A176280 -id:A176280 - cp's OEIS frontend

A176281 Hankel transform of A176280.

1, 3, 12, 56, 280, 1440, 7488, 39104, 204544, 1070592, 5604864, 29345792, 153653248, 804532224, 4212572160, 22057287680, 115493404672, 604731211776, 3166413520896, 16579556016128, 86811681488896, 454551863820288

Offset: 0

Paul Barry, Apr 14 2010

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-16,8).

Cf. A000045, A082761, A176280.

List([0..30], n-> 2^(n-1)*(Fibonacci(2*n+1) + 1)); # G. C. Greubel, Nov 24 2019

[2^(n-1)*(Fibonacci(2*n+1) + 1): n in [0..30]]; // G. C. Greubel, Nov 24 2019

with(combinat); seq(2^(n-1)*(fibonacci(2*n+1) + 1), n=0..30); # G. C. Greubel, Nov 24 2019

CoefficientList[Series[(1-5x+4x^2)/((1-2x)(1-6x+4x^2)),{x,0,40}],x] (* or *) LinearRecurrence[{8,-16,8},{1,3,12},40] (* Harvey P. Dale, Aug 14 2013 *)

vector(31, n, 2^(n-2)*(fibonacci(2*n-1) + 1)) \\ G. C. Greubel, Nov 24 2019

[2^(n-1)*(fibonacci(2*n+1) + 1) for n in (0..30)] # G. C. Greubel, Nov 24 2019

G.f.: (1-5*x+4*x^2)/(1-8*x+16*x^2-8*x^3) = (1-5*x+4*x^2)/((1-2*x)*(1-6*x+4*x^2)).
a(n) = 2^(n-1) + (3-sqrt(5))^n*((5-sqrt(5))/20) + (3+sqrt(5))^n*((5+sqrt(5))/20).
a(n) = 2^(n-1) + A082761(n)/2. - R. J. Mathar, Sep 30 2012
a(0)=1, a(1)=3, a(2)=12, a(n) = 8*a(n-1) - 16*a(n-2) + 8*a(n-3). - Harvey P. Dale, Aug 14 2013
a(n) = 2^(n-1)*(Fibonacci(2*n+1) + 1). - G. C. Greubel, Nov 24 2019

A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

Original entry on oeis.org

1, 2, 1, 6, 6, 1, 20, 30, 10, 1, 70, 140, 70, 14, 1, 252, 630, 420, 126, 18, 1, 924, 2772, 2310, 924, 198, 22, 1, 3432, 12012, 12012, 6006, 1716, 286, 26, 1, 12870, 51480, 60060, 36036, 12870, 2860, 390, 30, 1, 48620, 218790, 291720, 204204, 87516, 24310

Offset: 0

Wolfdieter Lang

Or, a triangle related to A000984 (central binomial) and A000302 (powers of 4).
This is an example of a Riordan matrix. See the Shapiro et al. reference quoted under A053121 and Notes 1 and 2 of the Wolfdieter Lang reference, p. 306.
As a number triangle, this is the Riordan array (1/sqrt(1-4x),x/(1-4x)). - Paul Barry, May 30 2005
The A- and Z- sequences for this Riordan matrix are (see the Wolfdieter Lang link under A006232 for the D. G. Rogers, D. Merlini et al. and R. Sprugnoli references on Riordan A- and Z-sequences with a summary): A-sequence [1,4,0,0,0,...] and Z-sequence 4+2*A000108(n)*(-1)^(n+1)=[2, 2, -4, 10, -28, 84, -264, 858, -2860, 9724, -33592, 117572, -416024, 1485800, -5348880, 19389690, -70715340, 259289580, -955277400, 3534526380], n >= 0. The o.g.f. for the Z-sequence is 4-2*c(-x) with the Catalan number o.g.f. c(x). - Wolfdieter Lang, Jun 01 2007
As a triangle, T(2n,n) is A001448. Row sums are A046748. Diagonal sums are A176280. - Paul Barry, Apr 14 2010
From Wolfdieter Lang, Aug 10 2017: (Start)
The row polynomials R(n, x) of Riordan triangles R = (G(x), F(x)), with F(x)= x*Fhat(x), belong to the class of Boas-Buck polynomials (see the reference). Hence they satisfy the Boas-Buck identity (we use the notation of Rainville, Theorem 50, p. 141):
  (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} (alpha(k)*1 + beta(k)*E_x)*R(n-1.k, x), for n >= 0, where E_x = x*d/dx (Euler operator). The Boas-Buck sequences are given by alpha(k) := [x^k] ((d/dx)log(G(x))) and beta(k) := [x^k] (d/dx)log(Fhat(x)).
This entails a recurrence for the sequence of column m of the Riordan triangle T, n > m >= 0: T(n, m) = (1/(n-m))*Sum_{k=m..n-1} (alpha(n-1-k) + m*beta(n-1-k))*T(k, m), with input T(m,m).
For the present case the Boas-Buck identity for the row polynomials is (E_x - n*1)*R(n, x) = -Sum_{k=0..n-1} 2^(2*k+1)*(1 + 2*E_x)*R(n-1-k, x), for n >= 0. For the ensuing recurrence for the columns m of the triangle T see the formula and example section. (End)
From Peter Bala, Mar 04 2018: (Start)
The following two remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
1) Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,4*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(4*x)/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,4*x). For example, when n = 3 we have exp(x)*(20 + 30*(4*x) + 10*(4*x)^2/2! + (4*x)^3/3!) = 20 + 140*x + 420*x^2/2! + 924*x^3/3! + 1716*x^4/4! + ....
2) Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. P(n,x) is the n-th degree Taylor polynomial of (1 + 4*x)^(n-1/2) about 0. For example, for n = 4 we have (1 + 4*x)^(7/2) = 70*x^4 + 140*x^3 + 70*x^2 + 14*x + 1 + O(x^5).
Let C(x) = (1 - sqrt(1 - 4*x))/(2*x) denote the o.g.f. of the Catalan numbers A000108. The derivatives of C(x) are determined by the identity (-1)^n * x^n/n! * (d/dx)^n(C(x)) = 1/(2*x)*( 1 - P(n,-x)/(1 - 4*x)^(n-1/2) ), n = 0,1,2,.... See Lang 2002. Cf. A283150 and A283151. (End)

			Array begins:
  1,  2,   6,  20,   70, ...
  1,  6,  30, 140,  630, ...
  1, 10,  70, 420, 2310, ...
  1, 14, 126, 924, 6006, ...
Recurrence from A-sequence: 140 = a(4,1) = 20 + 4*30.
Recurrence from Z-sequence: 252 = a(5,0) = 2*70 + 2*140 - 4*70 + 10*14 - 28*1.
From _Paul Barry_, Apr 14 2010: (Start)
As a number triangle, T(n, m) begins:
n\k       0      1       2       3      4      5     6    7   8  9 10 ...
0:        1
1:        2      1
2:        6      6       1
3:       20     30      10       1
4:       70    140      70      14      1
5:      252    630     420     126     18      1
6:      924   2772    2310     924    198     22     1
7:     3432  12012   12012    6006   1716    286    26    1
8:    12870  51480   60060   36036  12870   2860   390   30   1
9:    48620 218790  291720  204204  87516  24310  4420  510  34  1
10:  184756 923780 1385670 1108536 554268 184756 41990 6460 646 38  1
... [Reformatted and extended by _Wolfdieter Lang_, Aug 10 2017]
Production matrix begins
      2, 1,
      2, 4, 1,
     -4, 0, 4, 1,
     10, 0, 0, 4, 1,
    -28, 0, 0, 0, 4, 1,
     84, 0, 0, 0, 0, 4, 1,
   -264, 0, 0, 0, 0, 0, 4, 1,
    858, 0, 0, 0, 0, 0, 0, 4, 1,
  -2860, 0, 0, 0, 0, 0, 0, 0, 4, 1 (End)
Boas-Buck recurrence for column m = 2, and n = 4: T(4, 2) = (2*(2*2+1)/2) * Sum_{k=2..3} 4^(3-k)*T(k, 2) = 5*(4*1 + 1*10) = 70. - _Wolfdieter Lang_, Aug 10 2017
From _Peter Bala_, Feb 15 2018: (Start)
With C(x) = (1 - sqrt( 1 - 4*x))/(2*x),
-x^3/3! * (d/dx)^3(C(x)) = 1/(2*x)*( 1 - (1 - 10*x + 30*x^2 - 20*x^3)/(1 - 4*x)^(5/2) ).
x^4/4! * (d/dx)^4(C(x)) = 1/(2*x)*( 1 - (1 - 14*x + 70*x^2 - 140*x^3 + 70*x^4 )/(1 - 4*x)^(7/2) ). (End)

Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.

Muniru A Asiru, Antidiagonals n = 0..50, flattened
Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Paul Barry, Embedding structures associated with Riordan arrays and moment matrices, arXiv preprint arXiv:1312.0583 [math.CO], 2013.
Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
Wolfdieter Lang, First 10 rows.
Wolfdieter Lang, On polynomials related to derivatives of the generating function of Catalan numbers, Fib. Quart. 40,4 (2002) 299-313; T(n,m) is called B(n,m) there.
Benjamin Lovitz and Nathaniel Johnston, A hierarchy of eigencomputations for polynomial optimization on the sphere, arXiv:2310.17827 [math.OC], 2023. See pp. 35, 39. See also Math. Prog. (2025) Series A.
Helmut Prodinger, Some information about the binomial transform, The Fibonacci Quarterly, 32, 1994, 412-415.

Columns include: A000984 (m=0), A002457 (m=1), A002802 (m=2), A020918 (m=3), A020920 (m=4), A020922 (m=5), A020924 (m=6), A020926 (m=7), A020928 (m=8), A020930 (m=9), A020932 (m=10).
Row sums: A046748.
Cf. A001147, A006882, A007318, A068555, A111959, A283150, A283151.

Flat(List([0..9],n->List([0..n],m->Binomial(2*n,n)*Binomial(n,m)/Binomial(2*m,m)))); # Muniru A Asiru, Jul 19 2018

[Binomial(n+1,k+1)*Catalan(n)/Catalan(k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 28 2024

t[i_, j_] := If[i < 0 || j < 0, 0, (2*i + 2*j)!*i!/(2*i)!/(i + j)!/j!]; Flatten[Reverse /@ Table[t[n, k - n] , {k, 0, 9}, {n, k, 0, -1}]][[1 ;; 51]] (* Jean-François Alcover, Jun 01 2011, after PARI prog. *)

T(i,j)=if(i<0 || j<0,0,(2*i+2*j)!*i!/(2*i)!/(i+j)!/j!)

def A046521(n,k): return binomial(n+1, k+1)*catalan_number(n)/catalan_number(k)
flatten([[A046521(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Jul 28 2024

T(n, m) = binomial(2*n, n)*binomial(n, m)/binomial(2*m, m), n >= m >= 0.
G.f. for column m: ((x/(1-4*x))^m)/sqrt(1-4*x).
Recurrence from the A-sequence given above: a(n,m) = a(n-1,m-1) + 4*a(n-1,m), for n >= m >= 1.
Recurrence from the Z-sequence given above: a(n,0) = Sum_{j=0..n-1} Z(j)*a(n-1,j), n >= 1; a(0,0)=1.
As a number triangle, T(n,k) = C(2*n,n)*C(n,k)/C(2*k,k) = C(n-1/2,n-k)*4^(n-k). - Paul Barry, Apr 14 2010
From Peter Bala, Apr 11 2012: (Start):
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A007318 and A068555.
The triangular array equals exp(S), where the infinitesimal generator S has [2,6,10,14,18,...] on the main subdiagonal and zeros elsewhere.
Recurrence equation for the square array: T(n+1,k) = (k+1)/(4*n+2)*T(n,k+1). (End)
T(n,k) = 4^(n-k)*A006882(2*n - 1)/(A006882(2*n - 2*k)*A006882(2*k - 1)) = 4^(n-k)*(2*n - 1)!!/((2*n - 2*k)!*(2*k - 1)!!). - Peter Bala, Nov 07 2016
Boas-Buck recurrence for column m, m > n >= 0: T(n, m) = (2*(2*m+1)/(n-m))*Sum_{k=m..n-1} 4^(n-1-k)*T(k, m), with input T(n, n) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
From Peter Bala, Aug 13 2021: (Start)
Analogous to the binomial transform we have the following sequence transformation formula: g(n) = Sum_{k = 0..n} T(n,k)*b^(n-k)*f(k) iff f(n) = Sum_{k = 0..n} (-1)^(n-k)*T(n,k)*b^(n-k)*g(k). See Prodinger, bottom of p. 413, with b replaced with 4*b, c = 1 and d = 1/2.
Equivalently, if F(x) = Sum_{n >= 0} f(n)*x^n and G(x) = Sum_{n >= 0} g(n)*x^n are a pair of  formal power series then
G(x) = 1/sqrt(1 - 4*b*x) * F(x/(1 - 4*b*x)) iff F(x) = 1/sqrt(1 + 4*b*x) * G(x/(1 + 4*b*x)).
The m-th power of this array has entries m^(n-k)*T(n,k). (End)

A026671 Number of lattice paths from (0,0) to (n,n) with steps (0,1), (1,0) and, when on the diagonal, (1,1).

Original entry on oeis.org

1, 3, 11, 43, 173, 707, 2917, 12111, 50503, 211263, 885831, 3720995, 15652239, 65913927, 277822147, 1171853635, 4945846997, 20884526283, 88224662549, 372827899079, 1576001732485, 6663706588179, 28181895551161, 119208323665543, 504329070986033, 2133944799315027

Offset: 0

Clark Kimberling, Miklos Bona

1, 1, 3, 11, 43, 173, ... is the unique sequence for which both the Hankel transform of the sequence itself and the Hankel transform of its left shift are the powers of 2 (A000079). For example, det[{{1, 1, 3}, {1, 3, 11}, {3, 11, 43}}] = det[{{1, 3, 11}, {3, 11, 43}, {11, 43, 173}}] = 4. - David Callan, Mar 30 2007
From Paul Barry, Jan 25 2009: (Start)
a(n) is the image of F(2n+2) under the Catalan matrix (1,xc(x)) where c(x) is the g.f. of A000108.
The sequence 1,1,3,... is the image of A001519 under (1,xc(x)). This sequence has g.f. given by 1/(1-x-2x^2/(1-3x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction). (End)
Binomial transform of A111961. - Philippe Deléham, Feb 11 2009
From Paul Barry, Nov 03 2010: (Start)
The sequence 1,1,3,... has g.f. 1/(1-x/sqrt(1-4x)), INVERT transform of A000984.
It is an eigensequence of the sequence array for A000984. (End)

L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Jean-Christophe Aval, Adrien Boussicault and Sandrine Dasse-Hartaut, The tree structure in staircase tableaux, arXiv:1109.4907 [math.CO], 2011-2013.
Cyril Banderier, Markus Kuba, and Michael Wallner, Analytic Combinatorics of Composition schemes and phase transitions with mixed Poisson distributions, arXiv:2103.03751 [math.PR], 2021.
Miklos Bona, The permutation classes equinumerous to the smooth class, Electron. J. Combin., 5 (1998), no. 1, Research Paper 31, 12 pp.
David Callan and Toufik Mansour, Five subsets of permutations enumerated as weak sorting permutations, arXiv:1602.05182 [math.CO], 2016.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Milan Janjić, Pascal Matrices and Restricted Words, J. Int. Seq., Vol. 21 (2018), Article 18.5.2.
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.
Huyile Liang, Jeffrey Remmel and Sainan Zheng, Stieltjes moment sequences of polynomials, arXiv:1710.05795 [math.CO], 2017, see page 16.

a(n) = T(2n-1, n-1), T given by A026736.
a(n) = T(2n, n), T given by A026670.
a(n) = T(2n+1, n+1), T given by A026725.
Row sums of triangle A054335.
Cf. A026781, A001076, A176280.

a:=[3,11,43];; for n in [4..30] do a[n]:=(2*(4*n-3)*a[n-1] - 3*(5*n-8)*a[n-2] - 2*(2*n-3)*a[n-3])/n; od; Concatenation([1], a); # G. C. Greubel, Jul 16 2019

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 1/(Sqrt(1-4*x)-x) )); // G. C. Greubel, Jul 16 2019

Table[SeriesCoefficient[1/(Sqrt[1-4*x]-x),{x,0,n}],{n,0,30}] (* Vaclav Kotesovec, Oct 08 2012 *)

{a(n)= if(n<0, 0, polcoeff( 1/(sqrt(1 -4*x +x*O(x^n)) -x), n))} /* Michael Somos, Apr 20 2007 */

my(x='x+O('x^66)); Vec( 1/(sqrt(1-4*x)-x) ) \\ Joerg Arndt, May 04 2013

(1/(sqrt(1-4*x)-x)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Jul 16 2019

From Wolfdieter Lang, Mar 21 2000: (Start)
G.f.: 1/(sqrt(1-4*x)-x).
a(n) = Sum_{i=1..n} a(i-1)*binomial(2*(n-i), n-i) + binomial(2*n, n), n >= 1, a(0)=1. (End)
G.f.: 1/(1 -x -2*x*c(x)) where c(x) = g.f. for Catalan numbers A000108. - Michael Somos, Apr 20 2007
From Paul Barry, Jan 25 2009: (Start)
G.f.: 1/(1 - 3xc(x) + x^2*c(x)^2);
G.f.: 1/(1-3x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction).
a(0) = 1, a(n) = Sum_{k=0..n} (k/(2n-k))*C(2n-k,n-k)*F(2k+2). (End)
a(n) = Sum_{k=0..n} A039599(n,k) * A000045(k+2). - Philippe Deléham, Feb 11 2009
From Paul Barry, Feb 08 2009: (Start)
G.f.: 1/(1-x/(1-2x/(1-x/(1-x/(1-x/(1-x/(1-x/(1-... (continued fraction);
G.f. of 1,1,3,... is 1/(1-x-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) = the upper left term in M^n, M = the infinite square production matrix:
  3, 2, 0, 0, 0, 0, ...
  1, 1, 1, 0, 0, 0, ...
  1, 1, 1, 1, 0, 0, ...
  1, 1, 1, 1, 1, 0, ...
  1, 1, 1, 1, 1, 1, ...
  ... (End)
From Vaclav Kotesovec, Oct 08 2012: (Start)
D-finite with recurrence: n*a(n) = 2*(4*n-3)*a(n-1) - 3*(5*n-8)*a(n-2) - 2*(2*n-3)*a(n-3).
a(n) ~ (2+sqrt(5))^n/sqrt(5). (End)
a(n) = Sum_{k=0..n+1} 4^(n+1-k) * binomial(n-k/2,n+1-k). - Seiichi Manyama, Mar 30 2025
From Peter Luschny, Mar 30 2025: (Start)
a(n) = 4^n*(binomial(n-1/2, n)*hypergeom([1, (1-n)/2, -n/2], [1/2, 1/2-n], -1/4) + hypergeom([(1-n)/2, 1-n/2], [1-n], -1/4)/4) for n > 0.
a(n) = A001076(n) + A176280(n). (End)

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A176281 Hankel transform of A176280.

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A046521 Array T(i,j) = binomial(-1/2-i,j)*(-4)^j, i,j >= 0 read by antidiagonals going down.

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A026671 Number of lattice paths from (0,0) to (n,n) with steps (0,1), (1,0) and, when on the diagonal, (1,1).

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A338397 Triangle read by rows: T(n,m)= Sum_{k=0..m/2} C(n-k,m-2*k)*C(n-k,m-k)*C(n,k)/C(2*k,k).

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A338397 Triangle read by rows: T(n,m)= Sum_{k=0..m/2} C(n-k,m-2k)C(n-k,m-k)C(n,k)/C(2k,k).