A156644 -id:A156644 - cp's OEIS frontend

A167374 Triangle, read by rows, given by [ -1,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

1, -1, 1, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1

Offset: 0

Philippe Deléham, Nov 02 2009

Riordan array (1-x,1) read by rows; Riordan inverse is (1/(1-x),1). Columns have g.f. (1-x)x^k. Diagonal sums are A033999. Unsigned version in A097806.
Table T(n,k) read by antidiagonals. T(n,1) = 1, T(n,2) = -1, T(n,k) = 0, k > 2. - Boris Putievskiy, Jan 17 2013
Finite difference operator (pair difference): left multiplication by T of a sequence arranged as a column vector gives a running forward difference, a(k+1)-a(k), or first finite difference (modulo sign), of the elements of the sequence. T^n gives the n-th finite difference (mod sign). T is the inverse of the summation matrix A000012 (regarded as lower triangular matrices). - Tom Copeland, Mar 26 2014

			Triangle begins:
   1;
  -1,  1;
   0, -1,  1;
   0,  0, -1,  1;
   0,  0,  0, -1,  1;
   0,  0,  0,  0, -1,  1; ...
Row number r (r>4) contains (r-2) times '0', then '-1' and '1'.
From _Boris Putievskiy_, Jan 17 2013: (Start)
The start of the sequence as a table:
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  ...
(End)

Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.

Cf. A000012, A007318, A029653, A097805, A118800, A130595, A156644.

A167374 := proc(n,k)
    if k> n or k < n-1 then
        0;
    elif k = n then
        1;
    else
        -1 ;
    end if;
end proc: # R. J. Mathar, Sep 07 2016

Table[PadLeft[{-1, 1}, n], {n, 13}] // Flatten (* or *)
MapIndexed[Take[#1, First@ #2] &, CoefficientList[Series[(1 - x)/(1 - x y), {x, 0, 12}], {x, y}]] // Flatten (* Michael De Vlieger, Nov 16 2016 *)
T[n_, k_] := If[ k<0 || k>n, 0, Boole[n==k] - Boole[n==k+1]]; (* Michael Somos, Oct 01 2022 *)

{T(n, k) = if( k<0 || k>n, 0, (n==k) - (n==k+1))}; /* Michael Somos, Oct 01 2022 */

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A055268(n), A055276(n) for x = 1,2,3,4,5,6,7,8,9,10,11 respectively .
From Boris Putievskiy, Jan 17 2013: (Start)
a(n) = floor((A002260(n)+2)/(A003056(n)+2))*(-1)^(A002260(n)+A003056(n)+1),  n>0.
a(n) = floor((i+2)/(t+2))*(-1)^(i+t+1), n > 0, where
i = n - t*(t+1)/2,
t = floor((-1 + sqrt(8*n-7))/2). (End)
T*A000012 = Identity matrix. T*A007318 = A097805. T*(A007318)^(-1)= signed A029653. - Tom Copeland, Mar 26 2014
G.f.: (1-x)/(1-x*y). - R. J. Mathar, Aug 11 2015
T = A130595*A156644 = M*T^(-1)*M = M*A000012*M, where M(n,k) = (-1)^n A130595(n,k). Note that M = M^(-1). Cf. A118800 and A097805. - Tom Copeland, Nov 15 2016

A131428 a(n) = 2*C(n) - 1, where C(n) = A000108(n) are the Catalan numbers.

Original entry on oeis.org

1, 1, 3, 9, 27, 83, 263, 857, 2859, 9723, 33591, 117571, 416023, 1485799, 5348879, 19389689, 70715339, 259289579, 955277399, 3534526379, 13128240839, 48932534039, 182965127279, 686119227299, 2579808294647, 9723892802903, 36734706144303, 139067101832007

Offset: 0

Gary W. Adamson, Jul 10 2007

nonn

Starting (1, 3, 9, 27, 83, ...), = row sums of triangle A136522. - Gary W. Adamson, Jan 02 2008
Hankel transform is A171552. - Paul Barry, Dec 11 2009
Apparently, for n >= 1, the maximum peak height minus the maximum valley height summed over all Dyck n-paths (with max valley height deemed zero if no valleys). - David Scambler, Oct 05 2012
Apparently for n > 1 the number of fixed points in all Dyck (n-1)-paths. A fixed point occurs when a vertex of a Dyck k-path is also a vertex of the path U^kD^k. - David Scambler, May 01 2013

			a(3) = 9 = 2*C(3) - 1 = 2*5 - 1, where C refers to the Catalan numbers, A000108.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000108, A131427, A131429.
Cf. A131428, A118976, A136522.
Cf. A080233, A156644.

List([0..25], n-> 2*Binomial(2*n,n)/(n+1) - 1); # G. C. Greubel, Aug 12 2019

[2*Catalan(n) -1: n in [0..25]]; // G. C. Greubel, Aug 12 2019

seq(2*binomial(2*n,n)/(n+1)-1, n=0..25); # Emeric Deutsch, Jul 25 2007

2CatalanNumber[Range[0,25]]-1  (* Harvey P. Dale, Apr 17 2011 *)

vector(25, n, n--; 2*binomial(2*n,n)/(n+1) - 1) \\ G. C. Greubel, Aug 12 2019

[2*catalan_number(n) -1 for n in (0..25)] # G. C. Greubel, Aug 12 2019

Right border of triangle A131429.
From Emeric Deutsch, Jul 25 2007: (Start)
a(n) = 2*binomial(2*n,n)/(n+1) - 1.
G.f.: (1-sqrt(1-4*x))/x - 1/(1-x). (End)
(1, 3, 9, 27, 83, ...) = row sums of A118976. - Gary W. Adamson, Aug 31 2007
Row sums of triangle A131428 starting (1, 3, 9, 27, 83, ...). - Gary W. Adamson, Aug 31 2007
Starting with offset 1 = Narayana transform (A001263) of [1,2,2,2,...]. - Gary W. Adamson, Jul 29 2011
D-finite with recurrence (n+1)*a(n) +2*(-2*n+1)*a(n-1) +3*(-n+1)=0. - R. J. Mathar, Nov 22 2024
a(n) = Sum_{k=0..n} ( binomial(n,k) - binomial(n,k-1) )^2 = Sum_{k=0..n} A080233(n,k)^2 = Sum_{k=0..n} A156644(n,k)^2. - Seiichi Manyama, Mar 25 2025

More terms from Emeric Deutsch, Jul 25 2007

A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.

Original entry on oeis.org

1, 1, -1, 1, -3, 1, 1, -7, 5, -1, 1, -15, 17, -7, 1, 1, -31, 49, -31, 9, -1, 1, -63, 129, -111, 49, -11, 1, 1, -127, 321, -351, 209, -71, 13, -1, 1, -255, 769, -1023, 769, -351, 97, -15, 1, 1, -511, 1793, -2815, 2561, -1471, 545, -127, 17, -1, 1, -1023, 4097, -7423, 7937, -5503, 2561, -799, 161, -19, 1

Offset: 0

Paul D. Hanna, May 02 2006

Matrix inverse is triangle A118800. Row sums are: (1-n). Unsigned row sums equal A007051(n) = (3^n + 1)/2. Row squared sums equal A118802. Antidiagonal sums equal A080956(n) = (n+1)(2-n)/2. Unsigned antidiagonal sums form A024537 (with offset).
T = C^2*D^-1 where matrix product D = C^-1*T*C = T^-1*C^2 has only 2 nonzero diagonals: D(n,n)=-D(n+1,n)=(-1)^n, with zeros elsewhere. Also, [B^-1]*T*[B^-1] = B*[T^-1]*B forms a self-inverse matrix, where B^2 = C and B(n,k) = C(n,k)/2^(n-k). - Paul D. Hanna, May 04 2006
Riordan array ( 1/(1 - x), -x/(1 - 2*x) ) The matrix square is the Riordan array ( (1 - 2*x)/(1 - x)^2, x ), which belongs to the Appell subgroup of the Riordan group. See the Example section below. - Peter Bala, Jul 17 2013

			Formulas for initial columns are, for n>=0:
T(n+1,1) = 1 - 2^(n+1);
T(n+2,2) = 1 + 2^(n+1)*n;
T(n+3,3) = 1 - 2^(n+1)*(n*(n+1)/2 + 1);
T(n+4,4) = 1 + 2^(n+1)*(n*(n+1)*(n+2)/6 + n);
T(n+5,5) = 1 - 2^(n+1)*(n*(n+1)*(n+2)*(n+3)/24 + n*(n+1)/2 + 1).
Triangle begins:
1;
1,-1;
1,-3,1;
1,-7,5,-1;
1,-15,17,-7,1;
1,-31,49,-31,9,-1;
1,-63,129,-111,49,-11,1;
1,-127,321,-351,209,-71,13,-1;
1,-255,769,-1023,769,-351,97,-15,1;
1,-511,1793,-2815,2561,-1471,545,-127,17,-1;
1,-1023,4097,-7423,7937,-5503,2561,-799,161,-19,1; ...
The matrix square, T^2, starts:
1;
0,1;
-1,0,1;
-2,-1,0,1;
-3,-2,-1,0,1;
-4,-3,-2,-1,0,1; ...
where all columns are the same.
The matrix product C^-1*T*C = T^-1*C^2 is:
1;
-1,-1;
0, 1, 1;
0, 0,-1,-1;
0, 0, 0, 1, 1; ...
where C(n,k) = n!/(n-k)!/k!.

M. Shattuck and T. Waldhauser, Proofs of some binomial identities using the method of last squares, Fib. Q., 48 (2010), 290-297.

Cf. A118800 (inverse), A007051 (unsigned row sums), A118802 (Row squared sums), A080956 (antidiagonal sums), A024537 (unsigned antidiagonal sums).
A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011
Cf. A007318, A032807, A097805, A112857, A130595, A156644, A167374, A200139, A239473.

Table[(1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]) - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

{T(n,k)=if(n==0&k==0,1,1+(-1)^k*2^(n-k+1)*sum(j=0,k\2,binomial(n-2*j-2,k-2*j-1)))}

T(n,k) = 1 + (-1)^k*2^(n-k+1)*Sum_{j=0..[k/2]} C(n-2j-2,k-2j-1) for n>=k>=0 with T(0,0) = 1.
For k>0, T(n,k) = -T(n-1,k-1) + 2*T(n-1,k). - Gerald McGarvey, Aug 05 2006
O.g.f.: (1 - 2*t)/(1 - t) * 1/(1 + t*(x - 2)) = 1 + (1 - x)*t + (1 - 3*x + x^2)*t^2 + (1 - 7*x + 5*x^2 - x^3)*t^3 + .... - Peter Bala, Jul 17 2013
From Tom Copeland, Nov 17 2016: (Start)
Let M = A200139^(-1) = (unsigned A118800)^(-1) and NpdP be the signed padded Pascal matrix defined in A097805. Then T(n,k) = (-1)^n* M(n,k) and T = P*NpdP = (A239473)^(-1)*P^(-1) = P*A167374*P^(-1) = A156644*P^(-1), where P is the Pascal matrix A007318 with inverse A130595. Cf. A112857.
Signed P^2 = signed A032807 = T*A167374. (End)

A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

Original entry on oeis.org

1, 0, 1, 1, -1, 1, 0, 2, -2, 1, 1, -2, 4, -3, 1, 0, 3, -6, 7, -4, 1, 1, -3, 9, -13, 11, -5, 1, 0, 4, -12, 22, -24, 16, -6, 1, 1, -4, 16, -34, 46, -40, 22, -7, 1, 0, 5, -20, 50, -80, 86, -62, 29, -8, 1, 1, -5, 25, -70, 130, -166, 148, -91, 37, -9, 1, 0, 6, -30, 95, -200, 296, -314, 239, -128, 46, -10, 1

Offset: 0

Tom Copeland, Mar 19 2014

With T the lower triangular array above and the Laguerre polynomials L(k,x) = Sum_{j=0..k} (-1)^j binomial(k, j) x^j/j!, the following identities hold:
(A) Sum_{k=0..n} (-1)^k L(k,x) = Sum_{k=0..n} T(n,k) x^k/k!;
(B) Sum_{k=0..n} x^k/k! = Sum_{k=0..n} T(n,k) L(k,-x);
(C) Sum_{k=0..n} x^k = Sum_{k=0..n} T(n,k) (1+x)^k = (1-x^(n+1))/(1-x).
More generally, for polynomial sequences,
(D) Sum_{k=0..n} P(k,x) = Sum_{k=0..n} T(n,k) (1+P(.,x))^k,
where, e.g., for an Appell sequence, such as the Bernoulli polynomials, umbrally, (1+ Ber(.,x))^k = Ber(k,x+1).
Identity B follows from A through umbral substitution of j!L(j,-x) for x^j in A. Identity C, related to the cyclotomic polynomials for prime index, follows from B through the Laplace transform.
Integrating C gives Sum_{k=0..n} T(n,k) (2^(k+1)-1)/(k+1) = H(n+1), the harmonic numbers.
Identity A >= 0 for x >= 0 (see MathOverflow link for evaluation in terms of Hermite polynomials).
From identity C, W(m,n) = (-1)^n Sum_{k=0..n} T(n,k) (2-m)^k = number of walks of length n+1 between any two distinct vertices of the complete graph K_m for m > 2.
Equals A112468 with the first column of ones removed. - Georg Fischer, Jul 26 2023

			Triangle begins:
   1
   0    1
   1   -1    1
   0    2   -2    1
   1   -2    4   -3    1
   0    3   -6    7   -4    1
   1   -3    9  -13   11   -5    1
   0    4  -12   22  -24   16   -6    1
   1   -4   16  -34   46  -40   22   -7    1
   0    5  -20   50  -80   86  -62   29   -8    1
   1   -5   25  -70  130 -166  148  -91   37   -9    1

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
J. Adams, On the groups J(x)-II, Topology, Vol. 3, p. 137-171, Pergamon Press, (1965).
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mathoverflow, Inequality for Laguerre polynomials

For column 2: A001057, A004526, A008619, A140106.
Column 3: A002620, A087811.
Column 4: A002623, A173196.
Column 5: A001752.
Column 6: A001753.
Cf. Bottomley's cross-references in A059260.
Embedded in alternating antidiagonals of T are the reversals of arrays A071921 (A225010) and A210220.
Cf. A049310, A112468, A341091.

[[(&+[(-1)^(j+k)*Binomial(j,k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Feb 06 2018

A239473 := proc(n,k)
    add(binomial(j,k)*(-1)^(j+k),j=k..n) ;
end proc; # R. J. Mathar, Jul 21 2016

Table[Sum[(-1)^(j+k)*Binomial[j,k], {j,0,n}], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 06 2018 *)

for(n=0,10, for(k=0,n, print1(sum(j=0,n, (-1)^(j+k)*binomial(j, k)), ", "))) \\ G. C. Greubel, Feb 06 2018

Trow = lambda n: sum((x-1)^j for j in (0..n)).list()
for n in (0..10): print(Trow(n)) # Peter Luschny, Jul 09 2019

T(n, k) = Sum_{j=0..n} (-1)^(j+k) * binomial(j, k).
E.g.f: (exp(t) - (x-1)*exp((x-1)*t))/(2-x).
O.g.f. (n-th row): (1-(x-1)^(n+1))/(2-x).
Associated operator identities:
With D=d/dx, :xD:^n=x^n*D^n, and :Dx:^n=D^n*x^n, then bin(xD,n)= binomial(xD,n)=:xD:^n/n! and L(n,-:xD:)=:Dx:^n/n!=bin(xD+n,n)=(-1)^n bin(-xD-1,n),
A-o) Sum_{k=0..n} (-1)^k L(k,-:xD:) = Sum_{k=0..n} :-Dx:^k/k!
     = Sum_{k=0..n} T(n,k) :-xD:^k/k! = Sum_{k=0..n} (-1)^k T(n,k)bin(xD,k)
B-o) Sum_{k=0..n} :xD:^k/k! = Sum_{k=0..n}, T(n,k) L(k,-:xD:)
     = Sum_{k=0..n} T(n,k) :Dx:^k/k! = Sum_{k=0..n}, bin(xD,k).
Associated binomial identities:
A-b) Sum_{k=0..n} (-1)^k bin(s+k,k) = Sum_{k=0..n} (-1)^k T(n,k) bin(s,k)
     = Sum_{k=0..n} bin(-s-1,k) = Sum{k=0..n} T(n,k) bin(-s-1+k,k)
B-b) Sum_{k=0..n} bin(s,k) = Sum_{k=0..n} T(n,k) bin(s+k,k)
     = Sum_{k=0..n} (-1)^k bin(-s-1+k,k)
     = Sum_{k=0..n} (-1)^k T(n,k) bin(-s-1,k).
In particular, from B-b with s=n, Sum_{k=0..n} T(n,k) bin(n+k,k) = 2^n. From B-b with s=0, row sums are all 1.
From identity C with x=-2, the unsigned row sums are the Jacobsthal sequence, i.e., Sum_{k=0..n} T(n,k) (1+(-2))^k = (-1)^n A001045(n+1); for x=2, the Mersenne numbers A000225; for x=-3, A014983 or signed A015518; for x=3, A003462; for x=-4, A014985 or signed A015521; for x=4, A002450; for x=-5, A014986 or signed A015531; and for x=5, A003463; for x=-6, A014987 or signed A015540; and for x=6, A003464.
With -s-1 = m = 0,1,2,..., B-b gives finite differences (recursions):
Sum_{k=0..n} (-1)^k T(n,k) bin(m,k) = Sum_{k=0..n} (-1)^k bin(m+k,k) = T(n+m,m), i.e., finite differences of the columns of T generate shifted columns of T. The columns of T are signed, shifted versions of sequences listed in the cross-references. Since the finite difference is an involution, T(n,k) = Sum_{j=0..k} (-1)^j T(n+j,j) bin(k,j)}. Gauss-Newton interpolation can be applied to give a generalized T(n,s) for s noninteger.
From identity C, S(n,m) = Sum_{k=0..n} T(n,k) bin(k,m) = 1 for m < n+1 and 0 otherwise, i.e., S = T*P, where S = A000012, as a lower triangular matrix and P = Pascal = A007318, so T = S*P^(-1), where P^(-1) = A130595, the signed Pascal array (see A132440), the inverse of P, and T^(-1) = P*S^(-1) = P*A167374 = A156644.
U(n,cos(x)) = e^(-n*i*x)*Sum_{k=0..n} T(n,k)*(1+e^(2*i*x))^k = sin((n+1)x)/sin(x), where U is the Chebyschev polynomial of the second kind A053117 and i^2 = -1. - Tom Copeland, Oct 18 2014
From Tom Copeland, Dec 26 2015: (Start)
With a(n,x) = e^(nx), the partial sums are 1+e^x+...+e^(nx) = Sum_{k=0..n} T(n,k) (1+e^x)^k = [ x / (e^x-1) ] [ e^((n+1)x) -1 ] / x = [ (x / (e^x-1)) e^((n+1)x) -  (x / (e^x-1)) ] / x =  Sum_{k>=0} [ (Ber(k+1,n+1) - Ber(k+1,0)) / (k+1) ] * x^k/k!, where Ber(n,x) are the Bernoulli polynomials (cf. Adams p. 140). Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of the m-th powers of the integers, their binomial transforms, and the Bernoulli polynomials.
With a(n,x) = (-1)^n e^(nx), the partial sums are 1-e^x+...+(-1)^n e^(nx) = Sum_{k=0..n} T(n,k) (1-e^x)^k = [ (-1)^n e^((n+1)x) + 1 ] / (e^x+1) = [ (-1)^n (2 / (e^x+1)) e^((n+1)x) + (2 / (e^x+1)) ] / 2 = (1/2) Sum_{k>=0} [ (-1)^n Eul(k,n+1) + Eul(k,0) ] * x^k/k!, where Eul(n,x) are the Euler polynomials. Evaluating (d/dx)^m at x=0 of these expressions gives relations among the partial sums of signed m-th powers of the integers; their binomial transforms, related to the Stirling numbers of the second kind and face numbers of the permutahedra; and the Euler polynomials. (End)
As in A059260, a generator in terms of bivariate polynomials with the coefficients of this entry is given by (1/(1-y))*1/(1 + (y/(1-y))*x - (1/(1-y))*x^2) = 1 + y + (x^2 - x*y + y^2) + (2*x^2*y - 2*x*y^2 + y^3) + (x^4 - 2*x^3*y + 4*x^2*y^2 - 3*x*y^3 + y^4) + ... . This is of the form -h2 * 1 / (1 + h1*x + h2*x^2), related to the bivariate generator of A049310 with h1 = y/(1-y) and h2 = -1/(1-y) = -(1+h1). - Tom Copeland, Feb 16 2016
From Tom Copeland, Sep 05 2016: (Start)
Letting P(k,x) = x in D gives Sum_{k=0..n} T(n,k)*Sum_{j=0..k} binomial(k,j) = Sum_{k=0..n} T(n,k) 2^k = n + 1.
The quantum integers [n+1]q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)) = q^(-n)*(1 - q^(2*(n+1))) / (1 - q^2) = q^(-n)*Sum{k=0..n} q^(2k) = q^(-n)*Sum_{k=0..n} T(n,k)*(1 + q^2)^k. (End)
T(n, k) = [x^k] Sum_{j=0..n} (x-1)^j. - Peter Luschny, Jul 09 2019
a(n) = -n + Sum_{k=0..n} A341091(k). - Thomas Scheuerle, Jun 17 2022

Inverse array added by Tom Copeland, Mar 26 2014

Formula re Euler polynomials corrected by Tom Copeland, Mar 08 2024

A080233 Triangle T(n,k) obtained by taking differences of consecutive pairs of row elements of Pascal's triangle A007318.

Original entry on oeis.org

1, 1, 0, 1, 1, -1, 1, 2, 0, -2, 1, 3, 2, -2, -3, 1, 4, 5, 0, -5, -4, 1, 5, 9, 5, -5, -9, -5, 1, 6, 14, 14, 0, -14, -14, -6, 1, 7, 20, 28, 14, -14, -28, -20, -7, 1, 8, 27, 48, 42, 0, -42, -48, -27, -8, 1, 9, 35, 75, 90, 42, -42, -90, -75, -35, -9

Offset: 0

Paul Barry, Feb 10 2003

Row sums are 1,1,1,1,1,1 with g.f. 1/(1-x). Can also be obtained from triangle A080232 by taking sums of pairs of consecutive row elements.

Mirror image of triangle in A156644. - Philippe Deléham, Feb 14 2009

			Triangle begins as:
  1;
  1, 0;
  1, 1, -1;
  1, 2,  0, -2;
  1, 3,  2, -2, -3;
  1, 4,  5,  0, -5,  -4;
  1, 5,  9,  5, -5,  -9,  -5;
  1, 6, 14, 14,  0, -14, -14,  -6;
  1, 7, 20, 28, 14, -14, -28, -20,  -7;
  1, 8, 27, 48, 42,   0, -42, -48, -27,  -8;
  1, 9, 35, 75, 90,  42, -42, -90, -75, -35, -9;
  ...

Row sums give A000012.

Cf. A007318, A080232.

Table[Binomial[n, k] - Binomial[n, k - 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

{T(n, k) = if( n<0 || k>n, 0, binomial(n, k) - binomial(n, k-1))}; /* Michael Somos, Nov 25 2016 */

T(n, k) = if(k>n, 0, binomial(n, k)-binomial(n, k-1)).

A382435 a(n) = Sum_{k=0..n} ( binomial(n,k) - binomial(n,k-1) )^6.

Original entry on oeis.org

1, 1, 3, 129, 1587, 39443, 1125383, 30211457, 1107074979, 36214609683, 1433494688871, 54495716261011, 2275005440977063, 95146470595975399, 4170974287982618639, 185640304224109725569, 8492643748223480148419, 395051289603660979274339, 18726850582009755291702599

Offset: 0

Seiichi Manyama, Mar 25 2025

nonn

Cf. A131428, A382434.

Cf. A080233, A156644, A382433.

b:= proc(x, y) option remember; `if`(y<0 or y>x, 0,
     `if`(x=0, 1, add(b(x-1, y+j), j=[-1, 1])))
    end:
a:= n-> 2*add(b(n, n-2*j)^6, j=0..n/2)-1:
seq(a(n), n=0..18);  # Alois P. Heinz, Mar 25 2025

a(n) = sum(k=0, n, (binomial(n, k)-binomial(n, k-1))^6);

from math import comb
def A382435(n): return (sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**6 for j in range((n>>1)+1))<<1)-1 # Chai Wah Wu, Mar 25 2025

a(n) = Sum_{k=0..n} A080233(n,k)^6 = Sum_{k=0..n} A156644(n,k)^6.

a(n) = 2 * A382433(n) - 1.

A382434 a(n) = Sum_{k=0..n} ( binomial(n,k) - binomial(n,k-1) )^4.

Original entry on oeis.org

1, 1, 3, 33, 195, 1763, 15623, 156257, 1630947, 17911299, 203739015, 2389928995, 28749060871, 353362388551, 4424242664975, 56290517376737, 726355164976547, 9490129871680355, 125375330053632455, 1672895457018337859, 22522481793315373319, 305695116823973096519

Offset: 0

Seiichi Manyama, Mar 25 2025

nonn

Cf. A131428, A382435.

Cf. A080233, A129123, A156644.

b:= proc(x, y) option remember; `if`(y<0 or y>x, 0,
     `if`(x=0, 1, add(b(x-1, y+j), j=[-1, 1])))
    end:
a:= n-> 2*add(b(n, n-2*j)^4, j=0..n/2)-1:
seq(a(n), n=0..21);  # Alois P. Heinz, Mar 25 2025

a(n) = sum(k=0, n, (binomial(n, k)-binomial(n, k-1))^4);

from math import comb
def A382434(n): return (sum((comb(n,j)*(m:=n-(j<<1)+1)//(m+j))**4 for j in range((n>>1)+1))<<1)-1 # Chai Wah Wu, Mar 25 2025

a(n) = Sum_{k=0..n} A080233(n,k)^4 = Sum_{k=0..n} A156644(n,k)^4.
a(n) = 2 * A129123(n) - 1.
D-finite with recurrence n*(n+1)^3*a(n) -2*n*(11*n^3-17*n^2+5*n+5)*a(n-1) -4*(n-1)*(70*n^3-365*n^2+527*n-162)*a(n-2) +8*(n-2)*(584*n^3-5020*n^2+14111*n-13059)*a(n-3) +1344*(4*n-11)*(4*n-13)*(-3+n)^2*a(n-4) +9*(2875*n^4-33975*n^3+149945*n^2-293541*n+215336)=0. - R. J. Mathar, Mar 31 2025

A051632 Rows of triangle formed using Pascal's rule except we begin and end the n-th row with n-2.

Original entry on oeis.org

-2, -1, -1, 0, -2, 0, 1, -2, -2, 1, 2, -1, -4, -1, 2, 3, 1, -5, -5, 1, 3, 4, 4, -4, -10, -4, 4, 4, 5, 8, 0, -14, -14, 0, 8, 5, 6, 13, 8, -14, -28, -14, 8, 13, 6, 7, 19, 21, -6, -42, -42, -6, 21, 19, 7, 8, 26, 40, 15, -48, -84, -48, 15, 40, 26, 8, 9, 34, 66, 55, -33, -132, -132

Offset: 0

Asher Auel

Row sums are -2.

			Contribution from Roger L. Bagula, Feb 17 2009: (Start)
The rows of the triangle, negated, are:
{2},
{1, 1},
{0, 2, 0},
{-1, 2, 2, -1},
{-2, 1, 4, 1, -2},
{-3, -1, 5,5, -1, -3},
{-4, -4, 4, 10, 4, -4, -4},
{-5, -8, 0, 14, 14, 0, -8, -5},
{-6, -13, -8, 14, 28, 14, -8, -13, -6},
{-7, -19, -21, 6, 42,42, 6, -21, -19, -7},
{-8, -26, -40, -15, 48, 84, 48, -15, -40, -26, -8} (End)

Reinhard Zumkeller, Rows n=0..100 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

A156644 [From Roger L. Bagula, Feb 17 2009]

Cf. A007318.

a051632 n k = a051632_tabl !! n !! k
a051632_list = concat a051632_tabl
a051632_tabl = iterate (\rs -> zipWith (+) ([1] ++ rs) (rs ++ [1])) [-2]
-- Reinhard Zumkeller, Aug 21 2011

t[n_, k_] = ((2*k + 1 - n)/(k + 1))*Binomial[n, k] + ((1 - n + 2 (-k + n))/(1 - k + n)) Binomial[n, -k + n];
Table[Table[t[n, k], {k, 0, n}], {n, 0, 10}];
Flatten[%] (* Roger L. Bagula , Feb 17 2009 *)

-t(n,k)=((2*k + 1 - n)/(k + 1))*Binomial[n, k] + ((1 - n + 2 (-k + n))/(1 - k + n)) Binomial[n, -k + n]. [From Roger L. Bagula, Feb 17 2009]

cp's OEIS Frontend

A167374 Triangle, read by rows, given by [ -1,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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A131428 a(n) = 2*C(n) - 1, where C(n) = A000108(n) are the Catalan numbers.

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A118801 Triangle T that satisfies the matrix products: C*[T^-1]*C = T and T*[C^-1]*T = C, where C is Pascal's triangle.

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A239473 Triangle read by rows: signed version of A059260: coefficients for expansion of partial sums of sequences a(n,x) in terms of their binomial transforms (1+a(.,x))^n ; Laguerre polynomial expansion of the truncated exponential.

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A080233 Triangle T(n,k) obtained by taking differences of consecutive pairs of row elements of Pascal's triangle A007318.

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A382435 a(n) = Sum_{k=0..n} ( binomial(n,k) - binomial(n,k-1) )^6.

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A118801 Triangle T that satisfies the matrix products: C[T^-1]C = T and T[C^-1]T = C, where C is Pascal's triangle.