A100218 -id:A100218 - cp's OEIS frontend

A111125 Triangle read by rows: T(k,s) = ((2k+1)/(2s+1))binomial(k+s,2s), 0 <= s <= k.

1, 3, 1, 5, 5, 1, 7, 14, 7, 1, 9, 30, 27, 9, 1, 11, 55, 77, 44, 11, 1, 13, 91, 182, 156, 65, 13, 1, 15, 140, 378, 450, 275, 90, 15, 1, 17, 204, 714, 1122, 935, 442, 119, 17, 1, 19, 285, 1254, 2508, 2717, 1729, 665, 152, 19, 1, 21, 385, 2079, 5148, 7007, 5733, 2940, 952, 189, 21, 1

Offset: 0

N. J. A. Sloane, Oct 16 2005

Riordan array ((1+x)/(1-x)^2, x/(1-x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1-x), x/(1-x))(1+2*x, x*(1+x)). - Paul Barry, Oct 17 2005
Central coefficients of rows with odd numbers of term are A052227.
From Wolfdieter Lang, Jun 26 2011: (Start)
T(k,s) appears as T_s(k) in the Knuth reference, p. 285.
This triangle is related to triangle A156308(n,m), appearing in this reference as U_m(n) on p. 285, by T(k,s) - T(k-1,s) = A156308(k,s), k>=s>=1 (identity on p. 286). T(k,s) = A156308(k+1,s+1) - A156308(k,s+1), k>=s>=0 (identity on p. 286).
(End)
A111125 is jointly generated with A208513 as an array of coefficients of polynomials v(n,x): initially, u(1,x)= v(1,x)= 1; for n>1, u(n,x)= u(n-1,x) +x*(x+1)*v(n-1) and v(n,x)= u(n-1,x) +x*v(n-1,x) +1. See the Mathematica section. The columns of A111125 are identical to those of A208508. Here, however, the alternating row sums are periodic (with period 1,2,1,-1,-2,-1). - Clark Kimberling, Feb 28 2012
This triangle T(k,s) (with signs and columns scaled with powers of 5) appears in the expansion of Fibonacci numbers F=A000045 with multiples of odd numbers as indices in terms of odd powers of F-numbers. See the Jennings reference, p. 108, Theorem 1. Quoted as Lemma 3 in the Ozeki reference given in A111418. The formula is: F_{(2*k+1)*n} = Sum_{s=0..k} ( T(k,s)*(-1)^((k+s)*n)*5^s*F_{n}^(2*s+1) ), k >= 0, n >= 0. - Wolfdieter Lang, Aug 24 2012
From Wolfdieter Lang, Oct 18 2012: (Start)
This triangle T(k,s) appears in the formula x^(2*k+1) - x^(-(2*k+1)) = Sum_{s=0..k} ( T(k,s)*(x-x^(-1))^(2*s+1) ), k>=0. Prove the inverse formula (due to the Riordan property this will suffice) with the binomial theorem. Motivated to look into this by the quoted paper of Wang and Zhang, eq. (1.4).
Alternating row sums are A057079.
The Z-sequence of this Riordan array is A217477, and the A-sequence is (-1)^n*A115141(n). For the notion of A- and Z-sequences for Riordan triangles see a W. Lang link under A006232. (End)
The signed triangle ((-1)^(k-s))*T(k,s) gives the coefficients of (x^2)^s of the polynomials C(2*k+1,x)/x, with C the monic integer Chebyshev T-polynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1-x)/(1+x)^2, x/(1+x)^2). Proof by comparing the o.g.f. of the row polynomials where x is replaced by x^2 with the odd part of the bisection of the o.g.f. for C(n,x)/x. - Wolfdieter Lang, Oct 23 2012
From Wolfdieter Lang, Oct 04 2013: (Start)
The signed triangle S(k,s) := ((-1)^(k-s))*T(k,s) (see the preceding comment) is used to express in a (4*(k+1))-gon the length ratio s(4*(k+1)) = 2*sin(Pi/4*(k+1)) =  2*cos((2*k+1)*Pi/(4*(k+1))) of a side/radius as a polynomial in rho(4*(k+1)) = 2*cos(Pi/4*(k+1)), the length ratio (smallest diagonal)/side:
  s(4*(k+1)) = Sum_{s=0..k} ( S(k,s)*rho(4*(k+1))^(2*s+1) ).
This is to be computed modulo C(4*(k+1), rho(4*(k+1)) = 0, the minimal polynomial (see A187360) in order to obtain  s(4*(k+1)) as an integer in the algebraic number field Q(rho(4*(k+1))) of degree delta(4*(k+1)) (see A055034). Thanks go to Seppo Mustonen for asking me to look into the problem of the square of the total length in a regular n-gon, where this formula is used in the even n case. See A127677 for the formula in the (4*k+2)-gon. (End)
From Wolfdieter Lang, Aug 14 2014: (Start)
The row polynomials for the signed triangle (see the Oct 23 2012 comment above), call them todd(k,x) = Sum_{s=0..k} ( (-1)^(k-s)*T(k,s)*x^s ) = S(k, x-2) - S(k-1, x-2), k >= 0, with the Chebyshev S-polynomials (see their coefficient triangle (A049310) and S(-1, x) = 0), satisfy the recurrence todd(k, x) = (-1)^(k-1)*((x-4)/2)*todd(k-1, 4-x) + ((x-2)/2)*todd(k-1, x), k >= 1, todd(0, x) = 1. From the Aug 03 2014 comment on A130777.
This leads to a recurrence for the signed triangle, call it S(k,s) as in the Oct 04 2013 comment: S(k,s) = (1/2)*(1 + (-1)^(k-s))*S(k-1,s-1) + (2*(s+1)*(-1)^(k-s) - 1)*S(k-1,s) + (1/2)*(-1)^(k-s)*Sum_{j=0..k-s-2} ( binomial(j+s+2,s)*4^(j+2)* S(k-1, s+1+j) ) for k >= s >= 1, and S(k,s) = 0 if k < s and S(k,0) = (-1)^k*(2*k+1). Note that the recurrence derived from the Riordan A-sequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k-1,s-1+j), j=0..k-s), k >= s >=1.
(End)
From Tom Copeland, Nov 07 2015: (Start)
Rephrasing notes here: Append an initial column of zeros, except for a 1 at the top, to A111125 here. Then the partial sums of the columns of this modified entry are contained in A208513. Append an initial row of zeros to A208513. Then the difference of consecutive pairs of rows of the modified A208513 generates the modified A111125. Cf. A034807 and A127677.
For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 20 and 21) and Damianou and Evripidou (p. 7).
As suggested by the equations on p. 7 of Damianou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = (A(2*n+1, x) + 2)/x = (F(2*n+1, (2-x), 1, 0, 0, ... ) + 2)/x = F(2*n+1, -x, 2*x, -3*x, ..., (-1)^n n*x)/x = -F(2*n+1, x, 2*x, 3*x, ..., n*x)/x, where A(n,x) are the polynomials of A127677 and F(n, ...) are the Faber polynomials of A263196. Cf. A127672 and A127677.
(End)
The row polynomials P(k, x) of the signed triangle S(k, s) = ((-1)^(k-s))*T(k, s) are given from the row polynomials R(2*k+1, x) of triangle A127672 by
P(k, x) =  R(2*k+1, sqrt(x))/sqrt(x). - Wolfdieter Lang, May 02 2021

			Triangle T(k,s) begins:
k\s  0    1     2     3     4     5     6    7    8   9 10
0:   1
1:   3    1
2:   5    5     1
3:   7   14     7     1
4:   9   30    27     9     1
5:  11   55    77    44    11     1
6:  13   91   182   156    65    13     1
7:  15  140   378   450   275    90    15    1
8:  17  204   714  1122   935   442   119   17    1
9:  19  285  1254  2508  2717  1729   665  152   19   1
10: 21  385  2079  5148  7007  5733  2940  952  189  21  1
... Extended and reformatted by _Wolfdieter Lang_, Oct 18 2012
Application for Fibonacci numbers F_{(2*k+1)*n}, row k=3:
F_{7*n} = 7*(-1)^(3*n)*F_n + 14*(-1)^(4*n)*5*F_n^3 + 7*(-1)^(5*n)*5^2*F_n^5 + 1*(-1)^(6*n)*5^3*F_n^7, n>=0. - _Wolfdieter Lang_, Aug 24 2012
Example for the  Z- and A-sequence recurrences  of this Riordan triangle: Z = A217477 = [3,-4,12,-40,...]; T(4,0) = 3*7 -4*14 +12*7 -40*1 = 9. A =  [1, 2, -1, 2, -5, 14, ..]; T(5,2) = 1*30 + 2*27 - 1*9 + 2*1= 77. _Wolfdieter Lang_, Oct 18 2012
Example for the (4*(k+1))-gon length ratio s(4*(k+1))(side/radius) as polynomial in the ratio rho(4*(k+1)) ((smallest diagonal)/side): k=0, s(4) = 1*rho(4) = sqrt(2); k=1, s(8) = -3*rho(8) + rho(8)^3 = sqrt(2-sqrt(2)); k=2, s(12) = 5*rho(12) - 5*rho(12)^3 + rho(12)^5, and C(12,x) = x^4 - 4*x^2 + 1, hence rho(12)^5 = 4*rho(12)^3 - rho(12), and s(12) = 4*rho(12) - rho(12)^3 = sqrt(2 - sqrt(3)). - _Wolfdieter Lang_, Oct 04 2013
Example for the recurrence for the signed triangle S(k,s)= ((-1)^(k-s))*T(k,s) (see the Aug 14 2014 comment above):
S(4,1) = 0 + (-2*2 - 1)*S(3,1) - (1/2)*(3*4^2*S(3,2) + 4*4^3*S(3,3)) = - 5*14 - 3*8*(-7) - 128*1 = -30. The recurrence from the Riordan A-sequence A115141 is S(4,1) = -7 -2*14 -(-7) -2*1 = -30. - _Wolfdieter Lang_, Aug 14 2014

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
R. Andre-Jeannin, A generalization of Morgan-Voyce polynomials, The Fibonacci Quarterly 32.3 (1994): 228-31.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.
Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group and Chebyshev polynomials, arXiv:2502.13673 [math.CO], 2025.
K. Dilcher and K. B. Stolarsky, A Pascal-type triangle characterizing twin primes, Amer. Math. Monthly, 112 (2005), 673-681.
P. Damianou , On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014.
P. Damianou and C. Evripidou, Characteristic and Coxeter polynomials for affine Lie algebras, arXiv preprint arXiv:1409.3956 [math.RT], 2014.
D. Jennings, Some Polynomial Identities for the Fibonacci and Lucas Numbers, Fib. Quart., 31(2) (1993), 134-137.
D. E. Knuth, Johann Faulhaber and sums of powers, Math. Comp. 61 (1993), no. 203, 277-294.
Yidong Sun, Numerical triangles and several classical sequences, Fib. Quart., Nov. 2005, pp. 359-370.
T. Wang and W. Zhang, Some identities involving Fibonacci, Lucas polynomials and their applications, Bull. Math. Soc. Sci. Math. Roumanie, Tome 55(103), No.1, (2012) 95-103.
Eric Weisstein's World of Mathematics, Morgan-Voyce polynomials

Mirror image of A082985, which see for further references, etc.
Also closely related to triangles in A098599 and A100218.
Cf. A052227, A085478, A208513, A208508, A211957
Cf. A034807, A127672, A127677, A263196.

[((2*n+1)/(n+k+1))*Binomial(n+k+1, 2*k+1): k in [0..n], n in [0..12]];  // G. C. Greubel, Feb 01 2022

(* First program *)
u[1, x_]:=1; v[1, x_]:=1; z=16;
u[n_, x_]:= u[n-1, x] + x*v[n-1, x];
v[n_, x_]:= u[n-1, x] + (x+1)*v[n-1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A208513 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A111125 *) (* Clark Kimberling, Feb 28 2012 *)
(* Second program *)
T[n_, k_]:= ((2*n+1)/(2*k+1))*Binomial[n+k, 2*k];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 01 2022 *)

@CachedFunction
def T(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = 3*T(n-1,k) if n==1 else 2*T(n-1,k)
    return T(n-1,k-1) - T(n-2,k) - h
A111125 = lambda n,k: (-1)^(n-k)*T(n,k)
for n in (0..9): [A111125(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.
From Peter Bala, Apr 30 2012: (Start)
T(n,k) = binomial(n+k,2*k) + 2*binomial(n+k,2*k+1).
The row generating polynomials P(n,x) are a generalization of the Morgan-Voyce polynomials b(n,x) and B(n,x). They satisfy the recurrence equation P(n,x) = (x+2)*P(n-1,x) - P(n-2,x) for n >= 2, with initial conditions P(0,x) = 1, P(1,x) = x+r+1 and with r = 2. The cases r = 0 and r = 1 give the Morgan-Voyce polynomials A085478 and A078812 respectively. Andre-Jeannin has considered the case of general r.
P(n,x) = U(n+1,1+x/2) + U(n,1+x/2), where U(n,x) denotes the Chebyshev polynomial of the second kind - see A053117. P(n,x) = (2/x)*(T(2*n+2,u)-T(2*n,u)), where u = sqrt((x+4)/4) and T(n,x) denotes the Chebyshev polynomial of the first kind - see A053120. P(n,x) = Product_{k = 1..n} ( x + 4*(sin(k*Pi/(2*n+1)))^2 ). P(n,x) = 1/x*(b(n+1,x) - b(n-1,x)) and P(n,x) = 1/x*{(b(2*n+2,x)+1)/b(n+1,x) - (b(2*n,x)+1)/b(n,x)}, where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478. Cf. A211957.
(End)
From Wolfdieter Lang, Oct 18 2012 (Start)
O.g.f. column No. s: ((1+x)/(1-x)^2)*(x/(1-x)^2)^s, s >= 0. (from the Riordan data given in a comment above).
O.g.f. of the row polynomials R(k,x):= Sum_{s=0..k} ( T(k,s)*x^s ), k>=0: (1+z)/(1-(2+x)*z+z^2) (from the Riordan property).
(End)
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 12 2013

More terms from Paul Barry, Oct 17 2005

A100100 Triangle T(n,k) = binomial(2*n-k-1, n-k) read by rows.

Original entry on oeis.org

1, 1, 1, 3, 2, 1, 10, 6, 3, 1, 35, 20, 10, 4, 1, 126, 70, 35, 15, 5, 1, 462, 252, 126, 56, 21, 6, 1, 1716, 924, 462, 210, 84, 28, 7, 1, 6435, 3432, 1716, 792, 330, 120, 36, 8, 1, 24310, 12870, 6435, 3003, 1287, 495, 165, 45, 9, 1, 92378, 48620, 24310, 11440, 5005, 2002

Offset: 0

Paul Barry, Nov 08 2004

Sometimes called a Catalan triangle, although there are many other triangles that carry that name - see A009766, A008315, A028364, A033184, A053121, A059365, A062103.
Number of nodes of outdegree k in all ordered trees with n edges. Equivalently, number of ascents of length k in all Dyck paths of semilength n. Example: T(3,2) = 3 because the Dyck paths of semilength 3 are UDUDUD, UD(UU)DD, (UU)DDUD, (UU)DUDD and UUUDDD, where U = (1,1), D = (1,-1), the ascents of length 2 being shown between parentheses. - Emeric Deutsch, Nov 19 2006
Riordan array (f(x), x*g(x)) where f(x) is the g.f. of A088218 and g(x) is the g.f. of A000108. - Philippe Deléham, Jan 23 2010
T(n,k) is the number of nonnegative paths of upsteps U = (1,1) and downsteps D = (1,-1) of length 2*n with k returns to  ground level, the horizontal line through the initial vertex. Example: T(2,1) = 2 counts UDUU, UUDD. Also, T(n,k) = number of these paths whose last descent has length k, that is, k downsteps follow the last upstep. Example: T(2,1) = 2 counts UUUD, UDUD. - David Callan, Nov 21 2011
Belongs to the hitting-time subgroup of the Riordan group. Multiplying this triangle by the square Pascal matrix gives A092392 read as a square array. See the example below. - Peter Bala, Nov 03 2015

			From _Paul Barry_, Mar 15 2010: (Start)
Triangle begins in row n=0 with columns 0<=k<=n as:
    1;
    1,   1;
    3,   2,   1;
   10,   6,   3,  1;
   35,  20,  10,  4,  1;
  126,  70,  35, 15,  5, 1;
  462, 252, 126, 56, 21, 6, 1;
Production matrix begins
  1, 1;
  2, 1, 1;
  3, 1, 1, 1;
  4, 1, 1, 1, 1;
  5, 1, 1, 1, 1, 1;
  6, 1, 1, 1, 1, 1, 1;
  7, 1, 1, 1, 1, 1, 1, 1;
(End)
A092392 as a square array = A100100 * square Pascal matrix:
/1   1  1  1 ...\   / 1          \/1 1  1  1 ...\
|2   3  4  5 ...|   | 1 1        ||1 2  3  4 ...|
|6  10 15 21 ...| = | 3 2 1      ||1 3  6 10 ...|
|20 35 56 84 ...|   |10 6 3 1    ||1 4 10 20 ...|
|70 ...         |   |35 ...      ||1 ...        |
- _Peter Bala_, Nov 03 2015

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Gi-Sang Cheon, Hana Kim, and Louis W. Shapiro, An algebraic structure for Faber polynomials, Lin. Alg. Applic. 433 (2010) 1170-1179.
Tian-Xiao He and Renzo Sprugnoli, Sequence characterization of Riordan arrays, Discrete Math. 309 (2009), no. 12, 3962-3974.

Row sums are A000984. Equivalent to A092392, to which A088218 has been added as a first column. Columns include A088218, A000984, A001700, A001791, A002054, A002694. Diagonal sums are A100217. Matrix inverse is A100218.

Cf. A059481 (mirrored). Cf. A033184, A094527, A113955.

a100100 n k = a100100_tabl !! n !! n
a100100_row n = a100100_tabl !! n
a100100_tabl = [1] : f a092392_tabl where
   f (us : wss'@(vs : wss)) = (vs !! 1 : us) : f wss'
-- Reinhard Zumkeller, Jan 15 2014

/* As triangle */ [[Binomial(2*n - k - 1, n - k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Nov 21 2018

A100100 := proc(n,k)
    binomial(2*n-k-1,n-1) ;
end proc:
seq(seq(A100100(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Feb 06 2015

Flatten[Table[Binomial[2 n - k - 1, n - k], {n, 0, 11}, {k, 0, n}]] (* Vincenzo Librandi, Nov 21 2018 *)

T(n,k)=binomial(2*n-k-1,n-k) \\ Charles R Greathouse IV, Jan 16 2012

From Peter Bala, Sep 06 2015: (Start)
Matrix product A094527 * P^(-1) = A113955 * P^(-2), where P denotes Pascal's triangle A007318.
Essentially, the logarithmic derivative of A033184. (End)
Let column(k) = [T(n, k), n >= k], then the generating function for column(k) is (2/(sqrt(1-4*x)+1))^(k-1)/sqrt(1-4*x). - Peter Luschny, Mar 19 2021
O.g.f. row polynomials R(n, x) := Sum_{k=0..n} T(n, k)*x^k, i.e. o.g.f. of the triangle, G(z,x) = 1/((2 - c(z))*(1 - x*z*c(z))), with c the o.g.f. of A000108 (Catalan). See the Riordan coomment by Philippe Deléham above. - Wolfdieter Lang, Apr 06 2021

A100219 Expansion of (1-2x)/((1-x)(1-x+x^2)).

Original entry on oeis.org

1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0, 1, 0, -2, -3, -2, 0

Offset: 0

Paul Barry, Nov 08 2004

Row sums of number triangle A100218.

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-2,1).

Cf. A079757, A100218.

&cat[[1,0,-2,-3,-2,0]: n in [0..20]]; // G. C. Greubel, Mar 28 2024

PadRight[{}, 120, {1,0,-2,-3,-2,0}] (* or *) LinearRecurrence[{2,-2,1}, {1,0,-2}, 50] (* G. C. Greubel, Mar 13 2017; Mar 28 2024 *)
Table[Cos[Pi*n/3 + Pi/3] + Sqrt[3]*Sin[Pi*n/3 + Pi/3] - 1, {n, 0, 71}] (* Indranil Ghosh, Mar 13 2017 *)

my(x='x+O('x^50)); Vec((1-2*x)/((1-x)*(1-x+x^2))) \\ G. C. Greubel, Mar 13 2017

def A100219(n): return [1,0,-2,-3,-2,0][n%6]
[A100219(n) for n in range(121)] # G. C. Greubel, Mar 28 2024

a(n) = 2*a(n-1) - 2*a(n-2) + a(n-3).
a(n) = cos(Pi*n/3 + Pi/3) + sqrt(3)*sin(Pi*n/3 + Pi/3) - 1.
a(n) is the n-th order Taylor polynomial (centered at 0) of 1/c(x)^n evaluated at x = 1, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. of the Catalan numbers A000108. - Peter Bala, Apr 20 2024

A098599 Riordan array ((1+2*x)/(1+x), (1+x)).

Original entry on oeis.org

1, 1, 1, -1, 2, 1, 1, 0, 3, 1, -1, 0, 2, 4, 1, 1, 0, 0, 5, 5, 1, -1, 0, 0, 2, 9, 6, 1, 1, 0, 0, 0, 7, 14, 7, 1, -1, 0, 0, 0, 2, 16, 20, 8, 1, 1, 0, 0, 0, 0, 9, 30, 27, 9, 1, -1, 0, 0, 0, 0, 2, 25, 50, 35, 10, 1, 1, 0, 0, 0, 0, 0, 11, 55, 77, 44, 11, 1, -1, 0, 0, 0, 0, 0, 2, 36, 105, 112, 54, 12, 1, 1, 0, 0, 0, 0, 0, 0, 13, 91, 182, 156, 65, 13, 1

Offset: 0

Paul Barry, Sep 17 2004

			Triangle begins as:
   1;
   1, 1;
  -1, 2, 1;
   1, 0, 3, 1;
  -1, 0, 2, 4, 1;
   1, 0, 0, 5, 5,  1;
  -1, 0, 0, 2, 9,  6,  1;
   1, 0, 0, 0, 7, 14,  7,  1;
  -1, 0, 0, 0, 2, 16, 20,  8, 1;
   1, 0, 0, 0, 0,  9, 30, 27, 9, 1;

G. C. Greubel, Table of n, a(n) for n = 0..1325

Cf. A000007, A005408, A029635, A040000, A079757, A098601.
Row sums are A098600.
Diagonal sums are A098601.
Apart from signs, same as A100218.
Very similar to triangle A111125.

A098599:= func< n,k | n eq 0 select 1 else Binomial(k, n-k) + Binomial(k-1, n-k-1) >;
[A098599(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 27 2024

T[n_, k_]:= If[n==0, 1, Binomial[k,n-k] +Binomial[k-1,n-k-1]];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 27 2024 *)

def A098599(n,k): return 1 if n==0 else binomial(k, n-k) + binomial(k-1, n-k-1)
flatten([[A098599(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Mar 27 2024

Triangle: T(n, k) = binomial(k, n-k) + binomial(k-1, n-k-1), with T(0, 0) = 1.
Sum_{k=0..n} T(n, k) = A098600(n) (row sums).
T(n,k) = T(n-1,k-1) - T(n-1,k) + 2*T(n-2,k-1) + T(n-3,k-1), T(0,0)=1, T(1,0)=1, T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 09 2014
From G. C. Greubel, Mar 27 2024: (Start)
T(2*n, n) = A040000(n).
T(2*n+1, n) = A000007(n).
T(2*n-1, n) = A005408(n-1), n >= 1.
Sum_{k=0..n} (-1)^k*T(n, k) = A079757(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A098601(n). (End)

A098601 Expansion of (1+2x)/((1+x)(1-x^2-x^3)).

Original entry on oeis.org

1, 1, 0, 3, 0, 4, 2, 5, 5, 8, 9, 14, 16, 24, 29, 41, 52, 71, 92, 124, 162, 217, 285, 380, 501, 666, 880, 1168, 1545, 2049, 2712, 3595, 4760, 6308, 8354, 11069, 14661, 19424, 25729, 34086, 45152, 59816, 79237, 104969, 139052, 184207, 244020, 323260

Offset: 0

Paul Barry, Sep 17 2004

Diagonal sums of A098599.

The signed sequence 1,-1,0,-3,0,-4,... gives the diagonal sums of A100218. - Paul Barry, Nov 09 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,1,2,1).

Cf. A000931, A077883, A098599, A100218, A135364.

I:=[1,1,0,3]; [n le 4 select I[n] else -Self(n-1) +Self(n-2) +2*Self(n-3) +Self(n-4): n in [1..55]]; // G. C. Greubel, Mar 27 2024

CoefficientList[Series[(1+2x)/((1+x)(1-x^2-x^3)),{x,0,50}],x] (* or *) LinearRecurrence[{-1,1,2,1},{1,1,0,3},50] (* Harvey P. Dale, Dec 14 2011 *)

def A098601(n): return sum( binomial(k, n-2*k) + binomial(k-1, n-2*k-1) for k in range(1+n//2))
[A098601(n) for n in range(56)] # G. C. Greubel, Mar 27 2024

G.f.: x/((1+x)*(1-x^2-x^3)) + 1/(1-x^2-x^3).
a(n) = Sum_{k=0..floor(n/2)} (binomial(k, n-2*k) + binomial(k-1, n-2*k-1)).
a(n) = -a(n-1) + a(n-2) + 2*a(n-3) + a(n-4).
Inverse binomial transform of A135364. - Paul Curtz, Apr 25 2008

A111126 Triangle read by rows: T(k,s) = binomial(k+s,2s+1)(2k-1)(2k+1)/(2s+3), k >= 1, 0 <= s <= k-1.

Original entry on oeis.org

1, 10, 3, 35, 28, 5, 84, 126, 54, 7, 165, 396, 297, 88, 9, 286, 1001, 1144, 572, 130, 11, 455, 2184, 3510, 2600, 975, 180, 13, 680, 4284, 9180, 9350, 5100, 1530, 238, 15, 969, 7752, 21318, 28424, 20995, 9044, 2261, 304, 17, 1330, 13167, 45144, 76076, 72618

Offset: 1

N. J. A. Sloane, Oct 16 2005

			Triangle starts:
1;
10,3;
35,28,5;
84,126,54,7;
165,396,297,88,9;

K. Dilcher and K. B. Stolarsky, A Pascal-type triangle characterizing twin primes, Amer. Math. Monthly, 112 (2005), 673-681.

Mirror image of A111127. Cf. A111125, A082985, A100218, A098599.

T:=(k,s)->binomial(k+s,2*s+1)*(2*k-1)*(2*k+1)/(2*s+3): for k from 1 to 10 do seq(T(k,s),s=0..k-1) od; # yields sequence in triangular form; Emeric Deutsch, Feb 01 2006

More terms from Emeric Deutsch, Feb 01 2006

A247492 Triangle read by rows: T(n, k) = binomial(k-1, n-k)*(n+1)/(n+1-k), 0 <= k <= n.

Original entry on oeis.org

1, -1, 2, 1, 0, 3, -1, 0, 2, 4, 1, 0, 0, 5, 5, -1, 0, 0, 2, 9, 6, 1, 0, 0, 0, 7, 14, 7, -1, 0, 0, 0, 2, 16, 20, 8, 1, 0, 0, 0, 0, 9, 30, 27, 9, -1, 0, 0, 0, 0, 2, 25, 50, 35, 10, 1, 0, 0, 0, 0, 0, 11, 55, 77, 44, 11, -1, 0, 0, 0, 0, 0, 2, 36, 105, 112, 54, 12

Offset: 0

Peter Luschny, Oct 01 2014

			[0]  1;
[1] -1, 2;
[2]  1, 0, 3;
[3] -1, 0, 2, 4;
[4]  1, 0, 0, 5, 5;
[5] -1, 0, 0, 2, 9, 6;
[6]  1, 0, 0, 0, 7, 14, 7;
.
Taylor series: 1 + x*(2*y - 1) + x^2*(3*y^2 + 1) + x^3*(4*y^3 + 2*y^2 - 1) + x^4*(5*y^4 + 5*y^3 + 1) + O(x^5).

Cf. A001350 (row sums), A098599, A100218.

T := (n, k) -> (n+1)*binomial(k-1, n-k)/(n+1-k);
for n from 0 to 11 do seq(T(n,k), k=0..n) od;

Sum_{k = 0..n} T(n, k) = A001350(n+1).

G.f.: (x^2*y + 1)/((x^4 + 2*x^3 + x^2)*y^2 + (-x^3 - 3*x^2 - 2*x)*y + x + 1). Or: (x^2*y + 1)/((x + 1)*(x*y - 1)*(x^2*y + x*y - 1)). - Vladimir Kruchinin, Oct 23 2021

cp's OEIS Frontend

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A111126 Triangle read by rows: T(k,s) = binomial(k+s,2s+1)*(2k-1)*(2k+1)/(2s+3), k >= 1, 0 <= s <= k-1.

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A247492 Triangle read by rows: T(n, k) = binomial(k-1, n-k)*(n+1)/(n+1-k), 0 <= k <= n.

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