 3 (signs alternating) with sequential 4-tuples read by rows in Pascal's triangle A007318: T(n,k) is the linear combination of the 4-tuples (C(3,0), -C(3,1), C(3,2), -C(3,3)) and (C(n-1,k-3), C(n-1,k-2), C(n-1,k-1), C(n-1,k)), n >= 1, 0 <= k <= n+2.

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A168292 T(n,k) = 24A046802(n+1,k+1) - 9A008518(n,k) - 8*A007318(n,k), triangle read by rows (0 <= k <= n).

A168293 T(n,k) = 12A046802(n+1,k+1) - 9A008518(n,k) - 2*A007318(n,k), triangle read by rows (0 <= k <= n).