A090763 -id:A090763 - cp's OEIS frontend

A144485 a(n) = (3n + 2)*binomial(3n + 1,n).

2, 20, 168, 1320, 10010, 74256, 542640, 3922512, 28120950, 200300100, 1419269280, 10013421600, 70394353848, 493362138080, 3448674255840, 24051721745568, 167405449649550, 1163116182943260, 8068463611408200, 55891260077406600

Offset: 0

Roger L. Bagula, Oct 12 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Ömer Eğecioğlu, Timothy Redmond, and Charles Ryavec, Almost product evaluation of Hankel Determinants, The Electronic Journal of Combinatorics, Vol. 15, No. 1 (2008), #R6; arXiv preprint, arXiv:0704.3398 [math.CO], 2007.

Cf. A005809, A045721, A090763, A210453.

[(3*n+2)*Binomial(3*n+1, n): n in [0..20]]; // Vincenzo Librandi, Feb 14 2014

a:= proc(n) option remember; `if`(n=0, 2,
      3*(3*n+1)*(3*n+2)*a(n-1)/(2*n*(2*n+1)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Feb 01 2014

a[k_] = (3k + 2)Binomial[3k + 1, k]; Table[a[k], {k, 0, 30}]

a(n) = (3n+2)*A045721(n). - R. J. Mathar, Feb 01 2014
a(n) = 2*A090763(n). - Alois P. Heinz, Feb 01 2014
From Amiram Eldar, Dec 07 2024: (Start)
a(n) = 2 * (n+1) * A005809(n+1) / 3.
Sum_{n>=0} 1/a(n) = (3/2) * A210453. (End)

A096793 Triangle read by rows: a(n,k) is the number of Dyck n-paths containing k odd-length ascents.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 4, 0, 1, 3, 0, 10, 0, 1, 0, 21, 0, 20, 0, 1, 12, 0, 84, 0, 35, 0, 1, 0, 120, 0, 252, 0, 56, 0, 1, 55, 0, 660, 0, 630, 0, 84, 0, 1, 0, 715, 0, 2640, 0, 1386, 0, 120, 0, 1, 273, 0, 5005, 0, 8580, 0, 2772, 0, 165, 0, 1, 0, 4368, 0, 25025, 0, 24024, 0, 5148, 0, 220, 0, 1

Offset: 0

David Callan, Aug 17 2004

a(n,k)=0 unless k and n have the same parity and 0 <= k <= n.
From Emeric Deutsch, Oct 05 2008: (Start)
Sum_{k=0..n} k*a(n,k) = A014300(n).
For the case of even-length ascents see A143950. (End)

			Table begins
.
n |k = 0    1    2    3    4    5    6    7    8
--+---------------------------------------------
0 |    1
1 |    0,   1
2 |    1,   0,   1
3 |    0,   4,   0,   1
4 |    3,   0,  10,   0,   1
5 |    0,  21,   0,  20,   0,   1
6 |   12,   0,  84,   0,  35,   0,   1
7 |    0, 120,   0, 252,   0,  56,   0,   1
8 |   55,   0, 660,   0, 630,   0,  84,   0,   1
.
a(4,0)=3 because the Dyck 4-paths containing no odd-length ascents are UUUUDDDD,UUDUUDDD,UUDDUUDD.

The nonzero entries in column k=0 give A001764, in k=1 give A045721, in k=2 give A090763. The row sums are the Catalan numbers A000108.

Cf. A143950. - Emeric Deutsch, Oct 05 2008

bi[n_, k_] := If[IntegerQ[k], Binomial[n, k], 0]; TableForm[Table[bi[(n+k)/2, (n-k)/2]bi[(3n-k)/2+1, (n+k)/2]/((3n-k)/2+1), {n, 0, 10}, {k, 0, n}]]

a(n, k) = binomial((n+k)/2, (n-k)/2)*binomial((3n-k)/2+1, (n+k)/2)/((3n-k)/2+1).
Equivalently, a(2n+k, k) = binomial(3n+k, k)*T(n) where T(n) = binomial(3n, n)/(2n+1) is A001764. Proof: Given a Dyck (2n+k)-path with k ascents of odd length, delete the peaks (UD) that terminate odd-length ascents. This is a mapping to Dyck (2n)-paths all of whose ascents have even length; there are T[n] such paths. The mapping is clearly onto and is binomial(3n+k, k)-to-1 as follows. A Dyck (2n)-path all of whose ascents have even length has exactly 3n+1 vertices that are (i) not incident with an upstep, or (ii) incident with an upstep and at even distance (possibly 0) from the start of the ascent they lie in. The k deleted UDs can be inserted arbitrarily at these vertices, repetition allowed, to get the preimages -- binomial(3n+k, k) choices.
G.f.: G(z, t) + H(z, t) where G satisfies G^3*(t^2 - 1)*z^2 - G^2*t*z*(2 + t*z) + G*(1 + 2*t*z) - 1 = 0 and H satisfies H^3*(t^2 - 1)*z^2 + H^2*t*z*(2 + t*z) - H*t^2*(1 - t*z) + t^3*z = 0. Here z marks size (n) and t marks number of odd-length ascents (k). G is gf for paths that start with an even-length ascent and H is gf for paths that start with an odd-length ascent. - David Callan, Sep 03 2005
From Emeric Deutsch, Oct 05 2008: (Start)
G.f. G=G(t,z) satisfies G = 1 + zG(t + zG)/(1 - z^2*G^2).
The trivariate g.f. H=H(t,s,z), where t(s) marks odd-length (even-length) ascents satisfies H = 1 + zH(t+szH)/(1-z^2*H^2). (End)

A360560 Triangle read by rows. T(n, k) = (1/2) * C(n, k) * C(3*n - 1, n) for n > 0 and T(0, 0) = 1.

Original entry on oeis.org

1, 1, 1, 5, 10, 5, 28, 84, 84, 28, 165, 660, 990, 660, 165, 1001, 5005, 10010, 10010, 5005, 1001, 6188, 37128, 92820, 123760, 92820, 37128, 6188, 38760, 271320, 813960, 1356600, 1356600, 813960, 271320, 38760, 245157, 1961256, 6864396, 13728792, 17160990, 13728792, 6864396, 1961256, 245157

Offset: 0

Vladimir Kruchinin, Feb 11 2023

			Triangle begins:
     1;
     1,    1;
     5,   10,     5;
    28,   84,    84,    28;
   165,  660,   990,   660,  165;
  1001, 5005, 10010, 10010, 5005, 1001;

Cf. A025174, A090763, A360546.

T := (n, k) -> ifelse(n = 0, 1, binomial(n, k)*binomial(3*n - 1, n)/2):
for n from 0 to 6 do seq(T(n, k), k = 0..n) od;

T(n,m):=1/2*binomial(n+1,m)*binomial(3*n+2,n+1);

G.f.: 1/2 + x*sqrt(3 + 3*y)*cot(arcsin((3*sqrt(3*x*(y + 1)))/2)/3)/ (2*sqrt(4*x - 27*x^2*(y + 1))).

A378777 a(n) = n^2 * binomial(3*n, n).

Original entry on oeis.org

0, 3, 60, 756, 7920, 75075, 668304, 5697720, 47070144, 379632825, 3004501500, 23417943120, 180241588800, 1372689900036, 10360604899680, 77595170756400, 577241321893632, 4268838966063525, 31404136939468020, 229951212925133700, 1676737802322198000, 12180171012442098435

Offset: 0

Amiram Eldar, Dec 07 2024

Amiram Eldar, Table of n, a(n) for n = 0..500
Necdet Batir, On the series Sum_{k=1..oo} binomial(3k,k)^{-1} k^{-n} x^k, Proc. Indian Acad. Sci. (Math. Sci.), Vol. 115, No. 4 (2005), pp. 371-381; arXiv preprint, arXiv:math/0512310 [math.AC], 2005.

Cf. A005809, A090763, A144485.

a[n_] := n^2 * Binomial[3*n, n]; Array[a, 25, 0]

PARI
```
a(n) = n^2 * binomial(3*n, n);
```

a(n) = n^2 * A005809(n).
a(n) = 3 * n * A090763(n-1) = 3 * n * A144485(n) / 2.
Sum_{n>=1} 1/a(n) = 6 * arctan(sqrt(3)/(2*phi-1))^2 - log((phi^3+1)/(phi+1)^3)^2/2, where phi = ((25+3*sqrt(69))/2)^(1/3) (Batir, 2005, p. 378, eq. (3.1)).

A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 10, 15, 1, 21, 84, 84, 1, 36, 270, 660, 495, 1, 55, 660, 2860, 5005, 3003, 1, 78, 1365, 9100, 27300, 37128, 18564, 1, 105, 2520, 23800, 107100, 244188, 271320, 116280, 1, 136, 4284, 54264, 339150, 1139544, 2089164, 1961256, 735471, 1, 171, 6840, 111720, 921690, 4239774, 11306064

Offset: 0

Peter Bala, Feb 13 2024

Compare with A063007(n, k) = binomial(n, k)*binomial(n+k, k), the table of coefficients of the shifted Legendre polynomials P(n, 2*x + 1).

			Triangle begins
n\k| 0    1     2      3       4       5       6       7
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
 0 | 1
 1 | 1    3
 2 | 1   10    15
 3 | 1   21    84     84
 4 | 1   36   270    660     495
 5 | 1   55   660   2860    5005    3003
 6 | 1   78  1365   9100   27300   37128   18564
 7 | 1  105  2520  23800  107100  244188  271320  116280
 ...

A114496 (row sums), A000984 (alt. row sums unsigned), A005809 (main diagonal), A090763 (first subdiagonal), A014105 (column 1).

Cf. A026000, A063007, A102537, A110608, A339710.

seq(print(seq(binomial(n, k)*binomial(2*n+k, k), k = 0..n)), n = 0..10);

n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(2*n+k, k)*x^k = (1 + x)^n * Sum_{k = 0..n} binomial(n, k)*binomial(2*n, k)*(x/(1 + x))^k = Sum_{k = 0..n} A110608(n, n-k)*x^k*(1 + x)^(n-k).
(x - 1)^n * R(n, 1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)*binomial(2*n, n-k)*x^k = the n-th row polynomial of A110608.
R(n, x) = hypergeom([-n, 2*n + 1], [1], -x).
Second-order differential equation: ( (1 + x)^n * (x + x^2)*R(n, x)' )' = n*(2*n + 1)*(1 + x)^n * R(n, x), where the prime indicates differentiation w.r.t. x.
Equivalently, x*(1 + x)*R(n, x)'' + ((n + 2)*x + 1)*R(n, x)' - n*(2*n + 1)*R(n, x)' = 0.
Analog of Rodrigues' formula for the shifted Legendre polynomials:
R(n, x) = 1/(1 + x)^n * 1/n! * (d/dx)^n (x*(1 + x)^2)^n.
Analog of Rodrigues' formula for the Legendre polynomials:
R(n, (x-1)/2) = 1/(n!*2^n) * 1/(1 + x)^n *(d/dx)^n ((x - 1)*(x + 1)^2)^n.
Orthogonality properties:
Integral_{x = -1..0} (1 + x)^n * R(n, x) * R(m, x) dx = 0 for n > m.
Integral_{x = -1..0} (1 + x)^n * R(n, x)^2 dx = 1/(3*n + 1).
Integral_{x = -1..0} (1 + x)^(n+m) * R(n, x) * R(m, x) dx = 0 for m >= 2*n + 1 or m <= (n - 1)/2.
Integral_{x = -1..0} (1 + x)^k * R(n, x) dx = 0 for n <= k <= 2*n - 1;
Integral_{x = -1..0} (1 + x)^(2*n) * R(n, x) dx = (2*n)!*n!/(3*n+1)! = 1/A090816(n).
Recurrence for row polynomials:
2*n*(2*n - 1)*((9*n - 12)*x + 8*n - 11)*(1 + x)*R(n, x) = (9*(3*n - 1)*(3*n - 2)*(3*n - 4)*x^3 + 3*(3*n - 1)*(3*n - 2)*(20*n - 27)*x^2 + 6*(3*n - 2)*(20*n^2 - 34*n + 9)*x + 2*(32*n^3 - 76*n^2 + 50*n - 9))*R(n-1, x) - 2*(n - 1)*(2*n - 3)*((9*n - 3)*x + 8*n - 3)*R(n-2, x), with R(0, x) = 1, R(1, x) = 1 + 3*x.
Conjecture: exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + 3*t)*x + (1 + 8*t + 12*t^2)*x^2 + ... is the o.g.f. for A102537. If true, then it would follows that, for each integer t, the sequence u = {R(n,t) : n >= 0} satisfies the Gauss congruences u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all primes p and positive integers m and r.
R(n, 1) = A114496(n); R(n, -1) = (-1)^n * A000984(n).
R(n, 2) = A339710(n); R(n, -2) = (-1)^n * A026000(n).
(2^n)*R(n, -1/2) = A234839(n).

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A144485 a(n) = (3n + 2)*binomial(3n + 1,n).

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A096793 Triangle read by rows: a(n,k) is the number of Dyck n-paths containing k odd-length ascents.

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A360560 Triangle read by rows. T(n, k) = (1/2) * C(n, k) * C(3*n - 1, n) for n > 0 and T(0, 0) = 1.

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A378777 a(n) = n^2 * binomial(3*n, n).

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A370258 Triangle read by rows: T(n, k) = binomial(n, k)*binomial(2*n+k, k), 0 <= k <= n.

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A370258 Triangle read by rows: T(n, k) = binomial(n, k)binomial(2n+k, k), 0 <= k <= n.