A082985 -id:A082985 - 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

A084534 Triangle read by rows: row #n has n+1 terms. T(n,0)=1, T(n,n)=2, T(n,m) = T(n-1,m-1) + Sum_{k=0..m} T(n-1-k,m-k).

Original entry on oeis.org

1, 1, 2, 1, 4, 2, 1, 6, 9, 2, 1, 8, 20, 16, 2, 1, 10, 35, 50, 25, 2, 1, 12, 54, 112, 105, 36, 2, 1, 14, 77, 210, 294, 196, 49, 2, 1, 16, 104, 352, 660, 672, 336, 64, 2, 1, 18, 135, 546, 1287, 1782, 1386, 540, 81, 2, 1, 20, 170, 800, 2275, 4004, 4290, 2640, 825, 100, 2

Offset: 0

Gary W. Adamson, May 29 2003

Sum of row #n = A000204(2n). (But sum of row #0 = 1.)
Row #n has the unsigned coefficients of the monic polynomial whose roots are 2 cos(Pi*(2k-1)/(4n)) for k=1..2n. [Comment corrected by Barry Brent, Jan 03 2006]
The positive roots are some diagonal lengths of a regular (4n)-gon, inscribed in the unit circle.
Polynomial of row #n = Sum_{m=0..n} (-1)^m * T(n,m) x^(2*n-2*m).
This is the unsigned version of the coefficient table for scaled Chebyshev T(2*n,x) polynomials. - Wolfdieter Lang, Mar 07 2007
Reversed A127677 (cf. A156308, A217476, A263916). - Tom Copeland, Nov 07 2015

			First few Chebyshev T(2*n,x) polynomials:
  T(2*0,x) = 1;
  T(2*1,x) = x^2 -   2;
  T(2*2,x) = x^4 -   4*x^2 +  2;
  T(2*3,x) = x^6 -   6*x^4 +  9*x^2 -  2;
  T(2*4,x) = x^8 -   8*x^6 + 20*x^4 - 16*x^2 +  2;
  T(2*5,x) = x^10 - 10*x^8 + 35*x^6 - 50*x^4 + 25*x^2 - 2;
Triangle begins as:
  1;
  1,  2;
  1,  4,  2;
  1,  6,  9,   2;
  1,  8, 20,  16,   2;
  1, 10, 35,  50,  25,  2;
  1, 12, 54, 112, 105, 36, 2;

I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. See p. 118.
Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990. p. 37, eq.(1.96) and p. 4. eq.(1.10).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113-124. [Scan of annotated copy]

Row sums are A005248 for n > 0.
Companion triangle A082985.
Cf. A082985 (unsigned scaled coefficient table for Chebyshev's T(2*n+1, x) polynomials).
Cf. A127677, A156308, A217476, A263916.

A084534:= func< n,k | k eq 0 select 1 else 2*(n/k)*Binomial(2*n-k-1, k-1) >;
[A084534(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 02 2022

T := proc(n, m): if n=0 then 1 else binomial(2*n-m, m)*2*n/(2*n-m) fi: end: seq(seq(T(n,m),m=0..n),n=0..10); # Johannes W. Meijer, May 31 2018

a[n_, m_] := Binomial[2n-m, m]*2n/(2n-m); a[0, 0] = 1; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Apr 12 2016, after Wolfdieter Lang *)

T(n,m) = if(n==0, m==0, binomial(2*n-m, m)*2*n/(2*n-m)) \\ Andrew Howroyd, Dec 18 2017

def A084534(n,k): return 1 if (k==0) else 2*(n/k)*binomial(2*n-k-1, k-1)
flatten([[A084534(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Feb 02 2022

T(n,m) = binomial(2*n-m, m)*2*n/(2*n-m) for n > 0. - Andrew Howroyd, Dec 18 2017
Signed version from Wolfdieter Lang, Mar 07 2007: (Start)
a(n,m)=0 if na(n,m)=0 if na(n,m)=0 if nA127674(n,n-m)/2^(2*(n-m)-1) (scaled coefficients of Chebyshev's T(2*n,x), decreasing even powers). [corrected by Johannes W. Meijer, May 31 2018] (End)

Edited by Don Reble, Nov 12 2005

A172431 Even row Pascal-square read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 10, 4, 1, 8, 21, 20, 5, 1, 10, 36, 56, 35, 6, 1, 12, 55, 120, 126, 56, 7, 1, 14, 78, 220, 330, 252, 84, 8, 1, 16, 105, 364, 715, 792, 462, 120, 9, 1, 18, 136, 560, 1365, 2002, 1716, 792, 165, 10

Offset: 1

Mark Dols, Feb 02 2010

Apart from signs identical to A053123. Mirror of A078812.
As a triangle, row n consists of the coefficients of Morgan-Voyce polynomial B(n,x); e.g., B(3,x)=x^3+6x^2+10x+4. As a triangle, rows 0 to 4 are as follows: 1 1...2 1...4...3 1...6...10...4 1...8...21...20...5 See A054142 for coefficients of Morgan-Voyce polynomial b(n,x).
Scaled version of A119900. - Philippe Deléham, Feb 24 2012
A172431 is jointly generated with A054142 as an array of coefficients of polynomials v(n,x):  initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Mar 09 2012
Subtriangle of the triangle given by (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 22 2012

			Array begins:
  1,  2,  3,  4,  5,  6, ...
  1,  4, 10, 20, 35, ...
  1,  6, 21, 56, ...
  1,  8, 36, ...
  1, 10, ...
  1, ...
  ...
Example:
Starting with 1, every entry is twice the one to the left minus the second one to the left, plus the one above.
For n = 9 the a(9) = 10 solution is 2*4 - 1 + 3.
From _Philippe Deléham_, Feb 24 2012: (Start)
Triangle T(n,k) begins:
  1;
  1,   2;
  1,   4,   3;
  1,   6,  10,   4;
  1,   8,  21,  20,   5;
  1,  10,  36,  56,  35,   6;
  1,  12,  55, 120, 126,  56,   7; (End)
From _Philippe Deléham_, Mar 22 2012: (Start)
(1, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, ...) begins:
  1;
  1,   0;
  1,   2,   0;
  1,   4,   3,   0;
  1,   6,  10,   4,   0;
  1,   8,  21,  20,   5,   0;
  1,  10,  36,  56,  35,   6,   0;
  1,  12,  55, 120, 126,  56,   7,   0; (End)

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

Cf. A078812, A053123, A007318, A001906 (antidiagonals sums), A007685.

Cf. also A054142, A082985.

F:=Factorial;; Flat(List([1..15], n-> List([1..n], k-> Sum([0..Int((k-1)/2)], j-> (-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)) )))); # G. C. Greubel, Dec 15 2019

F:=Factorial; [ &+[(-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)): j in [0..Floor((k-1)/2)]]: k in [1..n], n in [1..15]]; // G. C. Greubel, Dec 15 2019

T := (n, k) -> simplify(GegenbauerC(k, n-k, 1)):
for n from 0 to 10 do seq(T(n,k), k=0..n-1) od; # Peter Luschny, May 10 2016

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];
v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];
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[%]    (* A054142 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A172431 *)
(* Clark Kimberling, Mar 09 2012 *)
Table[GegenbauerC[k-1, n-k+1, 1], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Dec 15 2019 *)

T(n,k) = sum(j=0, (k-1)\2, (-1)^j*(n-j-1)!*2^(k-2*j-1)/(j!*(n-k)!*(k-2*j-1)!) );
for(n=1, 10, for(k=1, n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 15 2019

[[gegenbauer(k-1, n-k+1, 1) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Dec 15 2019

As a decimal sequence: a(n)= 12*a(n-1)- a(n-2) with a(1)=1. [I interpret this remark as: 1, 12=1,2, 143=1,4,3, 1704=1,6,10,4,... taken from A004191 are decimals on the diagonal. - R. J. Mathar, Sep 08 2013]
As triangle T(n,k): T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2). - Philippe Deléham, Feb 24 2012
As DELTA-triangle T(n,k) with 0<=k<=n: G.f.: (1-y*x)^2/((1-y*x)^2-x). - Philippe Deléham, Mar 22 2012
T(n, k) = GegenbauerC(k, n-k, 1). - Peter Luschny, May 10 2016
As triangle T(n,k): Product_{k=1..n} T(n,k) = Product_{k=0..n-1} binomial(2*k,k) = A007685(n-1)  for n >= 1. - Werner Schulte, Apr 26 2017
As triangle T(n,k) with 1 <= k <= n: T(n,k) = binomial(2*n-k, k-1). - Paul Weisenhorn, Nov 25 2019

A157000 Triangle T(n,k) = (n/k)*binomial(n-k-1, k-1) read by rows.

Original entry on oeis.org

2, 3, 4, 2, 5, 5, 6, 9, 2, 7, 14, 7, 8, 20, 16, 2, 9, 27, 30, 9, 10, 35, 50, 25, 2, 11, 44, 77, 55, 11, 12, 54, 112, 105, 36, 2, 13, 65, 156, 182, 91, 13, 14, 77, 210, 294, 196, 49, 2, 15, 90, 275, 450, 378, 140, 15, 16, 104, 352, 660, 672, 336, 64, 2, 17, 119, 442, 935, 1122, 714, 204, 17

Offset: 2

Roger L. Bagula, Feb 20 2009

Row sums are A001610(n-1).
Triangle A034807 (coefficients of Lucas polynomials) with the first column omitted. - Philippe Deléham, Mar 17 2013
T(n,k) is the number of ways to select k knights from a round table of n knights, no two adjacent. - Bert Seghers, Mar 02 2014

			The table starts in row n=2, column k=1 as:
   2;
   3;
   4,  2;
   5,  5;
   6,  9,   2;
   7, 14,   7;
   8, 20,  16,   2;
   9, 27,  30,   9;
  10, 35,  50,  25,  2;
  11, 44,  77,  55, 11;
  12, 54, 112, 105, 36, 2;

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 199

G. C. Greubel, Rows n = 2..100 of triangle, flattened

Cf. A132460, A113279, A082985, A029635.

[[n*Binomial(n-k-1,k-1)/k: k in [1..Floor(n/2)]]: n in [2..20]]; // G. C. Greubel, Apr 25 2019

Table[(n/k)*Binomial[n-k-1, k-1], {n,2,20}, {k,1,Floor[n/2]}]//Flatten (* modified by G. C. Greubel, Apr 25 2019 *)

a(n,k)=n*binomial(n-k-1,k-1)/k; \\ Charles R Greathouse IV, Jul 10 2011

[[n*binomial(n-k-1,k-1)/k for k in (1..floor(n/2))] for n in (2..20)] # G. C. Greubel, Apr 25 2019

T(n,k) = binomial(n-k,k) + binomial(n-k-1,k-1). - Bert Seghers, Mar 02 2014

Offset 2, keyword:tabf, more terms by the Assoc. Eds. of the OEIS, Nov 01 2010

A127675 Coefficient table for Chebyshev's U(2*n,x) polynomials in decreasing powers of (1-x^2).

Original entry on oeis.org

1, -4, 3, 16, -20, 5, -64, 112, -56, 7, 256, -576, 432, -120, 9, -1024, 2816, -2816, 1232, -220, 11, 4096, -13312, 16640, -9984, 2912, -364, 13, -16384, 61440, -92160, 70400, -28800, 6048, -560, 15, 65536, -278528, 487424, -452608, 239360, -71808, 11424, -816, 17, -262144, 1245184

Offset: 0

Wolfdieter Lang, Mar 07 2007

This table gives therefore sin((2*n+1)*phi) in terms of falling odd powers of sin(phi).

The unsigned triangle with reversed rows is A084930 (the signs differ).

			[1];[ -4,3];[16,-20,5];[ -64,112,-56,7];[256,-576,432,-120,9]; ...
Row n=3: -64*(1-x^2)^3+ 112*(1-x^2)^2 -56*(1-x^2)^1 + 7 = 64*x^6 - 80*x^4 + 24* x^2 -1 =U(6,x).
Row n=3: sin(7*phi)=-64*sin(phi)^7 + 112*sin(phi)^5 - 56*sin(phi)^3 + 7*sin(phi).

W. Lang, First 15 rows and more.

Row sums (signed) A033999(n)=(-1)^n. Row sums (unsigned) A002315(n).

Cf. A082985 (scaled coefficient table).

a(n,m)=0 if n < m else a(n,m) = ((-4)^(n-m))*binomial(2n-m,m)*(2*n+1)/(2*(n-m)+1), n >= m >= 0. (Proof from the differential eq. for U(2*n,x): (1-x^2)*(d^2/dx^2)U(2*n,x) - 3*x*(d/dx)U(2*n,x) + 4*n*(n+1)*U(2*n,x) = 0.)

a(n,m)=0 if n < m else a(n,m) = Sum_{k=0..n-m} (binomial(m+k,k)*binomial(2*n+1,2*(m+k))*(-1)^(n-m)) (from de Moivre's formula for sin((2*n+1)*phi) after replacing cos(phi)^2 with 1-sin(phi)^2).

A207543 Triangle read by rows, expansion of (1+yx)/(1-2y*x+y(y-1)x^2).

Original entry on oeis.org

1, 0, 3, 0, 1, 5, 0, 0, 5, 7, 0, 0, 1, 14, 9, 0, 0, 0, 7, 30, 11, 0, 0, 0, 1, 27, 55, 13, 0, 0, 0, 0, 9, 77, 91, 15, 0, 0, 0, 0, 1, 44, 182, 140, 17, 0, 0, 0, 0, 0, 11, 156, 378, 204, 19, 0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21, 0

Offset: 0

Philippe Deléham, Feb 24 2012

Previous name was: "A scaled version of triangle A082985."

Triangle, read by rows, given by (0, 1/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (3, -4/3, 1/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

			Triangle begins :
1
0, 3
0, 1, 5
0, 0, 5, 7
0, 0, 1, 14, 9
0, 0, 0, 7, 30, 11
0, 0, 0, 1, 27, 55, 13
0, 0, 0, 0, 9, 77, 91, 15
0, 0, 0, 0, 1, 44, 182, 140, 17
0, 0, 0, 0, 0, 11, 156, 378, 204, 19
0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21
0, 0, 0, 0, 0, 0, 13, 275, 1122, 1254, 385, 23

R. Zielinski, Induction and Analogy in a Problem of Finite Sums arXiv:1608.04006 [math.GM], 2016.

Cf. A082985 which is another version of this triangle.

Cf. Diagonals : A005408, A000330, A005585, A050486, A054333, A057788. Cf. A119900.

s := (1+y*x)/(1-2*y*x+y*(y-1)*x^2): t := series(s,x,12):
seq(print(seq(coeff(coeff(t,x,n),y,m),m=0..n)),n=0..11); # Peter Luschny, Aug 17 2016

T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 3.
G.f.: (1+y*x)/(1-2*y*x+y*(y-1)*x^2).
Sum_{i, i>=0} T(n+i,n) = A000204(2*n+1).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A078069(n), A000007(n), A003945(n), A111566(n) for x = -1, 0, 1, 2 respectively.
Sum_{k, 0<=k<=n} T(n,k)*x^(n-k) = A090131(n), A005408(n), A003945(n), A078057(n), A028859(n), A000244(n), A063782(n), A180168(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively.
From Peter Bala, Aug 17 2016: (Start)
Let S(k,n) = Sum_{i = 1..n} i^k. Calculations in Zielinski 2016 suggest the following identity holds involving the p-th row elements of this triangle:
Sum_{k = 0..p} T(p,k)*S(2*k,n) = 1/2*(2*n + 1)*(n*(n + 1))^p.
For example, for row 6 we find S(6,n) + 27*S(8,n) + 55*S(10,n) + 13*S(12,n) = 1/2*(2*n + 1)*(n*(n + 1))^6.
There appears to be a similar result for the odd power sums S(2*k + 1,n) involving A119900. (End)

New name using a formula of the author from Peter Luschny, Aug 17 2016

A356777 G.f.: Sum_{n=-oo..+oo} x^(n^2) * C(x)^(2n-1), where C(x) = 1 + xC(x)^2 is a g.f. of the Catalan numbers (A000108).

Original entry on oeis.org

1, 1, -3, 0, 1, -5, 5, 0, 0, 1, -7, 14, -7, 0, 0, 0, 1, -9, 27, -30, 9, 0, 0, 0, 0, 1, -11, 44, -77, 55, -11, 0, 0, 0, 0, 0, 1, -13, 65, -156, 182, -91, 13, 0, 0, 0, 0, 0, 0, 1, -15, 90, -275, 450, -378, 140, -15, 0, 0, 0, 0, 0, 0, 0, 1, -17, 119, -442, 935, -1122, 714, -204, 17, 0, 0, 0, 0, 0, 0, 0, 0, 1, -19, 152, -665, 1729, -2717, 2508, -1254, 285, -19

Offset: 0

Paul D. Hanna, Sep 08 2022

sign

			G.f.: A(x) = 1 + x - 3*x^2 + x^4 - 5*x^5 + 5*x^6 + x^9 - 7*x^10 + 14*x^11 - 7*x^12 + x^16 - 9*x^17 + 27*x^18 - 30*x^19 + 9*x^20 + x^25 - 11*x^26 + 44*x^27 - 77*x^28 + 55*x^29 - 11*x^30 + x^36 - 13*x^37 + 65*x^38 - 156*x^39 + 182*x^40 - 91*x^41 + 13*x^42 + x^49 - 15*x^50 + 90*x^51 - 275*x^52 + 450*x^53 - 378*x^54 + 140*x^55 - 15*x^56 + ...
such that
A(x) = ... + x^16/C(x)^9 + x^9/C(x)^7 + x^4/C(x)^5 + x/C(x)^3 + 1/C(x) + x*C(x) + x^4*C(x)^3 + x^9*C(x)^5 + x^16*C(x)^7 + x^25*C(x)^9 + ... + x^(n^2)*C^(2*n-1) + ...
where the Catalan function C(x) = (1 - sqrt(1-4*x))/(2*x) begins
C(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + 1430*x^8 + 4862*x^9 + ... + A000108(n)*x^n + ...
RELATED TABLE.
This sequence may be written in the form of an irregular triangle:
1,
1, -3, 0,
1, -5, 5, 0, 0,
1, -7, 14, -7, 0, 0, 0,
1, -9, 27, -30, 9, 0, 0, 0, 0,
1, -11, 44, -77, 55, -11, 0, 0, 0, 0, 0,
1, -13, 65, -156, 182, -91, 13, 0, 0, 0, 0, 0, 0,
1, -15, 90, -275, 450, -378, 140, -15, 0, 0, 0, 0, 0, 0, 0,
1, -17, 119, -442, 935, -1122, 714, -204, 17, 0, 0, 0, 0, 0, 0, 0, 0,
1, -19, 152, -665, 1729, -2717, 2508, -1254, 285, -19, 0, 0, 0, 0, 0, 0, 0, 0, 0,
...
Compare the above construction to triangle A082985.

Paul D. Hanna, Table of n, a(n) for n = 0..2555

Cf. A355341, A355342, A355345, A356778, A082985, A000108.

/* By Definition: */
{a(n) = my(A, C = 1/x*serreverse(x-x^2 +O(x^(n+2))), M=ceil(sqrt(n+1)));
A = sum(m=-M, M, x^(m^2) * C^(2*m-1) ); polcoeff(A, n)}
for(n=0, 90, print1(a(n), ", "))

/* Without Using Catalan Series */
{a(n) = my(A, M=ceil(sqrt(n+1)));
A = sum(m=0, M, sum(k=0, 2*m, (-1)^k*binomial(2*m-k, k)*(2*m+1)/(2*m-2*k+1) * x^(m^2 + k) ) +x*O(x^n)); polcoeff(A, n)}
for(n=0, 90, print1(a(n), ", "))

G.f. A(x) = Sum_{n>=0} a(n)*x^n may be obtained from the following expressions; here, C(x) = 1 + x*C(x)^2 is the g.f. of the Catalan numbers (A000108).
(1) A(x) = Sum_{n=-oo..+oo} x^(n^2) * C(x)^(2*n-1).
(2) A(x) = 1/C(x) * Product_{n>=1} (1 + x^(2*n-1)*C(x)^2) * (1 + x^(2*n-1)/C(x)^2) * (1 - x^(2*n)), by the Jacobi triple product identity.
(3) A(x) = 1/C(x) + Sum_{n>=1} x^(n^2) * (C(x)^(2*n-1) + 1/C(x)^(2*n+1)).
(4) A(x) = Sum_{n>=0} Sum_{k=0..n} (-1)^k * binomial(2*n-k, k) * (2*n+1)/(2*n-2*k+1) * x^(n^2 + k).

Showing 1-10 of 12 results. Next

cp's OEIS Frontend

A125175 Triangle T(n,k) = |A053123(n/2+k/2,k)| for even n+k, T(n,k)= A082985((n+k-1)/2,k) for odd n+k; read by rows, 0<=k<=n.

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A084533 Duplicate of A082985.

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A230069 Numerators of inverse of triangle A082985(n).

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

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A084534 Triangle read by rows: row #n has n+1 terms. T(n,0)=1, T(n,n)=2, T(n,m) = T(n-1,m-1) + Sum_{k=0..m} T(n-1-k,m-k).

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A172431 Even row Pascal-square read by antidiagonals.

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A157000 Triangle T(n,k) = (n/k)*binomial(n-k-1, k-1) read by rows.

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

A207543 Triangle read by rows, expansion of (1+yx)/(1-2y*x+y(y-1)x^2).

A356777 G.f.: Sum_{n=-oo..+oo} x^(n^2) * C(x)^(2n-1), where C(x) = 1 + xC(x)^2 is a g.f. of the Catalan numbers (A000108).