A026637 -id:A026637 - cp's OEIS frontend

A026970 a(n) = Sum_{k=0..n} (k+1) * A026637(n,k).

1, 3, 8, 25, 60, 147, 336, 765, 1700, 3751, 8184, 17745, 38220, 81915, 174752, 371365, 786420, 1660239, 3495240, 7340025, 15379100, 32156323, 67108848, 139810125, 290805060, 603979767, 1252698776, 2594876065, 5368709100

Offset: 0

Clark Kimberling

nonn

Conjectures from Colin Barker, Feb 18 2016: (Start)
a(n) = -((-1)^n-2^(n+2)+3)*(n+2)/6 for n>0.
a(n) = 4*a(n-1)-2*a(n-2)-8*a(n-3)+7*a(n-4)+4*a(n-5)-4*a(n-6) for n>6.
G.f.: (1-x-2*x^2+7*x^3-7*x^4-4*x^5+4*x^6) / ((1-x)^2*(1+x)^2*(1-2*x)^2).
(End)

A026647 a(n) = Sum_{k=0..floor(n/2)} A026637(n-k, k).

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 17, 27, 45, 73, 119, 192, 312, 505, 818, 1323, 2142, 3466, 5609, 9075, 14685, 23761, 38447, 62208, 100656, 162865, 263522, 426387, 689910, 1116298, 1806209, 2922507, 4728717, 7651225, 12379943, 20031168

Offset: 0

Clark Kimberling

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,0,1,-1,-1).

Cf. A026637, A026638, A026639, A026640, A026641, A026642, A026643.
Cf. A026644, A026645, A026646, A026966, A026967, A026968, A026969.
Cf. A026970.

[1] cat [n le 5 select Binomial(n, Floor(n/2)) else Self(n-2) +Self(n-3) +2*Self(n-4) +Self(n-5) +3: n in [1..40]]; // G. C. Greubel, Jul 01 2024

a[n_]:= a[n]= If[n<6, Binomial[n, Floor[n/2]], a[n-2] +a[n-3] +2*a[n- 4] +a[n-5] +3]; (* a = A026647 *)
Table[a[n], {n,0,40}] (* G. C. Greubel, Jul 01 2024 *)
LinearRecurrence[{1,1,0,1,-1,-1},{1,1,2,3,6,10,17},40] (* Harvey P. Dale, Jan 19 2025 *)

@CachedFunction
def a(n): # a = A026647
    if n<6: return binomial(n, n//2)
    else: return a(n-2) + a(n-3) + 2*a(n-4) + a(n-5) + 3
[a(n) for n in range(41)] # G. C. Greubel, Jul 01 2024

G.f.: (1 + x^5 + x^6)/((1-x^4)*(1-x-x^2)).
From G. C. Greubel, Jul 01 2024: (Start)
a(n) = [n=0] - (3/4) + (1/4)*(-1)^n - (1/10)*2^((1-(-1)^n)/2)*(-1)^floor((n+1)/2) + (3/5)*LucasL(n+1).
a(n) = (1/20)*( 12*LucasL(n+1) + 5*(-1)^n - 15 - 2*cos(n*Pi/2) + 4*sin(n*Pi/2) ) + [n=0].
a(n) = a(n-2) + a(n-3) + 2*a(n-4) + a(n-5) + 3, with a(0) = a(1) = 1, a(2) = 2, a(3) = 3, a(4) = 6, a(5) = 10. (End)

A026645 a(n) = Sum_{k=0..floor(n/2)} A026637(n, k).

Original entry on oeis.org

1, 1, 3, 5, 14, 21, 55, 85, 216, 341, 848, 1365, 3340, 5461, 13191, 21845, 52208, 87381, 206968, 349525, 821514, 1398101, 3264044, 5592405, 12979006, 22369621, 51642594, 89478485, 205592744, 357913941, 818848135, 1431655765, 3262611696, 5726623061, 13003800704, 22906492245

Offset: 0

Clark Kimberling

nonn

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

Cf. A026637, A026638, A026639, A026640, A026642, A026643, A026644.

Cf. A026646, A026647, A026966, A026967, A026968, A026969, A026970.

T[n_, k_]:= T[n, k]= If[k==0 || k==n, 1, If[k==1 || k==n-1, Floor[(3*n- 1)/2], T[n-1,k] + T[n-1,k-1] ]];
A026645[n_]:= Sum[T[n, k], {k, 0, Floor[n/2]}];
Table[A026645[n], {n,0,40}] (* G. C. Greubel, Jun 29 2024 *)

@CachedFunction
def T(n,k): # T = A026637
    if k==0 or k==n: return 1
    elif k==1 or k==n-1: return ((3*n-1)//2)
    else: return T(n-1, k) + T(n-1, k-1)
def A026645(n): return sum(T(n,k) for k in range((n//2)+1))
[A026645(n) for n in range(41)] # G. C. Greubel, Jun 29 2024

A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

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

N. J. A. Sloane, Jan 23 2001

Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Diagonal sums give A008346. - Paul Barry, Sep 23 2004
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
From Robert Coquereaux, Oct 01 2014: (Start)
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
Cf. A193844. (End)
From Peter Bala, Feb 07 2024: (Start)
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
From Miquel A. Fiol, Oct 04 2024: (Start)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0). (End)

			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;
Sequences obtained with _Miquel A. Fiol_'s Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6):
A000007  A000012  A000027  A025747  A000292* A000332* A000389*
A059841  A008619  A087811* A002623  A001752  A001753  A001769
A193356  A008794* A005993  A005994  -------  -------  -------
-------  -------  -------  A005995  A018210  -------  A052267
-------  -------  -------  -------  A018211  A018212  -------
-------  -------  -------  -------  -------  A018213  A018214
-------  -------  -------  -------  -------  -------  A062136
*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
     22--12---24---14---44
          | \    / |
         23   11   34
            \  |  /
              13
               |
              33

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
Joseph Briggs, Alex Parker, Coy Schwieder, and Chris Wells, Frogs, hats and common subsequences, arXiv:2404.07285 [math.CO], 2024. See p. 28.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, J. of Geometry and Physics, 1995, Vol 15, pp 333-352.
Robert Coquereaux and Éric Ragoucy, Currents on Grassmann algebras, arXiv:hep-th/9310147, 1993.
Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. II. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 8.
R. H. Hammack and G. D. Smith, Cycle bases of reduced powers of graphs, Ars Math. Contemp. 12 (2017) 183-203.
Christian Kassel, A Künneth formula for the cyclic cohomology of Z2-graded algebras, Math. Ann. 275 (1986) 683.
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, Numerical Algorithms, June 2012, Volume 60, Issue 2, pp 297-314. - From _N. J. A. Sloane_, Oct 12 2012
Ana Filipa Loureiro and Pascal Maroni, Polynomial sequences associated with the classical linear functionals, preprint, Centro de Matemática da Universidade do Porto.
MathOverflow, Cyclotomic Polynomials in Combinatorics
Mark Norfleet, Characterization of second-order strong divisibility sequences of polynomials, The Fibonacci Quarterly, 43(2) (2005), 166-169.

Cf. A059259. Row sums give A001045.
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001
Cf. A026641, A014300.
Cf. A007318, A097805, A097808, A130595, A200139, A062135.

read transforms; 1/(1-y-x*y-x^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);

t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)

T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
for(n=0, 12, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 11 2017

from sympy import binomial
def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017

def A059260_row(n):
    @cached_function
    def prec(n, k):
        if k==n: return 1
        if k==0: return 0
        return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
    return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
for n in (1..9): print(A059260_row(n)) # Peter Luschny, Mar 16 2016

G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson, Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Feb 17 2007
Equals A130595*A097805*A007318 = (inverse Pascal matrix)*(padded Pascal matrix)*(Pascal matrix) = A130595*A200139. Inverse is A097808 = A130595*(padded A130595)*A007318. - Tom Copeland, Nov 14 2016
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
From Miquel A. Fiol, Sep 30 2024: (Start)
The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below. (End)

Formula corrected by Philippe Deléham, Jan 11 2014

A026644 a(n) = a(n-1) + 2*a(n-2) + 2, for n>=3, where a(0)= 1, a(1)= 2, a(2)= 4.

Original entry on oeis.org

1, 2, 4, 10, 20, 42, 84, 170, 340, 682, 1364, 2730, 5460, 10922, 21844, 43690, 87380, 174762, 349524, 699050, 1398100, 2796202, 5592404, 11184810, 22369620, 44739242, 89478484, 178956970, 357913940, 715827882, 1431655764, 2863311530, 5726623060

Offset: 0

Clark Kimberling

Number of moves to solve Chinese rings puzzle.
a(n-1) (with a(0):=0) enumerates all sequences of length m=1,2,...,floor(n/2) with nonzero integer entries n_i satisfying sum |n_i| <= n-m. Rephrasing K. A. Meissner's example p. 6. Example n=4: from length m=1: [1], [2], [3], each in 2 signed versions; from m=2: [1,1] in 2^2 = 4 signed versions. Hence a(3) = a(4-1) = 3*2 + 1*4 = 10.
Also the number of different 3-colorings (out of 4 colors) for the vertices of all triangulated planar polygons on a base with n+1 vertices if the colors of the two base vertices are fixed. - Patrick Labarque, Mar 23 2010
For n > 0, also the total distance that the disks travel from the leftmost peg to the rightmost peg in the Tower of Hanoi puzzle, in the unique solution with 2^n-1 moves (see links). - Sela Fried, Dec 17 2023

Richard I. Hess, Compendium of Over 7000 Wire Puzzles, privately printed, 1991.
Richard I. Hess, Analysis of Ring Puzzles, booklet distributed at 13th International Puzzle Party, Amsterdam, Aug 20 1993.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Thomas Baruchel, Properties of the cumulated deficient binary digit sum, arXiv:1908.02250 [math.NT], 2019.
Sela Fried, Economically solving the Tower of Hanoi puzzle.
Nicolas Gastineau and O. Togni, On S-packing edge-colorings of cubic graphs, arXiv preprint arXiv:1711.10906 [cs.DM], 2017.
Lee Hae-hwang, Illustration of initial terms in terms of rosemary plants
Krzysztof A. Meissner, Black hole entropy in Loop Quantum Gravity, arXiv:gr-qc/0407052, 2004.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Row sums of A026637.
For n >= 1, equals twice A000975, also A001045 - 1.
A167030 is an essentially identical sequence.

[n eq 0 select 1 else (2^(n+2) -3-(-1)^n)/3 : n in [0..40]]; // G. C. Greubel, Jun 28 2024

f:=n-> if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3; fi;

Join[{1}, Floor[(2^Range[3, 40] - 2)/3]] (* or *) LinearRecurrence[{2,1,-2},{1,2,4,10},40] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2012 *)
CoefficientList[Series[(1-x^2+2x^3)/((1-x)(1-x-2x^2)),{x,0,1001}],x] (* Vincenzo Librandi, Apr 04 2012 *)

Vec((1-x^2+2*x^3)/(1-x)/(1-x-2*x^2)+O(x^99)) \\ Charles R Greathouse IV, Apr 04 2012

def A026644(n): return ((4<Chai Wah Wu, Apr 17 2025

[(2^(n+2)-3-(-1)^n)/3 + int(n==0) for n in range(41)] # G. C. Greubel, Jun 28 2024

a(2*k) = 2*a(2*k-1), a(2*k+1) = 2*a(2*k) + 2. - Peter Shor, Apr 11 2002
For n>0: a(n+1) = a(n) + 2*b(n+1) + 4*b(n), where b(k) = A001045(k). - N. J. A. Sloane, May 16 2003
For n>0: if n mod 2 = 0 then (2^(n+2)-4)/3 else (2^(n+2)-2)/3. - Richard Hess
a(2*n) = 2*n-1 + Sum_{k=0..2*n-1} a(k), n>0; a(2*n+1) = 2*n+1 + Sum_{k=0..n} a(k). - Lee Hae-hwang, Sep 17 2002; corrected by R. J. Mathar, Oct 21 2008
a(n) = 2*n + 2*Sum_{k=1..n-2} a(k), n>0. - Lee Hae-hwang, Sep 19 2002; corrected by R. J. Mathar, Oct 21 2008
From Paul Barry, Oct 24 2007: (Start)
G.f.: (1 - x^2 + 2*x^3)/((1 - x)*(1 - x - 2*x^2)).
a(n) = J(n+2) - 1 + 0^n, where J(n) = A001045(n) (Jacobsthal numbers).
a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3).
a(n) = 0^n + Sum_{k=0..n} (2 - 2*0^(n-k))*J(k+1). (End)
a(n) = A052953(n+1) - 2, n>0. [Moved from A020988, R. J. Mathar, Oct 21 2008]
a(n) = floor(A097074(n+1)/2), n>0. - Gary Detlefs, Dec 19 2010
a(n) = A169969(2*n-1) - 1, n>=2; a(n) = 3*2^(n-1) - 1 - A169969(2*n-7), n>=5 . - Yosu Yurramendi, Jul 05 2016
a(n+3) = 3*2^(n+2) - 2 - a(n), n>=1, a(1)=2, a(2)=4, a(3)=10 . - Yosu Yurramendi, Jul 05 2016
a(n) + A084170(n) = 3*2^n - 2, n>=1. - Yosu Yurramendi, Jul 05 2016
E.g.f: (3 - 4*cosh(x) + 4*cosh(2*x) - 2*sinh(x) + 4*sinh(2*x))/3. - Ilya Gutkovskiy, Jul 05 2016
a(n+3) = 9*2^n + A084170(n), n>=0. - Yosu Yurramendi, Jul 07 2016
a(n) = A000975(n+1) - A000035(n+1), n>0, a(0)=1. - Yuchun Ji, Aug 05 2020

Recurrence in definition line found by Lee Hae-hwang, Apr 03 2002

A228053 A triangle formed like Pascal's triangle, but with (-1)^(n+1) on the borders instead of 1.

Original entry on oeis.org

-1, 1, 1, -1, 2, -1, 1, 1, 1, 1, -1, 2, 2, 2, -1, 1, 1, 4, 4, 1, 1, -1, 2, 5, 8, 5, 2, -1, 1, 1, 7, 13, 13, 7, 1, 1, -1, 2, 8, 20, 26, 20, 8, 2, -1, 1, 1, 10, 28, 46, 46, 28, 10, 1, 1, -1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1, 1, 1, 13, 49, 112, 166, 166, 112

Offset: 0

T. D. Noe, Aug 07 2013

This sequence is almost the same as A026637.

T(n,k) = A026637(n-2,k-1) for n > 3, 1 < k < n-1. - Reinhard Zumkeller, Aug 08 2013

			Triangle begins:
  -1,
  1, 1,
  -1, 2, -1,
  1, 1, 1, 1,
  -1, 2, 2, 2, -1,
  1, 1, 4, 4, 1, 1,
  -1, 2, 5, 8, 5, 2, -1,
  1, 1, 7, 13, 13, 7, 1, 1,
  -1, 2, 8, 20, 26, 20, 8, 2, -1,
  1, 1, 10, 28, 46, 46, 28, 10, 1, 1,
  -1, 2, 11, 38, 74, 92, 74, 38, 11, 2, -1

T. D. Noe, Rows n = 0..50 of triangle, flattened

Cf. A007318 (Pascal's triangle), A026637 (many terms in common).
Cf. A051601 (n on the borders), A137688 (2^n on borders).
Cf. A097073 (row sums).
Cf. A227074 (4^n edges), A227075 (3^n edges), A227076 (5^n edges).

a228053 n k = a228053_tabl !! n !! k
a228053_row n = a228053_tabl !! n
a228053_tabl = iterate (\row@(i:_) -> zipWith (+)
   ([- i] ++ tail row ++ [0]) ([0] ++ init row ++ [- i])) [- 1]
  -- Reinhard Zumkeller, Aug 08 2013

t = {}; Do[r = {}; Do[If[k == 0 || k == n, m = (-1)^(n+1), m = t[[n, k]] + t[[n, k + 1]]]; r = AppendTo[r, m], {k, 0, n}]; AppendTo[t, r], {n, 0, 10}]; t = Flatten[t]

cp's OEIS Frontend

A026970 a(n) = Sum_{k=0..n} (k+1) * A026637(n,k).

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A026647 a(n) = Sum_{k=0..floor(n/2)} A026637(n-k, k).

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A026645 a(n) = Sum_{k=0..floor(n/2)} A026637(n, k).

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A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

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A026644 a(n) = a(n-1) + 2*a(n-2) + 2, for n>=3, where a(0)= 1, a(1)= 2, a(2)= 4.

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A228053 A triangle formed like Pascal's triangle, but with (-1)^(n+1) on the borders instead of 1.

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