A113187 -id:A113187 - 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

A111418 Right-hand side of odd-numbered rows of Pascal's triangle.

Original entry on oeis.org

1, 3, 1, 10, 5, 1, 35, 21, 7, 1, 126, 84, 36, 9, 1, 462, 330, 165, 55, 11, 1, 1716, 1287, 715, 286, 78, 13, 1, 6435, 5005, 3003, 1365, 455, 105, 15, 1, 24310, 19448, 12376, 6188, 2380, 680, 136, 17, 1, 92378, 75582, 50388

Offset: 0

Philippe Deléham, Nov 13 2005

Riordan array (c(x)/sqrt(1-4*x),x*c(x)^2) where c(x) is g.f. of A000108. Unsigned version of A113187. Diagonal sums are A014301(n+1).
Triangle T(n,k),0<=k<=n, read by rows defined by :T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=3*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+2*T(n-1,k)+T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 22 2007
Reversal of A122366. - Philippe Deléham, Mar 22 2007
Column k has e.g.f. exp(2x)(Bessel_I(k,2x)+Bessel_I(k+1,2x)). - Paul Barry, Jun 06 2007
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0)-> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
Diagonal sums are A014301(n+1). - Paul Barry, Mar 08 2011
This triangle T(n,k) appears in the expansion of odd powers of Fibonacci numbers F=A000045 in terms of F-numbers with multiples of odd numbers as indices. See the Ozeki reference, p. 108, Lemma 2. The formula is: F_l^(2*n+1) = sum(T(n,k)*(-1)^((n-k)*(l+1))* F_{(2*k+1)*l}, k=0..n)/5^n, n >= 0, l >= 0. - Wolfdieter Lang, Aug 24 2012
Central terms give A052203. - Reinhard Zumkeller, Mar 14 2014
This triangle appears in the expansion of (4*x)^n in terms of the polynomials Todd(n, x):= T(2*n+1, sqrt(x))/sqrt(x) = sum(A084930(n,m)*x^m), n >= 0. This follows from the inversion of the lower triangular Riordan matrix built from A084930 and comparing the g.f. of the row polynomials. - Wolfdieter Lang, Aug 05 2014
From Wolfdieter Lang, Aug 15 2014: (Start)
This triangle is the inverse of the signed Riordan triangle (-1)^(n-m)*A111125(n,m).
This triangle T(n,k) appears in the expansion of x^n in terms of the polynomials todd(k, x):= T(2*k+1, sqrt(x)/2)/(sqrt(x)/2) = S(k, x-2) - S(k-1, x-2) with the row polynomials T and S for the triangles A053120 and A049310, respectively: x^n = sum(T(n,k)*todd(k, x), k=0..n). Compare this with the preceding comment.
The A- and Z-sequences for this Riordan triangle are [1, 2, 1, repeated 0] and [3, 1, repeated 0]. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. This corresponds to the recurrences given in the Philippe Deléham, Mar 22 2007 comment above. (End)

			From _Wolfdieter Lang_, Aug 05 2014: (Start)
The triangle T(n,k) begins:
n\k      0      1      2      3     4     5    6    7   8  9  10 ...
0:       1
1:       3      1
2:      10      5      1
3:      35     21      7      1
4:     126     84     36      9     1
5:     462    330    165     55    11     1
6:    1716   1287    715    286    78    13    1
7:    6435   5005   3003   1365   455   105   15    1
8:   24310  19448  12376   6188  2380   680  136   17   1
9:   92378  75582  50388  27132 11628  3876  969  171  19  1
10: 352716 293930 203490 116280 54264 20349 5985 1330 210 21   1
...
Expansion examples (for the Todd polynomials see A084930 and a comment above):
(4*x)^2 = 10*Todd(n,  0) + 5*Todd(n, 1) + 1*Todd(n, 2) = 10*1 + 5*(-3 + 4*x) + 1*(5 - 20*x + 16*x^2).
(4*x)^3 =  35*1 + 21*(-3 + 4*x) + 7*(5 - 20*x + 16*x^2) + (-7 + 56*x - 112*x^2 +64*x^3)*1. (End)
---------------------------------------------------------------------
Production matrix is
3, 1,
1, 2, 1,
0, 1, 2, 1,
0, 0, 1, 2, 1,
0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 0, 1, 2, 1,
0, 0, 0, 0, 0, 0, 0, 1, 2, 1
- _Paul Barry_, Mar 08 2011
Application to odd powers of Fibonacci numbers F, row n=2:
F_l^5 = (10*(-1)^(2*(l+1))*F_l + 5*(-1)^(1*(l+1))*F_{3*l} + 1*F_{5*l})/5^2, l >= 0. - _Wolfdieter Lang_, Aug 24 2012

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Paul Barry, On the Hurwitz Transform of Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.8.7.
E. Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Applied Mathematics, 34 (2005) pp. 101-122.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Article 12.3.3, 2012.
A. Nkwanta, A. Tefera, Curious Relations and Identities Involving the Catalan Generating Function and Numbers, Journal of Integer Sequences, 16 (2013), #13.9.5.
K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
Sun, Yidong; Ma, Luping Minors of a class of Riordan arrays related to weighted partial Motzkin paths. Eur. J. Comb. 39, 157-169 (2014), Table 2.2.

Cf. A000108, A113187.

Columns are : A001700, A002054, A003516, A030053, A030054, A030055, A030056.

a111418 n k = a111418_tabl !! n !! k
a111418_row n = a111418_tabl !! n
a111418_tabl = map reverse a122366_tabl
-- Reinhard Zumkeller, Mar 14 2014

Table[Binomial[2*n+1, n-k], {n,0,10}, {k,0,n}] (* G. C. Greubel, May 22 2017 *)
T[0, 0, x_, y_] := 1; T[n_, 0, x_, y_] := x*T[n - 1, 0, x, y] + T[n - 1, 1, x, y]; T[n_, k_, x_, y_] := T[n, k, x, y] = If[k < 0 || k > n, 0,
T[n - 1, k - 1, x, y] + y*T[n - 1, k, x, y] + T[n - 1, k + 1, x, y]];
Table[T[n, k, 3, 2], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, May 22 2017 *)

T(n, k) = C(2*n+1, n-k).
Sum_{k=0..n} T(n, k) = 4^n.
Sum_{k, 0<=k<=n}(-1)^k *T(n,k) = binomial(2*n,n) = A000984(n). - Philippe Deléham, Mar 22 2007
T(n,k) = sum{j=k..n, C(n,j)*2^(n-j)*C(j,floor((j-k)/2))}. - Paul Barry, Jun 06 2007
Sum_{k, k>=0} T(m,k)*T(n,k) = T(m+n,0)= A001700(m+n). - Philippe Deléham, Nov 22 2009
G.f. row polynomials: ((1+x) - (1-x)/sqrt(1-4*z))/(2*(x - (1+x)^2*z))
(see the Riordan property mentioned in a comment above). - Wolfdieter Lang, Aug 05 2014

A003516 Binomial coefficients C(2n+1, n-2).

Original entry on oeis.org

1, 7, 36, 165, 715, 3003, 12376, 50388, 203490, 817190, 3268760, 13037895, 51895935, 206253075, 818809200, 3247943160, 12875774670, 51021117810, 202112640600, 800472431850, 3169870830126, 12551759587422

Offset: 2

N. J. A. Sloane

a(n) is the number of royal paths (A006318) from (0,0) to (n,n) with exactly one diagonal step off the line y=x. - David Callan, Mar 25 2004

a(n) is the total number of DDUU's in all Dyck (n+2)-paths. - David Scambler, May 03 2013

			For n=4, C(2*4+1,4-2) = C(9,2) = 9*8/2 = 36, so a(4) = 36. - _Michael B. Porter_, Sep 10 2016

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

G. C. Greubel, Table of n, a(n) for n = 2..1000
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Heidi Goodson, An Identity for Vertically Aligned Entries in Pascal's Triangle, arXiv:1901.08653 [math.CO], 2019.
Milan Janjic, Two Enumerative Functions.
Toufik Mansour and Mark Shattuck, Counting occurrences of subword patterns in non-crossing partitions, Art Disc. Appl. Math. (2022).
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Vol. 15 (2012), Article 12.3.3. - From _N. J. A. Sloane_, Sep 16 2012
Daniel W. Stasiuk, An Enumeration Problem for Sequences of n-ary Trees Arising from Algebraic Operads, Master's Thesis, University of Saskatchewan-Saskatoon (2018).

Diagonal 6 of triangle A100257.
Third unsigned column (s=2) of A113187. - Wolfdieter Lang, Oct 18 2012
Cf. triangle A114492 - Dyck paths with k DDUU's.
Cf. A001622, A115139.
Cf. binomial(2*n+m, n): A000984 (m = 0), A001700 (m = 1), A001791 (m = 2), A002054 (m = 3), A002694 (m = 4), A002696 (m = 6), A030053 - A030056, A004310 - A004318.

List([2..25], n-> Binomial(2*n+1, n-2)); # G. C. Greubel, Mar 21 2019

[Binomial(2*n+1,n-2): n in [2..25]]; // Vincenzo Librandi, Apr 13 2011

CoefficientList[ Series[ 32/(((Sqrt[1 - 4 x] + 1)^5)*Sqrt[1 - 4 x]), {x, 0, 25}], x] (* Robert G. Wilson v, Aug 08 2011 *)
Table[Binomial[2*n +1,n-2], {n,2,25}] (* G. C. Greubel, Jan 23 2017 *)

{a(n) = binomial(2*n+1, n-2)}; \\ G. C. Greubel, Mar 21 2019

[binomial(2*n+1, n-2) for n in (2..25)] # G. C. Greubel, Mar 21 2019

G.f.: 32*x^2/(sqrt(1-4*x)*(sqrt(1-4*x)+1)^5). - Marco A. Cisneros Guevara, Jul 18 2011
a(n) = Sum_{k=0..n-2} binomial(n+k+2,k). - Arkadiusz Wesolowski, Apr 02 2012
D-finite with recurrence (n+3)*(n-2)*a(n) = 2*n*(2*n+1)*a(n-1). - R. J. Mathar, Oct 13 2012
G.f.: x^2*c(x)^5/sqrt(1-4*x) = ((-1 + 2*x) + (1 - 3*x + x^2) * c(x))/(x^2*sqrt(1-4*x)), with c(x) the o.g.f. of the Catalan numbers A000108. See the W. Lang link under A115139 for powers of c. - Wolfdieter Lang, Sep 10 2016
a(n) ~ 2^(2*n+1)/sqrt(Pi*n). - Ilya Gutkovskiy, Sep 10 2016
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=2} 1/a(n) = 4 - 14*Pi/(9*sqrt(3)).
Sum_{n>=2} (-1)^n/a(n) = 228*log(phi)/(5*sqrt(5)) - 134/15, where phi is the golden ratio (A001622). (End)
G.f.: 2F1([7/2,3],[6],4*x). - Karol A. Penson, Apr 24 2024
a(n) = Integral_{x = 0..4} x^n * w(x) dx, where the weight function w(x) = 1/(2*Pi) * sqrt(x)*(x^2 - 5*x + 5)/sqrt(4 - x). - Peter Bala, Oct 13 2024

A030053 a(n) = binomial(2n+1,n-3).

Original entry on oeis.org

1, 9, 55, 286, 1365, 6188, 27132, 116280, 490314, 2042975, 8436285, 34597290, 141120525, 573166440, 2319959400, 9364199760, 37711260990, 151584480450, 608359048206, 2438362177020, 9762479679106, 39049918716424, 156077261327400, 623404249591760

Offset: 3

N. J. A. Sloane

Number of UUUUUU's in all Dyck (n+3)-paths. - David Scambler, May 03 2013

			G.f. = x^3 + 9*x^4 + 55*x^5 + 286*x^6 + 1365*x^7 + 6188*x68 + ...

T. D. Noe, Table of n, a(n) for n = 3..200
Milan Janjic, Two Enumerative Functions.
Asamoah Nkwanta and Earl R. Barnes, Two Catalan-type Riordan Arrays and their Connections to the Chebyshev Polynomials of the First Kind, Journal of Integer Sequences, Vol. 15 (2012), Article 12.3.3. - From _N. J. A. Sloane_, Sep 16 2012.

Diagonal 8 of triangle A100257.
Cf. A001622, A113187 (unsigned fourth column).
Cf. binomial(2*n+m, n): A000984 (m = 0), A001700 (m = 1), A001791 (m = 2), A002054 (m = 3), A002694 (m = 4), A003516 (m = 5), A002696 (m = 6), A030054 - A030056, A004310 - A004318.

[Binomial(2*n+1,n-3): n in [3..30]]; // Vincenzo Librandi, Aug 11 2015

Table[Binomial[2*n + 1, n - 3], {n, 3, 20}] (* T. D. Noe, Apr 03 2014 *)
Rest[Rest[Rest[CoefficientList[Series[128 x^3 / ((1 - Sqrt[1 - 4 x])^7 Sqrt[1 - 4 x]) + (-1 / x^4 + 5 / x^3 - 6 / x^2 + 1 / x), {x, 0, 40}], x]]]] (* Vincenzo Librandi, Aug 11 2015 *)

a(n) = binomial(2*n+1,n-3); \\ Joerg Arndt, May 08 2013

G.f.: x^3*128/((1-sqrt(1-4*x))^7*sqrt(1-4*x))+(-1/x^4+5/x^3-6/x^2+1/x). - Vladimir Kruchinin, Aug 11 2015
D-finite with recurrence: -(n+4)*(n-3)*a(n) +2*n*(2*n+1)*a(n-1)=0. - R. J. Mathar, Jan 28 2020
G.f.: x^3* 2F1(4,9/2;8;4x). - R. J. Mathar, Jan 28 2020
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=3} 1/a(n) = 22*Pi/(9*sqrt(3)) - 33/10.
Sum_{n>=3} (-1)^(n+1)/a(n) = 852*log(phi)/(5*sqrt(5)) - 1073/30, where phi is the golden ratio (A001622). (End)
From Peter Bala, Oct 13 2024: (Start)
a(n) = Integral_{x = 0..4} x^n * w(x) dx, where the weight function w(x) = (1/(2*Pi)) * sqrt(x)*(x^3 - 7*x^2 + 14*x - 7)/sqrt((4 - x)).
G.f. x^3 * B(x) * C(x)^7, where B(x) = 1/sqrt(1 - 4*x) is the g.f. of the central binomial coefficients A000984 and C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. (End)

A030054 a(n) = binomial(2n+1,n-4).

Original entry on oeis.org

1, 11, 78, 455, 2380, 11628, 54264, 245157, 1081575, 4686825, 20030010, 84672315, 354817320, 1476337800, 6107086800, 25140840660, 103077446706, 421171648758, 1715884494940, 6973199770790, 28277527346376, 114456658306760, 462525733568080, 1866442158555975

Offset: 4

N. J. A. Sloane

nonn

Robert Israel, Table of n, a(n) for n = 4..1661
Milan Janjic, Two Enumerative Functions.

Diagonal 10 of triangle A100257.
Fifth unsigned column (s=4) of A113187. - Wolfdieter Lang, Oct 19 2012
Cf. A001622.
Cf. binomial(2*n+m, n): A000984 (m = 0), A001700 (m = 1), A001791 (m = 2), A002054 (m = 3), A002694 (m = 4), A003516 (m = 5), A002696 (m = 6), A030053 - A030056, A004310 - A004318.

seq(binomial(2*n+1,n-4),n=4..50); # Robert Israel, Jun 11 2019

Table[Binomial[2n+1,n-4],{n,4,40}]  (* Harvey P. Dale, Mar 31 2011 *)

vector(30, n, m=n+4; binomial(2*m+1,m-4)) \\ Michel Marcus, Aug 11 2015

G.f.: x^4*512/((1-sqrt(1-4*x))^9*sqrt(1-4*x))+(-1/x^5+7/x^4-15/x^3+10/x^2-1/x). - Vladimir Kruchinin, Aug 11 2015
From Robert Israel, Jun 11 2019: (Start)
(54 + 36*n)*a(n) + (-438 - 129*n)*a(n + 1) + (714 + 138*n)*a(n + 2) + (-432 - 63*n)*a(n + 3) + (110 + 13*n)*a(n + 4) + (-10 - n)*a(n + 5) = 0.
a(n) ~ 2^(2*n+1)/sqrt(n*Pi). (End)
From Amiram Eldar, Jan 24 2022: (Start)
Sum_{n>=4} 1/a(n) = 317/210 - 2*Pi/(9*sqrt(3)).
Sum_{n>=4} (-1)^n/a(n) = 2908*log(phi)/(5*sqrt(5)) - 8697/70, where phi is the golden ratio (A001622). (End)
G.f.: 2F1([11/2,5],[10],4*x). - Karol A. Penson, Apr 24 2024
From Peter Bala, Oct 13 2024: (Start)
a(n) = Integral_{x = 0..4} x^n * w(x) dx, where the weight function w(x) = 1/(2*Pi) * sqrt(x)*(x^4 - 9*x^3 + 27*x^2 - 30*x + 9)/sqrt((4 - x)).
G.f. x^4 * B(x) * C(x)^9, where B(x) = 1/sqrt(1 - 4*x) is the g.f. of the central binomial coefficients A000984 and C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers A000108. (End)
D-finite with recurrence -(n+5)*(n-4)*a(n) +2*n*(2*n+1)*a(n-1)=0. - R. J. Mathar, Nov 22 2024

A217478 Triangle of coefficients of polynomials providing the second term of the numerator for the generating function for odd powers (2*m+1) of Chebyshev S-polynomials. The present polynomials are called P(m;1,x^2).

Original entry on oeis.org

-2, 3, -4, -4, 10, -6, 5, -20, 21, -8, -6, 35, -56, 36, -10, 7, -56, 126, -120, 55, -12, -8, 84, -252, 330, -220, 78, -14, 9, -120, 462, -792, 715, -364, 105, -16, -10, 165, -792, 1716, -2002, 1365, -560, 136, -18, 11, -220, 1287, -3432, 5005, -4368, 2380, -816, 171, -20

Offset: 1

Wolfdieter Lang, Nov 14 2012

The o.g.f. for S(n,x)^(2*m+1), m >= 0, with Chebyshev's S-polynomials (see A049310), is
  G(m;z,x) := sum(S(n,x)^(2*m+1)*z^n, n=0..infinity)= sum(T(m,k)*S(2*k,x)/(1-z*x*tau(k,x)+z^2), k=0..m)/(x^2-4)^m, with the signed Riordan triangle
  T(m,k) = binomial(2*m+1,m-k)*(-1)^(m-k) = A113187(m,k),
  and tau(k,x):= R(2*k+1,x)/x with R the monic integer Chebyshev T-polynomials (see A127672). The proof uses the de Moivre-Binet formula: S(n,x) = (q^(n+1) - 1/q^(n+1))/(q-1/q), with q:=(x+sqrt(x^2-4))/2, and the one for tau(n,x) = (q^(2*n+1) + 1/q^(2*n+1))/(q+1/q). This can be written as  G(m;z,x) = Z(m;z,x)/N(m;z,x) with N(m;z,x) = product((1+z^2) - z*x*tau(k,x),k=0..m), and Z(m;z,x) = sum((1+z^2)^(m-l)*(-z*x)^l*P(m;l,x^2),l=0..m), where P(m;l,x^2) = sum(T(m,k)*S(2*k,x)*sigma(m;k,l,x^2), k=0..m)/(x^2-4)^m, with sigma(m;k,l,x^2) the elementary symmetric function of a product of l factors from tau(j,x), for j=0..m, with tau(k,x) missing. E.g., sigma(3;1,2,x^2) = tau(0,x)*tau(2,x) + tau(0,x)*tau(3,x) + tau(2,x)*tau(3,x), (tau(1,x) is missing). P(m;0,x^2) = 1 due to the identity Id(0;m,x^2) := sum(T(m,k)*S(2*k,x), k=0..m) = (x^2-4)^m (proof by using the de Moivre-Binet formula and the formula mentioned in a comment on A113187). Also the other P(m;l,x^2) turn out to be polynomials in x^2.
The present triangle a(m,k) provides the P(m;1,x^2) coefficients: P(m;1,x^2) = sum(a(m,k)*(x^2)^k, k=0..m-1), m>=1.
Using inclusion-exclusion one can write (x^2-4)^m*P(m;1,x^2) = sum(T(m,k)*S(2*k,x)*(sum(tau(k,x),k=0..m) - tau(k,x)), k=0..m) = sum(tau(k,x),k=0..m)*(x^2-4)^m - sum(T(m,k)*S(2*k,x)* tau(k,x),k=0..m), using the mentioned identity Id(0;m,x^2). In the second term S(2*k,x)*tau(k,x) = S(4*k+1,x)/x (de Moivre-Binet formulas for S and tau). This leads to the l=1 identity Id(1;m,x^2) := sum(T(m,k)*S(4*k+1,x)/x,k=0..m) =
  ((x^2-4)*x^2)^m, using again de Moivre-Binet and the identity
  given in a comment on A113187. Therefore, after dividing by (x^2-4)^m,  P(m;1,x^2) = sum(tau(k,x),k=0..m) - x^(2*m).

			The triangle a(m,k) begins:
m\k   0    1    2     3     4      5     6     7    8    9 ...
1:   -2
2:    3   -4
3:   -4   10   -6
4:    5  -20   21    -8
5:   -6   35  -56    36   -10
6:    7  -56  126  -120    55    -12
7:    8   84 -252   330  -220     78   -14
8:    9 -120  462  -792   715   -364   105   -16
9:  -10  165 -792  1716 -2002   1365  -560   136  -18
10:  11 -220 1287 -3432  5005  -4368  2380  -816  171  -20
...
P(2;1,x^2) = 3 - 4*x^2, appears in the second term of the numerator of the o.g.f. for S(n,x)^5 which is  Z(2;z,x) = (1+z^2)^2 + (1+z^2)*(-x*z)*(3-4*x^2) + ((-x*z)^2)*2*(-4 +3*x^2). The last term is taken from row m=2 of A217479. The denominator is  N(2;z,x) = product((1+z^2)-z*x*tau(k,x), k=0..2). This checks with [1,x^5,-1+5*x^2-10*x^4+10*x^6-5*x^8
+x^10,-32*x^5+80*x^7-80*x^9+40*x^11-10* x^13+x^15,...] for S(n,x)^5, n=0,1,2,3,...

Cf. A049310, A217479 (P(m;2,x^2)).

a(m,k) = [x^(2*k)] P(m;1,x^2) = [x^(2*k)](sum(tau(k,x),k=0..m) - x^(2*m)) (see the comment above), m>=1, k = 0..m-1.

a(m,k) = (-1)^(m-k)*binomial(m+k+1,2*k+1). For the proof one uses the identity sum(tau(j,x),j=0..m) = S(m,x^2-2) which holds by comparing the o.g.f.s of both sides (see a Nov 13 2012 comment on Riordan A053122 where tau is called r).

A217479 Array of coefficients of polynomials providing the third term of the numerator of the generating function for odd powers (2*m+1) of Chebyshev S-polynomials. The present polynomials are called P(m;2,x^2), m >= 2.

Original entry on oeis.org

-8, 6, -27, 65, -56, 15, -61, 260, -469, 415, -176, 28, -114, 736, -2104, 3214, -2838, 1456, -400, 45, -190, 1714, -6988, 15699, -21461, 18760, -10614, 3768, -760, 66, -293, 3507, -19195, 58807, -112123, 141441, -122168, 73185, -30077, 8107, -1288, 91

Offset: 2

Wolfdieter Lang, Nov 14 2012

The row length of this irregular triangle is 2*(m-1), m >= 2.
For the o.g.f. of S(m,x)^(2*m+1), m>=0, with Chebyshev's S-polynomials (coefficient triangle A049310) see the comment on A217478. G(m;z,x) = Z(m;z,x)/N(m;z,x) with N(m;z,x) = product((1+z^2) - z*x*tau(k,x),k=0..m), and Z(m;z,x) = sum((1+z^2)^(m-l)*(-z*x)^l*P(m;l,x^2),l=0..m), where P(m,l,x^2) = sum(T(m,k)*S(2*k,x)*sigma(m;k,l,x^2), k=0..m)/(x^2-4)^m, with sigma(m;k,l,x^2) the elementary symmetric function of a product of l factors from tau(j,x), for j=0..m, with tau(k,x) missing. Here tau(j,x):= 2*T(2*j+1,x/2)/x = R(2*j+1,x)/x (see A127672 for the coefficients of R(n,x)).
The present array a(m,k) provides the P(m;2,x^2) coefficients, and m >= 2: P(m;2,x^2) = sum(a(m,k)*x^2,k=0..(2*m-3)).
Using inclusion-exclusion one can write (x^2-4)^m*P(m;2,x^2) =
sum(T(m,k)*S(2*k,x)*(sigma(m+1;2,x^2) - sum(tau(j,x),j=0..m)* tau(k,x) + tau(k,x)^2),k=0..m), with sigma(m+1;2,x^2) the elementary symmetric function of 2 factors from tau(j,x), for j=0,1,...,m. E.g.,m=2:  sigma(2+1;2,x^2) = tau(0,x)*tau(1,x) + tau(0,x)*tau(2,x) + tau(1,x)*tau(2,x). The identities Id(0;m,x^2) and Id(1;m,x^2) (given in the comment on A217478) together with the new identity Id(2;m,x^2) := sum(T(m,k)*S(2*k,x)*tau(k,x)^2,k=0..m) = (x^2-4)^m*((x^2-1)^(2*m+1) + 1)/x^2 are now used. The new identity is obtained from the de Moivre-Binet formula for S and tau using first twice the identity mentioned in a Nov 14 2012 comment on A113187, and then the identity q^3 - 1/q^3 = sqrt(x^2-4)*(x^2-1) (see the instance k=1 of the formula given in a Oct 18 2012 comment on A111125 with x -> q which is defined by  (x+sqrt(x^2-4))/2). This yields, after division by (x^2-4)^m, finally the polynomial P(2;m,x^2) = sigma(m+1;2,x^2) - sum(tau(j,x),j=0..m)*x^(2*m) + ((x^2-1)^(2*m+1) + 1)/x^2, for m >= 2.

			The array a(m,k) starts:
m\k   0    1     2     3      4     5      6    7    8  9 ...
2:   -8    6
3:  -27   65   -56    15
4:  -61  260  -469   415   -176    28
5: -114  736 -2104  3214  -2838  1456   -400   45
6: -190 1714 -6988 15699 -21461 18760 -10614 3768 -760 66
...
Row m=7:  -293, 3507, -19195, 58807, -112123, 141441, -122168, 73185, -30077, 8107, -1288, 91.
Row m=8: -427, 6536, -46102, 183762, -461654, 780716, -926345, 790773, -491397, 221760, -71139, 15405, -2016, 120.
Row 9: -596, 11346, -100077, 502036, -1600280, 3470116, -5352805, 6051236, -5110145, 3256825, -1568416, 564980, -148176, 26770, -2976, 153.
m=2: P(2;2,x^2) = tau(0,x)*tau(1,x) + tau(0,x)*tau(2,x) + tau(1,x)*tau(2,x) - (tau(0,x)+tau(1,x)+tau(2,x))*x^4 + (5 -10*x^2 + 10*x^4 - 5*x^6 + x^8)  = -8 + 6*x^2 = 2*(-4 + 3*x^2).
  The numerator of the o.g.f. for S(n,x)^5 is Z(2;z,x) = (1+z^2)^2 + (1+z^2)*(-x*z)*(3-4*x^2) + (-x*z)^2*2*(-4 + 3*x^2), where the last bracket in the second term comes from row m=2 of A217478. The denominator is N(2;z,x) = product((1+z^2)-z*x*tau(k,x), k=0..2). See the example of A217478.

Cf. A217478.

a(m,k) = [x^(2*k)] P(2;m,x^2), m >= 2, k = 0..(2*m-3), with P(2;m,x^2) given in the comment above.

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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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A111418 Right-hand side of odd-numbered rows of Pascal's triangle.

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A003516 Binomial coefficients C(2n+1, n-2).

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A030053 a(n) = binomial(2n+1,n-3).

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A030054 a(n) = binomial(2n+1,n-4).

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A217478 Triangle of coefficients of polynomials providing the second term of the numerator for the generating function for odd powers (2*m+1) of Chebyshev S-polynomials. The present polynomials are called P(m;1,x^2).

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A217479 Array of coefficients of polynomials providing the third term of the numerator of the generating function for odd powers (2*m+1) of Chebyshev S-polynomials. The present polynomials are called P(m;2,x^2), m >= 2.

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