A109188 -id:A109188 - cp's OEIS frontend

A037965 a(n) = nbinomial(2n-2, n-1).

0, 1, 4, 18, 80, 350, 1512, 6468, 27456, 115830, 486200, 2032316, 8465184, 35154028, 145608400, 601749000, 2481880320, 10218366630, 42004911960, 172427570700, 706905276000, 2894777105220, 11841673237680, 48394276165560, 197602337462400, 806190092077500

Offset: 0

N. J. A. Sloane

a(n+1) is the convolution of A000984 and A081294. - Paul Barry, Sep 18 2008

The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1972.

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A000984, A001622, A081294, A109188 (inverse binomial transform).

Cf. A000108, A100071, A134757.

[0] cat [n^2*Catalan(n-1): n in [1..30]]; // G. C. Greubel, Jun 19 2022

a[n_]:= n*Binomial[2*n-2, n-1]; Array[a, 30, 0] (* Amiram Eldar, Mar 10 2022 *)

a(n) = n*binomial(2*n-2, n-1); \\ Joerg Arndt, Sep 04 2017

[n^2*catalan_number(n-1) for n in (0..30)] # G. C. Greubel, Jun 19 2022

Assuming offset -1 here and offset 0 in A134757, A134757 is the inverse binomial transform of this sequence. - Gary W. Adamson, Nov 08 2007
G.f.: Hypergeometric2F1([1/2, 2], [1], 4*x). - Paul Barry, Sep 03 2008
From Paul Barry, Sep 18 2008: (Start)
G.f.: x*(1-2*x)/(1-4*x)^(3/2);
a(n+1) = Sum_{k=0..n} binomial(2*k,k)*(4^(n-k) + 0^(n-k))/2. (End)
D-finite with recurrence (n-1)*a(n) - 2*(3*n-4)*a(n-1) + 4*(2*n-5)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012
E.g.f.: x*exp(2*x)*BesselI(0,2*x). - Ilya Gutkovskiy, Aug 22 2018
a(n) = n*A000984(n-1). - R. J. Mathar, Nov 08 2021
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=1} 1/a(n) = 4*Pi/(3*sqrt(3)) - Pi^2/9.
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*log(phi)/sqrt(5) - 4*log(phi)^2, where phi is the golden ratio (A001622). (End)

More terms from Zerinvary Lajos, Oct 02 2007

A331431 Triangle read by rows: T(n,k) = (-1)^(n+k)(n+k+1)binomial(n,k)*binomial(n+k,k) for n >= k >= 0.

Original entry on oeis.org

1, -2, 6, 3, -24, 30, -4, 60, -180, 140, 5, -120, 630, -1120, 630, -6, 210, -1680, 5040, -6300, 2772, 7, -336, 3780, -16800, 34650, -33264, 12012, -8, 504, -7560, 46200, -138600, 216216, -168168, 51480, 9, -720, 13860, -110880, 450450, -1009008, 1261260, -823680, 218790

Offset: 0

N. J. A. Sloane, Jan 17 2020

Tables I, III, IV on pages 92 and 93 of Ser have integer entries and are A331430, A331431 (the present sequence), and A331432.

Given the system of equations 1 = Sum_{j=0..n} H(i, j) * x(j) for i = 2..n+2 where H(i,j) = 1/(i+j-1) for 1 <= i,j <= n is the n X n Hilbert matrix, then the solutions are x(j) = T(n, j). - Michael Somos, Mar 20 2020 [Corrected by Petros Hadjicostas, Jul 09 2020]

			Triangle begins:
   1;
  -2,    6;
   3,  -24,    30;
  -4,   60,  -180,     140;
   5, -120,   630,   -1120,     630;
  -6,  210, -1680,    5040,   -6300,     2772;
   7, -336,  3780,  -16800,   34650,   -33264,   12012;
  -8,  504, -7560,   46200, -138600,   216216, -168168,   51480;
   9, -720, 13860, -110880,  450450, -1009008, 1261260, -823680, 218790;
  ...

J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 93. See Table III.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
A. Buhl, Book review: J. Ser - Les calculs formels des séries de factorielles, L'Enseignement Mathématique, 32 (1933), p. 275.
L. A. MacColl, Review: J. Ser, Les calculs formels des séries de factorielles, Bull. Amer. Math. Soc., 41(3) (1935), p. 174.
L. M. Milne-Thomson, Review of Les calculs formels des séries de factorielles. By J. Ser. Pp. vii, 98. 20 fr. 1933. (Gauthier-Villars), The Mathematical Gazette, Vol. 18, No. 228 (May, 1934), pp. 136-137.
J. Ser, Les Calculs Formels des Séries de Factorielles, Gauthier-Villars, Paris, 1933 [Local copy].
J. Ser, Les Calculs Formels des Séries de Factorielles (Annotated scans of some selected pages.)

Columns 1 is A331433 or equally A007531, column 2 is A331434 or equally A054559; the last three diagonals are A002738, A002736, A002457.

Cf. A000290 (row sums), A002457,, A100071, A108666 (alternating row sums), A109188 (diagonal sums), A331322, A331323, A331430, A331432.

[(-1)^(n+k)*(k+1)*(2*k+1)*Binomial(n+k+1,n-k)*Catalan(k): k in [0..n], n in [0..15]]; // G. C. Greubel, Mar 22 2022

gf := k -> (1+x)^(-2*(k+1)): ser := k -> series(gf(k), x, 32):
T := (n, k) -> ((2*k+1)!/(k!)^2)*coeff(ser(k), x, n-k):
seq(seq(T(n,k), k=0..n),n=0..7); # Peter Luschny, Jan 18 2020
S:=(n,k)->(-1)^(n+k)*(n+k+1)!/((k!)^2*(n-k)!);
rho:=n->[seq(S(n,k),k=0..n)];
for n from 0 to 14 do lprint(rho(n)); od: # N. J. A. Sloane, Jan 18 2020

Table[(-1)^(n+k)*(n+k+1)*Binomial[2*k,k]*Binomial[n+k,n-k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 22 2022 *)

flatten([[(-1)^(n+k)*(2*k+1)*binomial(2*k,k)*binomial(n+k+1,n-k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Mar 22 2022

T(n, 0) = (-1)^n*A000027(n+1).
T(n, 1) = A331433(n-1) = (-1)^(n+1)*A007531(n+2).
T(n, 2) = A331434(n-2) = (-1)^n*A054559(n+3).
T(n, n-2) = A002738(n-2).
T(n, n-1) = (-1)*A002736(n).
T(n, n) = A002457(n).
T(2*n, n) = (-1)^n*(3*n+1)!/(n!)^3 = (-1)^n*A331322(n).
Sum_{k=0..n} T(n, k) = A000290(n+1) (row sums).
Sum_{k=0..n} (-1)^k*T(n, k) = (-1)^n*A108666(n+1) (alternating row sums).
Sum_{k=0..n} T(n-k, k) = (-1)^n*A109188(n+1) (diagonal sums).
2^n*Sum_{k=0..n} T(n, k)/2^k = (-1)^floor(n/2)*A100071(n+1) (positive half sums).
(-2)^n*Sum_{k=0..n} T(n, k)/(-2)^k = A331323(n) (negative half sums).
T(n, k) = ((2*k+1)!/(k!)^2)*[x^(n-k)] (1+x)^(-2*(k+1)). - Georg Fischer and Peter Luschny, Jan 18 2020
T(n,k) = (-1)^(n+k)*(n+k+1)!/((k!)^2*(n-k)!), for n >= k >= 0. - N. J. A. Sloane, Jan 18 2020
From Petros Hadjicostas, Jul 09 2020: (Start)
Michael Somos's formulas above can be restated as
Sum_{k=0..n} T(n,k)/(i+k) = 1 for i = 1..n+1.
These are special cases of the following formula that is alluded to (in some way) in Ser's book:
1 - Sum_{k=0..n} T(n,k)/(x + k) = (x-1)*...*(x-(n + 1))/(x*(x+1)*...*(x+n)).
Because T(n,k) = (-1)^(n+1)*(n + k + 1)*A331430(n,k) and Sum_{k=0..n} A331430(n,k) = (-1)^(n+1), one may derive this formula from Ser's second formula stated in A331430. (End)
T(2*n+1, n) = (-2)*(-27)^n*Pochhammer(4/3, n)*Pochhammer(5/3, n)/(n!*(n+1)!). - G. C. Greubel, Mar 22 2022

Several typos in the data corrected by Georg Fischer and Peter Luschny, Jan 18 2020

Definition changed by N. J. A. Sloane, Jan 18 2020

A109187 Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k (1,0)-steps.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 6, 0, 1, 6, 0, 12, 0, 1, 0, 30, 0, 20, 0, 1, 20, 0, 90, 0, 30, 0, 1, 0, 140, 0, 210, 0, 42, 0, 1, 70, 0, 560, 0, 420, 0, 56, 0, 1, 0, 630, 0, 1680, 0, 756, 0, 72, 0, 1, 252, 0, 3150, 0, 4200, 0, 1260, 0, 90, 0, 1, 0, 2772, 0, 11550, 0, 9240, 0, 1980, 0, 110, 0, 1

Offset: 0

Emeric Deutsch, Jun 21 2005

A Grand Motzkin path is a path in the half-plane x >= 0, starting at (0,0), ending at (n,0) and consisting of steps u=(1,1), d=(1,-1) and h=(1,0).
From Peter Bala, Feb 11 2017: (Start)
Consider an infinite 1-dimensional integer lattice with an oriented self-loop at each vertex. Then T(n,k) equals the number of walks of length n from a vertex to itself having k loops. There is a bijection between such walks and Grand Motzkin paths which takes a right step and a left step on the lattice to an up step U and a down step D of a Grand Motzkin path respectively, and takes traversing a loop on the lattice to the horizontal step H. See A282252 for the corresponding triangle of walks on a 2-dimensional lattice with self-loops. (End)

			T(3,1)=6 because we have hud,hdu,udh,duh,uhd,dhu, where u=(1,1),d=(1,-1), h=(1,0).
Triangle begins:
n\k   [0]  [1]   [2]   [3]   [4]   [5]   [6]  [7]  [8]  [9] [10]
[0]    1;
[1]    0,   1;
[2]    2,   0,    1;
[3]    0,   6,    0,    1;
[4]    6,   0,   12,    0,    1;
[5]    0,  30,    0,   20,    0,    1;
[6]   20,   0,   90,    0,   30,    0,    1;
[7]    0, 140,    0,  210,    0,   42,    0,   1;
[8]   70,   0,  560,    0,  420,    0,   56,   0,   1;
[9]    0, 630,    0, 1680,    0,  756,    0,  72,   0,   1;
[10] 252,   0, 3150,    0, 4200,    0, 1260,   0,  90,   0,   1;
[11] ...
From _Peter Bala_, Feb 11 2017: (Start)
The infinitesimal generator begins
      0
      0    0
      2    0     0
      0    6     0     0
     -6    0    12     0     0
      0  -30     0    20     0   0
     80    0   -90     0    30   0   0
      0  560     0  -210     0  42   0  0
  -2310    0  2240     0  -420   0  56  0  0
  ....
and equals the generalized exponential Riordan array [log(Bessel_I(0,2x)),x], and so has integer entries. (End)

Gheorghe Coserea, Rows n = 0..100, flattened
Paul Barry, On the duals of the Fibonacci and Catalan-Fibonacci polynomials and Motzkin paths, arXiv:2101.10218 [math.CO], 2021.

Diagonal of rational function R(x, y, t) = 1/(1 - (x^2 + t*x*y + y^2)) with respect to x,y, i.e., T(n,k) = [(xy)^n*t^k] R(x,y,t). For t=0..7 we have the diagonals: A126869(t=0, column 0), A002426(t=1, row sums), A000984(t=2), A026375(t=3), A081671(t=4), A098409(t=5), A098410(t=6), A104454(t=7).

Cf. A089627, A109188, A105868, A171128, A108666, A282252, A298608.

G:=1/sqrt((1-t*z)^2-4*z^2):Gser:=simplify(series(G,z=0,15)): P[0]:=1: for n from 1 to 13 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 13 do seq(coeff(t*P[n],t^k),k=1..n+1) od;
with(PolynomialTools): CL := p -> CoefficientList(simplify(p), x):
C := (n,x) -> binomial(2*n,n)*hypergeom([-n,-n],[-n+1/2],1/2-x/4):
seq(print(CL(C(n,x))), n=0..11); # Peter Luschny, Jan 23 2018

p[0] := 1; p[n_] := GegenbauerC[n, -n , -x/2];
Flatten[Table[CoefficientList[p[n], x], {n, 0, 11}]] (* Peter Luschny, Jan 23 2018 *)

T(n,k) = if ((n-k)%2, 0, binomial(n,k)*binomial(n-k, (n-k)/2));
concat(vector(12, n, vector(n, k, T(n-1, k-1)))) \\ Gheorghe Coserea, Sep 06 2018

G.f.: 1/sqrt((1-tz)^2-4z^2).
Row sums yield the central trinomial coefficients (A002426).
T(2n+1, 0) = 0.
T(2n, 0) = binomial(2n,n) (A000984).
Sum_{k=0..n} k*T(n,k) = A109188(n).
Except for the order, same rows as those of A105868.
Column k has e.g.f. (x^k/k!)*Bessel_I(0,2x). - Paul Barry, Mar 11 2006
T(n,k) = binomial((n+k)/2,k)*binomial(n,(n+k)/2)*(1+(-1)^(n-k))/2. - Paul Barry, Sep 18 2007
Coefficient array of the polynomials P(n,x) = x^n*hypergeom([1/2-n/2,-n/2], [1], 4/x^2). - Paul Barry, Oct 04 2008
G.f.: 1/(1-xy-2x^2/(1-xy-x^2/(1-xy-x^2/(1-xy-x^2/(1-.... (continued fraction). - Paul Barry, Jan 28 2009
From Paul Barry, Apr 21 2010: (Start)
Exponential Riordan array [Bessel_I(0,2x), x].
Coefficient array of the polynomials P(n,x) = Sum_{k=0..floor(n/2)} C(n,2k)*C(2k, k)*x^(n - 2k).
Diagonal sums are the aerated central Delannoy numbers (A001850 with interpolated zeros). (End)
From Peter Bala, Feb 11 2017: (Start)
T(n,k) = binomial(n,k)*binomial(n-k,floor((n-k)/2))*(1 + (-1)^(n-k))/2.
T(n,k) = (n/k) * T(n-1,k-1).
T(n,k) = the coefficient of H^k in the expansion of (H + U + 1/U)^n.
n-th row polynomial R(n,t) = Sum_{k = 0..floor(n/2)} binomial(n,2*k) * binomial(2*k,k) * t^(n-2*k) = coefficient of x^n in the expansion of (1 + t*x + x^2)^n.
R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(2*k,k)*(t - 2)^(n-k).
d/dt(R(n,t)) = n*R(n-1,t).
R(n,t) = (1/Pi) * Integral_{x = 0..Pi} (t + 2*cos(x))^n dx.
Moment representation on a finite interval: R(n,t) = 1/Pi * Integral_{x = t-2 .. t+2} x^n/sqrt((t + 2 - x)*(x - t + 2)) dx.
Recurrence: n*R(n,t) = t*(2*n - 1)*R(n-1,t) - (t^2 - 4)*(n - 1)*R(n-2,t) with R(0,t) = 1 and R(1,t) = t.
R(n,t) = A002426 (t = 1), A000984 (t = 2), A026375 (t = 3), A081671 (t = 4), A098409 (t = 5), A098410 (t = 6) and A104454(t = 7).
The zeros of the row polynomials appear to lie on the imaginary axis in the complex plane. Also, the zeros of R(n,t) and R(n+1,t) appear to interlace on the imaginary axis.
The polynomials R(n,1 + t) are the row polynomials of A171128. (End)
From Peter Luschny, Jan 23 2018: (Start)
These are the coefficients of the polynomials G(n, -n , -x/2) where G(n, a, x) denotes the n-th Gegenbauer polynomial.
These polynomials can also be expressed as C(n, x) = binomial(2*n,n)*hypergeom([-n, -n], [-n+1/2], 1/2-x/4). (End)

A132885 Triangle read by rows: T(n,k) is the number of paths in the right half-plane from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k H=(2,0) steps (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 3, 1, 7, 2, 19, 9, 1, 51, 28, 3, 141, 95, 18, 1, 393, 306, 70, 4, 1107, 987, 285, 30, 1, 3139, 3144, 1071, 140, 5, 8953, 9963, 3948, 665, 45, 1, 25653, 31390, 14148, 2856, 245, 6, 73789, 98483, 49815, 11844, 1330, 63, 1, 212941, 307836, 172645, 47160

Offset: 0

Emeric Deutsch, Sep 03 2007

Row n has 1+floor(n/2) terms. T(n,0)=A002426(n) (the central trinomial coefficients). T(n,1)=A109188(n-1). Row sums yield A059345. See A132280 for the same statistic on paths restricted to the first quadrant.

			T(4,1)=9 because we have hhH, hHh, Hhh, HUD, UDH, UHD, HDU, DUH and DHU.
Triangle starts:
                     1;
                     1;
                 3,      1;
                 7,      2;
            19,      9,      1;
            51,     28,      3;
       141,     95,     18,      1;
       393,    306,     70,      4;
  1107,    987,    285,     30,      1;
  3139,   3144,   1071,    140,      5;

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Columns k=0..3 give A002426, A109188(n-1), A373651(n-4), A375260(n-6).
Row sums gives A059345.
Cf. A132280.

G:=1/sqrt((1+z-t*z^2)*(1-3*z-t*z^2)): Gser:=simplify(series(G,z=0,18)): for n from 0 to 13 do P[n]:=sort(coeff(Gser,z,n)) end do: for n from 0 to 13 do seq(coeff(P[n],t,j),j=0..floor((1/2)*n)) end do; # yields sequence in triangular form
A132885 := (n,k) -> binomial(n-k,k)*hypergeom([k-n/2,k-n/2+1/2], [1], 4): seq(print(seq(round(evalf(A132885(n,k))),k=0..iquo(n,2))),n=0..9); # Peter Luschny, Sep 18 2014

T[n_, k_] := Binomial[n - k, k]*Hypergeometric2F1[k - n/2, k - n/2 + 1/2, 1, 4]; Table[T[n, k], {n,0,10}, {k, 0, Floor[n/2]}] // Flatten  (* G. C. Greubel, Mar 01 2017 *)

G.f.: 1/sqrt((1+z-t*z^2)*(1-3*z-t*z^2)).

T(n,k) = C(n-k,k)*hypergeom([k-n/2,k-n/2+1/2], [1], 4). - Peter Luschny, Sep 18 2014

A147685 Squares and centered square numbers interleaved.

Original entry on oeis.org

0, 1, 1, 5, 4, 13, 9, 25, 16, 41, 25, 61, 36, 85, 49, 113, 64, 145, 81, 181, 100, 221, 121, 265, 144, 313, 169, 365, 196, 421, 225, 481, 256, 545, 289, 613, 324, 685, 361, 761, 400, 841, 441, 925, 484, 1013, 529, 1105, 576, 1201, 625, 1301, 676, 1405, 729, 1513

Offset: 0

R. J. Mathar, Nov 10 2008

This could be called the inverse Motzkin transform of A109188 since the substitution x -> x/(1+x+x^2) in the independent variable of the g.f. A109188(x) yields this sequence here.

			G.f. = x + x^2 + 5*x^3 + 4*x^4 + 13*x^5 + 9*x^6 + 25*x^7 + 16*x^8 + 41*x^9 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 3, 0, -3, 0, 1).

Cf. A000290, A001844, A109188, A120328.

A147685:=n->n^2*(1+(-1)^n)/8+(n^2+1)*(1-(-1)^n)/4: seq(A147685(n), n=0..70); # Wesley Ivan Hurt, Sep 06 2015

CoefficientList[Series[x (1 + x + x^2) (1 + x^2)/((1 - x)^3 (1 + x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 07 2014 *)

{a(n) = if( n%2, (n^2 + 1) / 2, n^2 / 4)}; /* Michael Somos, Aug 07 2014 */

a(2*n) = A000290(n),  a(2*n+1) = A001844(n).
O.g.f.: x*(1+x+x^2)*(1+x^2)/((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6), n>5.
Euler transform of length 4 sequence [ 1, 4, -1, -1]. - Michael Somos, Aug 07 2014
a(2n+1) = a(2n) + a(2n+2) for all n in Z. - Michael Somos, Aug 07 2014
A120328(n-1) = 3*n^2 + 2 = a(2*n + 1) - a(2*n)+ a(2*n - 1) for all n in Z. - Michael Somos, Aug 07 2014
a(n) = n^2*(1+(-1)^n)/8+(n^2+1)*(1-(-1)^n)/4. - Wesley Ivan Hurt, Sep 06 2015

A194586 Triangle read by rows, T(n,k) the coefficients of the polynomials Sum_{k=0..n} binomial(n,k)A056040(k)(k mod 2)*q^k.

Original entry on oeis.org

0, 0, 1, 0, 2, 0, 0, 3, 0, 6, 0, 4, 0, 24, 0, 0, 5, 0, 60, 0, 30, 0, 6, 0, 120, 0, 180, 0, 0, 7, 0, 210, 0, 630, 0, 140, 0, 8, 0, 336, 0, 1680, 0, 1120, 0, 0, 9, 0, 504, 0, 3780, 0, 5040, 0, 630, 0, 10, 0, 720, 0, 7560, 0, 16800, 0, 6300, 0, 0, 11, 0, 990, 0, 13860, 0, 46200, 0, 34650, 0, 2772, 0, 12

Offset: 0

Peter Luschny, Aug 29 2011

Substituting q^k -> 1/(floor(k/2)+1) in the polynomials gives the complementary Motzkin numbers A005717. (See A089627 for the Motzkin numbers and A163649 for the extended Motzkin numbers.)

			               0
              0, 1
            0, 2, 0
           0, 3, 0, 6
         0, 4, 0, 24, 0
       0, 5, 0, 60, 0, 30
    0, 6, 0, 120, 0, 180, 0
  0, 7, 0, 210, 0, 630, 0, 140
                0
                q
               2 q
            3 q + 6 q^3
           4 q + 24 q^3
       5 q + 60 q^3  + 30 q^5
      6 q + 120 q^3  + 180 q^5
  7 q + 210 q^3  + 630 q^5  + 140 q^7

Peter Luschny, The lost Catalan numbers.

Row sums are A109188. Cf. A056040, A005717, A163649, A089627.

A194586 := proc(n,k) local j, swing; swing := n -> n!/iquo(n,2)!^2:
add(binomial(n,j)*swing(j)*q^j*(j mod 2),j=0..n); coeff(%,q,k) end:
seq(print(seq(A194586(n,k),k=0..n)),n=0..8);

sf[n_] := n!/Quotient[n, 2]!^2;
row[n_] := Sum[Binomial[n, j] sf[j] q^j Mod[j, 2], {j, 0, n}] // CoefficientList[#, q]& // PadRight[#, n+1]&;
Table[row[n], {n, 0, 12}] (* Jean-François Alcover, Jun 26 2019 *)

egf(x,y) = x*y*exp(x)*BesselI(0,2*x*y).

A375248 Expansion of (1 - x)/(1 - 2x - 3x^2)^(7/2).

Original entry on oeis.org

1, 6, 35, 168, 756, 3192, 12936, 50688, 193479, 722722, 2651649, 9581936, 34176324, 120526056, 420852204, 1456709328, 5002984791, 17062825626, 57827993685, 194871361608, 653285629920, 2179701604080, 7241015510820, 23958512912880, 78978801164445

Offset: 0

Seiichi Manyama, Aug 07 2024

nonn

First differences of A374506.
Cf. A046717, A109188, A371408.
Cf. A014531, A375253.

a[n_]:=(1+n)(2+n)(3+n)(4+n)(5+n)Hypergeometric2F1[(1-n)/2,-n/2,3,4]/120; Array[a,25,0] (* Stefano Spezia, Aug 07 2024 *)

my(N=30, x='x+O('x^N)); Vec((1-x)/(1-2*x-3*x^2)^(7/2))

a(n) = (binomial(n+5,3)/10) * Sum_{k=0..floor(n/2)} binomial(n+2,n-2*k) * binomial(2*k+2,k).
a(n) = (binomial(n+5,3)/10) * A014531(n+1).
a(n) = ((n+5)/(n*(n+4))) * ((2*n+3)*a(n-1) + 3*(n+4)*a(n-2)).
a(n) = (1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*hypergeom([(1-n)/2, -n/2], [3], 4)/120. - Stefano Spezia, Aug 07 2024

cp's OEIS Frontend

A037965 a(n) = n*binomial(2*n-2, n-1).

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A331431 Triangle read by rows: T(n,k) = (-1)^(n+k)*(n+k+1)*binomial(n,k)*binomial(n+k,k) for n >= k >= 0.

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A109187 Triangle read by rows: T(n,k) is number of Grand Motzkin paths of length n having k (1,0)-steps.

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A132885 Triangle read by rows: T(n,k) is the number of paths in the right half-plane from (0,0) to (n,0), consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), having k H=(2,0) steps (0 <= k <= floor(n/2)).

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A147685 Squares and centered square numbers interleaved.

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A194586 Triangle read by rows, T(n,k) the coefficients of the polynomials Sum_{k=0..n} binomial(n,k)*A056040(k)*(k mod 2)*q^k.

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A037965 a(n) = nbinomial(2n-2, n-1).

A331431 Triangle read by rows: T(n,k) = (-1)^(n+k)(n+k+1)binomial(n,k)*binomial(n+k,k) for n >= k >= 0.

A194586 Triangle read by rows, T(n,k) the coefficients of the polynomials Sum_{k=0..n} binomial(n,k)A056040(k)(k mod 2)*q^k.

A375248 Expansion of (1 - x)/(1 - 2x - 3x^2)^(7/2).