A000894 -id:A000894 - cp's OEIS frontend

A228330 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(4).

1, 20, 362, 6504, 114686, 1992536, 34231540, 583027920, 9862508790, 165918037560, 2778642667020, 46358257249200, 770951008563372, 12785838603285104, 211540243555702376, 3492587812271418784, 57557091526140668070, 946970607665938615032, 15557339429900195819164, 255246113991506558429936

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

G. C. Greubel, Table of n, a(n) for n = 0..825
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega4(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492 (h(0)), A000894 (h(1)), A228329 (h(2)), A000515 (h(3)), this sequence (h(4)), A228331 (h(5)), A228332 (h(6)), A228333 (h(7)).

List([0..20], n-> Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))); # G. C. Greubel, Mar 02 2019

[Binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)): n in [0..20]]; // G. C. Greubel, Mar 02 2019

Table[4*Sum[(k+1)^6*(Binomial[2n+1, n-k]/(n+k+2))^2, {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

vector(20, n, n--; binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1))) \\ G. C. Greubel, Mar 02 2019

[binomial(4*n,2*n)*(15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)) for n in (0..20)] # G. C. Greubel, Mar 02 2019

Conjecture: n*(2*n+1)*(3467*n-4029)*a(n) + 8*(-36721*n^3 + 109040*n^2 - 137926*n + 69822)*a(n-1) + 4*(4*n-9)*(45706*n-7907)*(4*n-7)*a(n-2) = 0. - R. J. Mathar, Sep 08 2013
Recurrence: n*(2*n+1)*(15*n^3 - 30*n^2 + 16*n - 2)*a(n) = 2*(4*n-5)*(4*n-3)*(15*n^3 + 15*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
From Vaclav Kotesovec, Dec 08 2013: (Start)
a(n) = binomial(4*n,2*n) * (15*n^3+15*n^2+n-1)/((2*n+1)*(4*n-1)).
a(n) = 4*Sum_{k=0..n} (k+1)^6*(binomial(2*n+1, n-k)/(n+k+2))^2. (End)

A228331 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(5).

Original entry on oeis.org

1, 36, 780, 16240, 321300, 6131664, 114017904, 2079380160, 37356642180, 663144710800, 11657925495216, 203295462691776, 3521108298744400, 60632838691387200, 1038859802556120000, 17721669103065158400, 301147406355880764900, 5099997408534884394000, 86106549929771707182000

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega5. Remark 3 p. 1882.
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017, 2013
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228332, A228333.

Table[Sum[(k+1)^5*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Conjecture: n^2*a(n) +4*(2*n^2-22*n+11)*a(n-1) +16*(-7*n^2-54*n+166)*a(n-2) -1088*(2*n-3)*(2*n-7)*a(n-3)=0. - R. J. Mathar, Sep 08 2013
Recurrence: n^2*(3*n^2 - 5*n + 1)*a(n) = 4*(2*n-3)*(2*n+1)*(3*n^2 + n - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(2*n,n)^2 * (2*n+1)*(3*n^2+n-1)/(2*n-1). - Vaclav Kotesovec, Dec 08 2013
G.f.: ((28*x+3)*hypergeom([1/2, 5/2],[1],16*x)+20*x*(1-16*x)*hypergeom([3/2, 7/2],[2],16*x))/3. - Mark van Hoeij, Apr 12 2014

A228333 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(7).

Original entry on oeis.org

1, 132, 4260, 120400, 3017700, 69776784, 1524611088, 31951782720, 648578888100, 12837530477200, 248966505964176, 4747739344525632, 89267646282614800, 1658349027407016000, 30489930211792680000, 555544747397829254400, 10042477557290424843300, 180267292319119226298000, 3215718323211443887530000

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Omega7. Remark 3 p. 1882.
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228332.

Table[Sum[(k+1)^7*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Conjecture: n^2*(304*n-411)*a(n) + 4*(-1814*n^3+2554*n^2-4776*n+7567)*a(n-1) + 32*(2*n-5)*(2*n-1)*(299*n-176)*a(n-2) = 0. - R. J. Mathar, Dec 04 2013
Recurrence: n^2*(6*n^3 - 12*n^2 + 6*n - 1)*a(n) = 4*(2*n-3)*(2*n+1)*(6*n^3 + 6*n^2 - 1)*a(n-1). - Vaclav Kotesovec, Dec 08 2013
a(n) = binomial(2*n,n)^2 * (2*n+1)*(6*n^3+6*n^2-1)/(2*n-1). - Vaclav Kotesovec, Dec 08 2013
G.f.: ((256*x+3)*hypergeom([1/2, 5/2],[1],16*x)+80*(38*x+1)*x*hypergeom([3/2, 7/2],[2],16*x))/3.  - Mark van Hoeij, Apr 12 2014

A241530 a(n) = binomial(n,floor(n/2))binomial(n+1,floor(n/2+1/2))(1+floor(n/2))/(1+2*floor(n/2)).

Original entry on oeis.org

1, 2, 4, 12, 36, 120, 400, 1400, 4900, 17640, 63504, 232848, 853776, 3171168, 11778624, 44169840, 165636900, 625739400, 2363904400, 8982836720, 34134779536, 130332794592, 497634306624, 1907598175392, 7312459672336, 28124844893600, 108172480360000

Offset: 0

Peter Luschny, Apr 25 2014

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000894, A002894, A005566.

Row n=3 of A275784.

A241530 := n -> binomial(n,iquo(n,2))*binomial(n+1,iquo(n+1,2))
*(1+iquo(n,2))/(1+2*iquo(n,2)); seq(A241530(n), n=0..26);
# second Maple program:
a:= proc(n) option remember; `if`(n<2, 2^n,
     ((8*n-4)*a(n-1)+16*(n-1)*(n-2)*a(n-2))/(n*(n+1)))
    end:
seq(a(n), n=0..30);  # Alois P. Heinz, Aug 10 2016

CoefficientList[Series[(-EllipticE[16 x^2] + (1 + 4 x) EllipticK[16 x^2])/(2Pi x), {x, 0, 20}], x] (* Benedict W. J. Irwin, Aug 15 2016 *)
Table[Binomial[n, #] Binomial[n + 1, Floor[(n + 1)/2]] (1 + #)/(1 + 2 #) &@ Floor[n/2], {n, 0, 26}] (* Michael De Vlieger, Aug 15 2016 *)

a(n) = ((8*n-4)*a(n-1)+16*(n-1)*(n-2)*a(n-2))/(n*(n+1)) for n>=2, a(n) = 2^n for n<2. - Alois P. Heinz, Apr 25 2014
G.f.: ((1+4*x)*K(4*x) - E(4*x))/(2*Pi*x), where K and E are the complete elliptic integrals of the first and second kind, respectively, with modulus k = 4*x. - Benedict W. J. Irwin, Aug 15 2016
From Wolfdieter Lang, Sep 06 2016 (Start):
The preceding g.f. can be rewritten as ((1+4*x)*F(1/2,1/2;1;(4*x)^2) -
   F(-1/2,1/2;1;(4*x)^2))/(4*x), where F is the hypergyometric function F(a,b;c;z).
This leads to the bisection a(2*k) = ((2*k)!)^2/k!^4 = A002894(k) and a(2*k+1) = 2*(2*k)!*(2*k+1)!/((k+1)*k!^4) = 2*A000894(k), for k >= 0.
(End)

A248045 (2(n-1))! (2n-1)! / (n (n-1)!^3).

Original entry on oeis.org

1, 6, 120, 4200, 211680, 13970880, 1141620480, 111307996800, 12614906304000, 1629845894476800, 236475822507724800, 38072607423743692800, 6735922851893114880000, 1299070835722243584000000, 271245990498804460339200000, 60962536364606302461235200000

Offset: 1

Reinhard Zumkeller, Sep 30 2014

nonn

Central terms in triangles of Lah numbers: a(n) = - A008297(2*n-1,n) = A105278(2*n-1,n) = A000891(n-1)*A000142(n) = A000894(n-1)*A000142(n-1).

a(n) = n * A204515(n-1). - Reinhard Zumkeller, Oct 19 2014

Reinhard Zumkeller, Table of n, a(n) for n = 1..250

Cf. A000142, A000891, A000894, A008297, A105278, A204515.

Cf. A187535 (Central Lah numbers).

Haskell
```
a248045 n = a000891 (n - 1) * a000142 n
```

n*a(n) = 4*(2*n-1)*(2*n-3)*a(n-1). - R. J. Mathar, Oct 07 2014

A228332 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(6).

Original entry on oeis.org

1, 68, 1778, 43080, 958430, 20119736, 405350788, 7921691280, 151231519350, 2834134359000, 52320693313020, 953960351550960, 17212782834351468, 307826474156801840, 5462948893700675720, 96303960593503261984, 1687752152779483045542, 29424712141610821296408, 510621541414656188646220

Offset: 0

N. J. A. Sloane, Aug 26 2013

nonn

Pedro J. Miana and Natalia Romero, Moments of combinatorial and Catalan numbers, Journal of Number Theory, Volume 130, Issue 8, August 2010, Pages 1876-1887. See Remark 3 p. 1882. Omega6(n) = a(n-1).
Yidong Sun and Fei Ma, Four transformations on the Catalan triangle, arXiv preprint arXiv:1305.2017 [math.CO], 2013.
Yidong Sun and Fei Ma, Some new binomial sums related to the Catalan triangle, Electronic Journal of Combinatorics 21(1) (2014), #P1.33.

Cf. A000108, A039598, A024492, A000894, A228329, A000515, A228330, A228331, A228333.

Table[Sum[(k+1)^6*(Binomial[2n+1, n-k]*2*(k+1)/(n+k+2))^2,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Dec 08 2013 *)

Recurrence: n*(2*n+1)*(105*n^5 - 420*n^4 + 588*n^3 - 356*n^2 + 96*n - 10)*a(n) = 2*(4*n-7)*(4*n-5)*(105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3)*a(n-1). - Vaclav Kotesovec, Dec 08 2013

a(n) = binomial(4*n,2*n) * (105*n^5 + 105*n^4 - 42*n^3 - 62*n^2 - 7*n + 3) / ((2*n+1)*(4*n-3)*(4*n-1)). - Vaclav Kotesovec, Dec 08 2013

A082578 a(n) = Sum_{k=0..n} binomial(2k, k) binomial(2*k+1,k).

Original entry on oeis.org

1, 7, 67, 767, 9587, 126011, 1711595, 23796515, 336666215, 4828084575, 69994481871, 1023793569567, 15086216016367, 223704570996367, 3335098322412367, 49954148031128767, 751296616443141667

Offset: 0

Emanuele Munarini, May 07 2003

Old name was "A binomial sum".

Vincenzo Librandi, Table of n, a(n) for n = 0..200

Partial sums of A000894.

Table[Sum[Binomial[2k,k]*Binomial[2k+1,k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)

makelist(sum(binomial(2*k,k)*binomial(2*k+1,k),k,0,n),n,0,12);

Recurrence: (n+3)*(n+2)*a(n+2) - (17*n^2+69*n+66)*a(n+1) + (16*n^2+64*n+60)*a(n) = 0.

a(n) ~ 2^(4*n+5)/(15*Pi*n). - Vaclav Kotesovec, Oct 14 2012

Name changed by Wesley Ivan Hurt, Apr 18 2023

A002463 Coefficients of Legendre polynomials.

Original entry on oeis.org

1, 3, 30, 175, 4410, 29106, 396396, 2760615, 156434850, 1122854590, 16291599324, 119224885962, 3515605611700, 26077294372500, 388924218927000, 2913690606794775, 350671234206006450, 2647224022927695750, 40095381399899017500, 304513870316075169750

Offset: 1

N. J. A. Sloane

nonn

Apparently, a(n) divides A000894(n). - Ralf Stephan, Aug 05 2004
Coefficients of cos(x) term of the Tisserand functions of odd order for the planar case with the denominators factored out (see Table 1 from Laskar & Boué's paper) (cf A002462). - Michel Marcus, May 29 2013
Also cos(x) term of the Legendre polynomials of odd order when they are expressed in terms of the cosine function (see 22.3.13 from Abramowitz & Stegun) with the denominators factored out. - Michel Marcus, May 29 2013

A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 362.
G. Prévost, Tables de Fonctions Sphériques. Gauthier-Villars, Paris, 1933, pp. 156-157.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

lista(nn) = {forstep (n=1, nn, 2, lcmc = 1; for (m=0, n\2, lcmc = lcm(lcmc, denominator(binomial(2*n-2*m, n-m) * binomial(2*m, m)/4^n));); m = n\2; print1(lcmc*binomial(2*n-2*m, n-m) * binomial(2*m, m)/4^n, ", "););} \\ Michel Marcus, May 29 2013

from sympy import binomial as C, lcm
def a_list(nn):
    l = []
    for n in range(1, nn + 1, 2):
        lcmc = 1
        for m in range(n//2 + 1):
            lcmc = lcm(lcmc, (C(2*n - 2*m, n - m)*C(2*m, m)/4**n).denominator)
        m = n//2
        l.append(lcmc*C(2*n - 2*m, n - m)*C(2*m, m)//4**n)
    return l # Indranil Ghosh, Jul 02 2017, after PARI code by Michel Marcus

More terms from Michel Marcus, May 29 2013

A124051 Quasi-mirror of A062196 formatted as a triangular array.

Original entry on oeis.org

3, 6, 8, 10, 30, 15, 15, 80, 90, 24, 21, 175, 350, 210, 35, 28, 336, 1050, 1120, 420, 48, 36, 588, 2646, 4410, 2940, 756, 63, 45, 960, 5880, 14112, 14700, 6720, 1260, 80, 55, 1485, 11880, 38808, 58212, 41580, 13860, 1980, 99, 66, 2200, 22275, 95040, 194040, 199584, 103950, 26400, 2970, 120

Offset: 0

Zerinvary Lajos, Nov 03 2006

			Triangle begins as:
   3;
   6,    8;
  10,   30,    15;
  15,   80,    90,    24;
  21,  175,   350,   210,     35;
  28,  336,  1050,  1120,    420,     48;
  36,  588,  2646,  4410,   2940,    756,     63;
  45,  960,  5880, 14112,  14700,   6720,   1260,    80;
  55, 1485, 11880, 38808,  58212,  41580,  13860,  1980,   99;
  66, 2200, 22275, 95040, 194040, 199584, 103950, 26400, 2970, 120;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000108, A000894, A062196, A286033.
Columns k: A000217(n+2) (k=0), A002417(n+1) (k=1), A001297(n) (k=2), A105946(n-2) (k=3), A105947(n-3) (k=4), A105948(n-4) (k=5), A107319(n-5) (k=6).
Diagonals: A005563(n+1) (k=n), A033487(n) (k=n-1), A027790(n) (k=n-2), A107395(n-3) (k=n-3), A107396(n-4) (k=n-4), A107397(n-5) (k=n-5), A107398(n-6) (k=n-6), A107399(n-7) (k=n-7).
Sums: A322938(n+1) (row).

A124051:= func< n,k | Binomial(n+1,n-k+1)*Binomial(n+3,n-k+1) >;
[A124051(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 07 2025

for n from 0 to 10 do seq(binomial(n,i-1)*binomial(n+2,n+1-i), i=1..n ) od;

A124051[n_, k_]:= Binomial[n+1,n-k+1]*Binomial[n+3,n-k+1];
Table[A124051[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 07 2025 *)

def A124051(n,k): return binomial(n+1,n-k+1)*binomial(n+3,n-k+1)
print(flatten([[A124051(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Feb 07 2025

From G. C. Greubel, Feb 07 2025: (Start)
T(n, k) = binomial(n+1, n-k+1)*binomial(n+3, n-k+1).
T(2*n, n) = (1/2)*A000894(n) + (5/2)*[n=0].
Sum_{k=0..n} (-1)^k*T(n, k) = (1/2)*( (1+(-1)^n)*(-1)^(n/2)*A286033((n+4)/2) + (1-(-1)^n)*((-1)^((n+1)/2)*A000108((n+1)/2) - 1) ). (End)

cp's OEIS Frontend

A228330 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(4).

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A228331 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(5).

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A228333 Let h(m) denote the sequence whose n-th term is Sum__{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(7).

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A241530 a(n) = binomial(n,floor(n/2))*binomial(n+1,floor(n/2+1/2))*(1+floor(n/2))/(1+2*floor(n/2)).

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A248045 (2*(n-1))! * (2*n-1)! / (n * (n-1)!^3).

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A228332 Let h(m) denote the sequence whose n-th term is Sum_{k=0..n} (k+1)^m*T(n,k)^2, where T(n,k) is the Catalan triangle A039598. This is h(6).

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A082578 a(n) = Sum_{k=0..n} binomial(2*k, k) * binomial(2*k+1,k).

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A002463 Coefficients of Legendre polynomials.

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A241530 a(n) = binomial(n,floor(n/2))binomial(n+1,floor(n/2+1/2))(1+floor(n/2))/(1+2*floor(n/2)).

A248045 (2(n-1))! (2n-1)! / (n (n-1)!^3).

A082578 a(n) = Sum_{k=0..n} binomial(2k, k) binomial(2*k+1,k).