A002457 -id:A002457 - cp's OEIS frontend

A090816 a(n) = (3n+1)!/((2n)! * n!).

1, 12, 105, 840, 6435, 48048, 352716, 2558160, 18386775, 131231100, 931395465, 6580248480, 46312074900, 324897017760, 2272989850440, 15863901576864, 110487596768703, 768095592509700, 5330949171823275, 36945070220658600, 255702514854135195, 1767643865751234240

Offset: 0

Al Hakanson (hawkuu(AT)excite.com), Feb 11 2004

			a(1) = 4!/(2!*1!) = 24/2 = 12.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Necdet Batir, On certain series involving reciprocals of binomial coefficients, Journal of Classical Analysis, Vol. 2, No. 1 (2013), pp. 1-8.

Cf. A002457, A005809, A016777, A045721, A090957, A090969.

Halfdiagonal of triangle A003506.

[Factorial(3*n+1)/(Factorial(n)*Factorial(2*n)): n in [0..20]]; // G. C. Greubel, Feb 03 2019

a:=n-> binomial(3*n+1,2*n)*(n+1): seq(a(n), n=0..20); # Zerinvary Lajos, Jul 31 2006

f[n_] := 1/Integrate[(x^2 - x^3)^n, {x, 0, 1}]; Table[ f[n], {n, 0, 19}] (* Robert G. Wilson v, Feb 18 2004 *)
Table[1/Beta[2*n+1,n+1], {n,0,20}] (* G. C. Greubel, Feb 03 2019 *)

a(n)=if(n<0,0,(3*n+1)!/(2*n)!/n!) /* Michael Somos, Feb 14 2004 */

a(n)=if(n<0,0,1/subst(intformal((x^2-x^3)^n),x,1)) /* Michael Somos, Feb 14 2004 */

[1/beta(2*n+1,n+1) for n in range(20)] # G. C. Greubel, Feb 03 2019

a(n) = A005809(n) * A016777(n).
a(n) = 1/(Integral_{x=0..1} (x^2 - x^3)^n dx).
G.f.: (((8 + 27*z)*(1/(4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(1/3) + 1/(4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(1/3)) - 3*i*sqrt(3)*sqrt(4 - 27*z)*sqrt(z)*(1/(4*sqrt(4 - 27*z) + 12*i*sqrt(3)*sqrt(z))^(1/3) - 1/(4*sqrt(4 - 27*z) - 12*i*sqrt(3)*sqrt(z))^(1/3)))*8^(1/3))/(2*(4 - 27*z)^(3/2)), where i is the imaginary unit. - Karol A. Penson, Feb 06 2024
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (2*n + 2*k + 1)*binomial(2*n+k, k).  This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (2*n + 2*k + 1) * binomial(2*n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+2)*n+1, 2*n). Cf. A002457 and A306290. - Peter Bala, Nov 02 2024
From Amiram Eldar, Dec 09 2024: (Start)
Sum_{n>=0} 1/a(n) = f(c) = 1.09422712102982285131..., where f(x) = (x*(x-1)/(3*x-1)) * ((3/2)*log(abs(x/(x-1))) + ((3*x-2)/sqrt(3*x^2-4*x)) * (arctan(x/sqrt(3*x^2-4*x)) + arctan((2-x)/sqrt(3*x^2-4*x)))), and c = 2/3 + (1/3)*((25+3*sqrt(69))/2)^(-1/3) + (1/3)*((25+3*sqrt(69))/2)^(1/3).
Sum_{n>=0} (-1)^n/a(n) = f(d) = 0.92513707957813718109..., where f(x) is defined above, and d = 2/3 - (1/3)*((29+3*sqrt(93))/2)^(-1/3) - (1/3)*((29+3*sqrt(93))/2)^(1/3).
Both formulas are from Batir (2013). (End)

New definition from Vladeta Jovovic, Feb 12 2004

A092443 Sequence arising from enumeration of domino tilings of Aztec Pillow-like regions.

Original entry on oeis.org

3, 12, 50, 210, 882, 3696, 15444, 64350, 267410, 1108536, 4585308, 18929092, 78004500, 320932800, 1318498920, 5409723510, 22169259090, 90751353000, 371125269900, 1516311817020, 6189965556060, 25249187564640, 102917884095000, 419218847880300, 1706543186909652

Offset: 1

Christopher Hanusa (chanusa(AT)math.washington.edu), Mar 24 2004

The sequence 1, 3, 12, 50, ... is ((n+2)/2)*C(2n,n) with g.f. F(1/2,3;2;4x). - Paul Barry, Sep 18 2008

			a(3) = 5!/2!2! + 6!/3!3! = 50.

James Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 13).

Michael De Vlieger, Table of n, a(n) for n = 1..1659
Sanjay Moudgalya, Abhinav Prem, Rahul Nandkishore, Nicolas Regnault, and B. Andrei Bernevig, Thermalization and its absence within Krylov subspaces of a constrained Hamiltonian, arXiv:1910.14048 [cond-mat.str-el], 2019.
James Propp, Publications and Preprints.
James Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics, Cambridge University Press, Cambridge, 1999, pp. 255-291.
Eric Rowland and Jason Wu, The entries of the Sinkhorn limit of an m X n matrix, arXiv:2409.02789 [math.NT], 2024. See p. 11.

Cf. A000984, A001622, A001700, A002457.

Cf. A092437, A092438, A092439, A092440, A092441, A092442.

Array[Binomial[2 # + 1, # + 1] &[# - 1]*(# + 2) &, 22] (* Michael De Vlieger, Dec 17 2017 *)

combinat::catalan(n) *binomial(n+2,2) $ n = 1..22 // Zerinvary Lajos, Feb 15 2007

a(n) = (n+2)*binomial(2*n-1, n); \\ Altug Alkan, Dec 17 2017

a(n) = (2*n-1)!/((n-1)!)^2+(2*n)!/(n!)^2 = A002457(n-1) + A000984(n).
a(n) = (n+2)*A001700(n-1). - Vladeta Jovovic, Jul 12 2004
n*a(n) + (-7*n+4)*a(n-1) + 6*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 30 2012
From Amiram Eldar, Jan 27 2024: (Start)
Sum_{n>=1} 1/a(n) = 4*Pi*(11*sqrt(3)-3*Pi)/9 - 13.
Sum_{n>=1} (-1)^(n+1)/a(n) = 8*log(phi)*(13*sqrt(5)-30*log(phi))/5 - 11, where phi is the golden ratio (A001622). (End)
From Peter Bala, Aug 02 2024: (Start)
a(n) = 1/(n + 1)^2 * Sum_{k = 1..n+1} (k^3)*binomial(n+1, k)^2 = hypergeom([2, -n, -n], [1, 1], 1).
a(n) = 2*(n + 2)*(2*n - 1)/(n*(n + 1)) * a(n-1) with a(1) = 3. (End)

A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.

Original entry on oeis.org

1, 18, 294, 4724, 75654, 1210716, 19372380, 309961512, 4959397062, 79350401612, 1269606610548, 20313706474200, 325019306291356, 5200308911062296, 83204942617113336, 1331279082028930896, 21300465313063974726, 340807445011357201836, 5452919120190790364676

Offset: 0

Alexander Adamchuk, Nov 10 2009

nonn

p^2 divides a((p-3)/2) for prime p of the form p = 6k + 1 (A002476).

The expression a(n) = B^n*Sum_{k=0..n} binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5), a(n) = A167713 (B=16).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Eric Weisstein's World of Mathematics, Central Binomial Coefficient.
Eric Weisstein's World of Mathematics, Binomial Sums.

Cf. A000984, A002476, A066796, A006134, A082590, A132310, A002457, A144635.

A167713 := proc(n) coeftayl( 1/(1-16*x)/sqrt(1-4*x),x=0,n) ; end proc: seq(A167713(n),n=0..40) ; # R. J. Mathar, Nov 13 2009

Table[ 16^n * Sum[ (2k)!/(k!)^2 / 16^k, {k,0,n} ], {n,0,20} ]
CoefficientList[Series[1 / ((1 - 16 x) Sqrt[1 - 4 x]), {x, 0, 20}], x] (* Vincenzo Librandi, May 27 2013 *)

a(n) = 16^n * Sum_{k=0..n} ((2k)!/(k!)^2) / 16^k.
a(n) = 16^n * Sum_{k=0..n} binomial(2k,k) / 16^k.
G.f.: 1/((1-16*x)*sqrt(1-4*x)). - R. J. Mathar, Nov 13 2009
From Vaclav Kotesovec, Oct 20 2012: (Start)
Recurrence: n*a(n) = 2*(10*n-1)*a(n-1) - 32*(2*n-1)*a(n-2).
a(n) ~ 2^(4*n+1)/sqrt(3). (End)

Extended by R. J. Mathar, Nov 13 2009

A331511 Square array T(n,k), n >= 0, k >= 0, read by descending antidiagonals, where column k is the expansion of (1 - (k-3)x)/(1 - 2(k-1)x + ((k-3)x)^2)^(3/2).

Original entry on oeis.org

1, 1, 0, 1, 2, -15, 1, 4, -6, 32, 1, 6, 9, -12, 105, 1, 8, 30, 16, 30, -576, 1, 10, 57, 140, 25, 60, 105, 1, 12, 90, 384, 630, 36, -140, 5760, 1, 14, 129, 772, 2505, 2772, 49, -280, -13167, 1, 16, 174, 1328, 6430, 16008, 12012, 64, 630, -30400

Offset: 0

Seiichi Manyama, Jan 18 2020

			Square array begins:
      1,   1,  1,    1,     1,     1, ...
      0,   2,  4,    6,     8,    10, ...
    -15,  -6,  9,   30,    57,    90, ...
     32, -12, 16,  140,   384,   772, ...
    105,  30, 25,  630,  2505,  6430, ...
   -576,  60, 36, 2772, 16008, 52524, ...
.
From _Peter Luschny_, Jan 20 2020: (Start)
Read by ascending antidiagonals gives:
[0]      1
[1]      0,    1
[2]    -15,    2,  1
[3]     32,   -6,  4,     1
[4]    105,  -12,  9,     6,     1
[5]   -576,   30, 16,    30,     8,    1
[6]    105,   60, 25,   140,    57,   10,    1
[7]   5760, -140, 36,   630,   384,   90,   12,   1
[8] -13167, -280, 49,  2772,  2505,  772,  129,  14,  1
[9] -30400,  630, 64, 12012, 16008, 6430, 1328, 174, 16, 1 (End)

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..5 give A331551, A331552, A000290(n+1), A002457, A108666(n+1), A331323.
T(n,n+3) gives A331512.
Cf. A307883, A331513, A331514.

T := (n, k) -> (n + 1)^2*hypergeom([-n, -n], [2], k - 2):
seq(lprint(seq(simplify(T(n,k)), k=0..7)), n=0..6) # Peter Luschny, Jan 20 2020

T[n_, k_] := (n + 1)^2 * HypergeometricPFQ[{-n, -n}, {2}, k - 2];  Table[Table[T[n, k - n], {n, 0, k}], {k, 0, 9}] //Flatten (* Amiram Eldar, Jan 20 2020 *)

T(n,k) = Sum_{j=0..n} (k-3)^(n-j) * (n+j+1) * binomial(n,j) * binomial(n+j,j).
T(n,k) = Sum_{j=0..n} (k-2)^j * (j+1) * binomial(n+1,j+1)^2.
T(n,k) = (n + 1)^2*hypergeom([-n, -n], [2], k - 2). - Peter Luschny, Jan 20 2020
n * (2*n-1) * T(n,k) = 2 * (2 * (k-1) * n^2 - k + 2) * T(n-1,k) - (k-3)^2 * n * (2*n+1) * T(n-2,k) for n>1. - Seiichi Manyama, Jan 25 2020

A337464 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1+2(k-4)x+((k+4)x)^2) (1-(k+4)x+sqrt(1+2(k-4)x+((k+4)x)^2)) )).

Original entry on oeis.org

1, 1, 6, 1, 5, 30, 1, 4, 11, 140, 1, 3, -6, -29, 630, 1, 2, -21, -120, -365, 2772, 1, 1, -34, -139, -266, -1409, 12012, 1, 0, -45, -92, 531, 2520, -155, 51480, 1, -1, -54, 15, 1654, 6489, 17380, 29485, 218790, 1, -2, -61, 176, 2755, 4828, -9723, -13104, 170035, 923780

Offset: 0

Seiichi Manyama, Aug 28 2020

			Square array begins:
     1,     1,    1,    1,    1,     1, ...
     6,     5,    4,    3,    2,     1, ...
    30,    11,   -6,  -21,  -34,   -45, ...
   140,   -29, -120, -139,  -92,    15, ...
   630,  -365, -266,  531, 1654,  2755, ...
  2772, -1409, 2520, 6489, 4828, -5853, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..4 give A002457, A337394, A337466, A337467, A337397.
Main diagonal gives A337465.
Cf. A337369, A337419.

T[n_, k_] := Sum[If[k == n-j == 0, 1, (-k)^(n-j)] * Binomial[2*j, j] * Binomial[2*n+1, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Apr 29 2021 *)

{T(n, k) = sum(j=0, n, (-k)^(n-j)*binomial(2*j, j)*binomial(2*n+1, 2*j))}

T(n,k) = Sum_{j=0..n} (-k)^(n-j) * binomial(2*j,j) * binomial(2*n+1,2*j).

T(0,k) = 1, T(1,k) = 6-k and n * (2*n+1) * (4*n-3) * T(n,k) = (4*n-1) * (-4*(k-4)*n^2+2*(k-4)*n+k-2) * T(n-1,k) - (k+4)^2 * (n-1) * (2*n-1) * (4*n+1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 29 2020

A062344 Triangle of binomial(2*n, k) with n >= k.

Original entry on oeis.org

1, 1, 2, 1, 4, 6, 1, 6, 15, 20, 1, 8, 28, 56, 70, 1, 10, 45, 120, 210, 252, 1, 12, 66, 220, 495, 792, 924, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 1, 16, 120, 560, 1820, 4368, 8008, 11440, 12870, 1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 48620

Offset: 0

Henry Bottomley, Jul 06 2001

From Wolfdieter Lang, Sep 19 2012: (Start)
The triangle a(n,k) appears in the formula F(2*l+1)^(2*n) = (sum(a(n,k)*L(2*(n-k)*(2*l+1)),k=0..n-1) + a(n,n))/5^n, n>=0, l>=0, with F=A000045 (Fibonacci) and L=A000032 (Lucas).
The signed triangle as(n,k):=a(n,k)*(-1)^k appears in the formula F(2*l)^(2*n) = (sum(as(n,k)*L(4*(n-k)*l),k=0..n-1) + as(n,n))/5^n, n>=0, l>=0. Proof with the Binet-de Moivre formula for F and L and the binomial formula. (End)

			Rows start
  (1),
  (1,2),
  (1,4,6),
  (1,6,15,20)
  etc.
Row n=2, (1,4,6):
F(2*l+1)^4 = (1*L(4*(2*l+1)) + 4*L(2*(2*l+1)) + 6)/25,
F(2*l)^4 = (1*L(8*l) - 4*L(4*l) + 6)/25, l>=0, F=A000045, L=A000032. See a comment above. - _Wolfdieter Lang_, Sep 19 2012

G. C. Greubel, Rows n = 0..100 of triangle, flattened
E. H. M. Brietzke, An identity of Andrews and a new method for the Riordan array proof of combinatorial identities, Discrete Math., 308 (2008), 4246-4262.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
Index entries for triangles and arrays related to Pascal's triangle

Columns include (sometimes truncated) A000012, A005843, A000384, A002492, A053134 etc. Right hand side includes A000984, A001791, A002694, A002696 etc. Row sums are A032443. Row alternate differences (e.g., 6-4+1=3 or 20-15+6-1=10) are A001700.
Cf. A122366.
a(2*n,n) gives A005810.

[[Binomial(2*n, k): k in [0..n]]: n in [0..20]]; // G. C. Greubel, Jun 28 2018

Flatten[Table[Binomial[2 n, k], {n, 0, 20}, {k, 0, n}]] (* G. C. Greubel, Jun 28 2018 *)

create_list(binomial(2*n,k),n,0,9,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

for(n=0, 20, for(k=0, n, print1(binomial(2*n, k), ", "))) \\ G. C. Greubel, Jun 28 2018

a(n,k) = a(n,k-1)*((2*n+1)/k-1) with a(n,0)=1.

G.f.: 1/((1-sqrt(1-4*x*y))^4/(16*x*y^2) + sqrt(1-4*x*y) - x). - Vladimir Kruchinin, Jan 26 2021

A093375 Array T(m,n) read by ascending antidiagonals: T(m,n) = m*binomial(n+m-2, n-1) for m, n >= 1.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 6, 1, 5, 16, 18, 8, 1, 6, 25, 40, 30, 10, 1, 7, 36, 75, 80, 45, 12, 1, 8, 49, 126, 175, 140, 63, 14, 1, 9, 64, 196, 336, 350, 224, 84, 16, 1, 10, 81, 288, 588, 756, 630, 336, 108, 18, 1, 11, 100, 405, 960, 1470, 1512, 1050, 480, 135, 20, 1, 12

Offset: 1

Ralf Stephan, Apr 28 2004

Number of n-long m-ary words avoiding the pattern 1-1'2'.
T(n,n+1) = Sum_{i=1..n} T(n,i).
Exponential Riordan array [(1+x)e^x, x] as a number triangle. - Paul Barry, Feb 17 2009
From Peter Bala, Jul 22 2014: (Start)
Call this array M and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0  M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. The infinite matrix product M(0)*M(1)*M(2)*..., which is clearly well-defined, is equal to A059298. (End)

			Array T(m,n) (with rows m >= 1 and columns n >= 1) begins as follows:
   1   1   1   1   1   1 ...
   2   4   6   8  10  12 ...
   3   9  18  30  45  63 ...
   4  16  40  80 140 224 ...
   5  25  75 175 350 630 ...
   ...
Triangle S(n,k) = T(n-k+1, k+1) begins
.n\k.|....0....1....2....3....4....5....6
= = = = = = = = = = = = = = = = = = = = =
..0..|....1
..1..|....2....1
..2..|....3....4....1
..3..|....4....9....6....1
..4..|....5...16...18....8....1
..5..|....6...25...40...30...10....1
..6..|....7...36...75...80...45...12....1
...

Muniru A Asiru, Antidiagonals, n=1..100 flattened
Yasemin Alp and E. Gokcen Kocer, Exponential Almost-Riordan Arrays, Results Math 79, 173 (2024). See page 7.
Sergey Kitaev and Toufik Mansour, Partially ordered generalized patterns and k-ary words, arXiv:math/0210023 [math.CO], 2002.
Wikipedia, Sheffer sequence.

Rows include A045943. Columns include A002411, A027810.
Main diagonal is A037965. Subdiagonals include A002457.
Antidiagonal sums are A001792.
See A103283 for a signed version.
Cf. A103406, A059298, A073107 (unsigned inverse).

nmax:=14;; T:=List([1..nmax],n->List([1..nmax],k->k*Binomial(n+k-2,n-1)));;
b:=List([2..nmax],n->OrderedPartitions(n,2));;
a:=Flat(List([1..Length(b)],i->List([1..Length(b[i])],j->T[b[i][j][1]][b[i][j][2]]))); # Muniru A Asiru, Aug 07 2018

nmax = 10;
T = Transpose[CoefficientList[# + O[z]^(nmax+1), z]& /@ CoefficientList[(1 - x z)/(1 - z - x z)^2 + O[x]^(nmax+1), x]];
row[n_] := T[[n+1, 1 ;; n+1]];
Table[row[n], {n, 0, nmax}] // Flatten (* Jean-François Alcover, Aug 07 2018 *)

# uses[riordan_array from A256893]
riordan_array((1+x)*exp(x), x, 8, exp=true) # Peter Luschny, Nov 02 2019

Triangle = P*M, the binomial transform of the infinite bidiagonal matrix M with (1,1,1,...) in the main diagonal and (1,2,3,...) in the subdiagonal, and zeros elsewhere. P = Pascal's triangle as an infinite lower triangular matrix. - Gary W. Adamson, Nov 05 2006
From Peter Bala, Sep 20 2012: (Start)
E.g.f. for triangle: (1 + z)*exp((1 + x)*z) = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2/2! + ....
O.g.f. for triangle: (1 - x*z)/(1 - z - x*z)^2 = 1 + (2 + x)*z + (3 + 4*x + x^2)*z^2 + ....
The n-th row polynomial R(n,x) of the triangle equals (1+x)^n + n*(1+x)^(n-1) for n >= 0 and satisfies d/dx(R(n,x)) = n*R(n-1,x), as well as R(n,x+y) = Sum_{k = 0..n} binomial(n,k)*R(k,x)*y^(n-k). The row polynomials are a Sheffer sequence of Appell type.
Matrix inverse of the triangle is a signed version of A073107. (End)
From Tom Copeland, Oct 20 2015: (Start)
With offset 0 and D = d/dx, the raising operator for the signed row polynomials P(n,x) is RP = x - d{log[e^D/(1-D)]}/dD = x - 1 - 1/(1-D) =  x - 2 - D - D^2 + ..., i.e., RP P(n,x) = P(n+1,x).
The e.g.f. for the signed array is (1-t) * e^(-t) * e^(x*t).
From the Appell formalism, the row polynomials PI(n,x) of A073107 are the umbral inverse of this entry's row polynomials; that is, P(n,PI(.,x)) = x^n = PI(n,P(.,x)) under umbral composition. (End)
From Petros Hadjicostas, Nov 01 2019: (Start)
As a triangle, we let S(n,k) = T(n-k+1, k+1) = (n-k+1)*binomial(n, k) for n >= 0 and 0 <= k <= n. See the example below.
As stated above by Peter Bala, Sum_{n,k >= 0} S(n,k)*z^n*x^k = (1 - x*z)/(1 - z -x*z)^2.
Also, Sum_{n, k >= 0} S(n,k)*z^n*x^k/n! = (1+z)*exp((1+x)*z).
As he also states, the n-th row polynomial is R(n,x) = Sum_{k = 0..n} S(n, k)*x^k = (1 + x)^n + n*(1 + x)^(n-1).
If we define the signed triangle S*(n,k) = (-1)^(n+k) * S(n,k) = (-1)^(n+k) * T(n-k+1, k+1), as Tom Copeland states, Sum_{n,k >= 0} S^*(n,k)*t^n*x^k/n! = (1-t)*exp((1-x)*(-t)) = (1-t) * e^(-t) * e^(x*t).
Apparently, S*(n,k) = A103283(n,k).
As he says above, the signed n-th row polynomial is P(n,x) =  (-1)^n*R(n,-x) = (x - 1)^n - n*(x - 1)^(n-1).
According to Gary W. Adamson, P(n,x) is "the monic characteristic polynomial of the n X n matrix with 2's on the diagonal and 1's elsewhere." (End)

A097070 Consider all compositions (ordered partitions) of n into n parts, allowing zeros. E.g., for n = 3 we get 300, 030, 003, 210, 120, 201, 102, 021, 012, 111. Then a(n) is the total number of 1's.

Original entry on oeis.org

1, 2, 9, 40, 175, 756, 3234, 13728, 57915, 243100, 1016158, 4232592, 17577014, 72804200, 300874500, 1240940160, 5109183315, 21002455980, 86213785350, 353452638000, 1447388552610, 5920836618840, 24197138082780, 98801168731200, 403095046038750, 1643337883690776, 6694900194799404

Offset: 1

Amy J. Kolan, Sep 15 2004

Number of compositions of n into n parts, allowing zeros = binomial(2*n-1,n) = A088218 = essentially A001700.

			The compositions for n=2 are 20, 02, 11. There are two 1's in these so a(2) = 2.
From _Robert G. Wilson v_, Sep 16 2004: (Start)
The case n = 5:
A. There are 5 combinations associated with the numbers 50000: 50000, 05000, 00500, 00050, 00005.
B. There are 20 combinations associated with the numbers 41000.
C. There are 20 combinations associated with 32000.
D. There are 30 combinations associated with 31100.
E. There are 30 combinations associated with 22100.
F. There are 20 combinations associated with 21110.
G. There is one combinations associated with 11111.
The number of 1's associated with A is 0, with B 20, with C 0, with D 60, with E 30, with F 60 and with G 5. 0 + 20 + 0 + 60 + 30 + 60 + 5 = 175.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
G.-S. Cheon, H. Kim, and L. W. Shapiro, Mutation effects in ordered trees, arXiv preprint arXiv:1410.1249 [math.CO], 2014.

Cf. A037965.
Cf. A000984, A001622, A088218.
cf. A005430, A002740, A002457, A000917, A110609, A097613, A001791, A110609.

List([1..30], n-> n*Binomial(2*n-3, n-1)); # G. C. Greubel, Jul 27 2019

[n*Binomial(2*n-3, n-1): n in [1..30]]; // Vincenzo Librandi, Jul 13 2019

A097070 := n -> ifelse(n=1, 1, 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3)):
seq(simplify(A097070(n)), n = 1..28);  # Peter Luschny, Jan 22 2025

Table[n*Binomial[2n-3, n-1], {n, 30}] (* Robert G. Wilson v, Sep 17 2004 *)

a(n) = n*binomial(2*n-3, n-1); \\ Joerg Arndt, Feb 17 2015

[n*binomial(2*n-3, n-1) for n in (1..30)] # G. C. Greubel, Jul 27 2019

a(n) = n*binomial(2*n-3, n-1).
More generally, total number of k's (k>=0) in all ordered partitions of n into n parts, allowing zeros, is n*binomial(2*n-k-2, n-2) if n >= k, 0 otherwise.
Total number of 0's is given by A005430.
From Vladeta Jovovic, Sep 17 2004: (Start)
a(n) = Sum_{k=0..n} k*binomial(n, k)*binomial(n-2, k-2).
G.f.: x*(1 -2*x +(1-4*x)^(3/2))/(2*(1-4*x)^(3/2)).
E.g.f.: (x/2)*(exp(2*x)*BesselI(0, 2*x)+1). (End)
a(n) = A014107(n)*A000108(n-2). - Philippe Deléham, Apr 12 2007
a(n) = n*A088218(n-1) for n > 0. - Werner Schulte, Jan 22 2017
From Bruce J. Nicholson, Jul 11 2019: (Start)
a(n) = A002740(n) + A097613(n).
a(n) = A110609(n-1) - A002457(n-2) + A097613(n).
a(n) = A005430(n-1) - A000917(n-3) for n > 1.
a(n) = A002457(n-1) - A037965(n) - A000917(n-3) for n > 1.
a(n) = A037965(n)/2.
a(n) = A001700(n-2)*n.
a(n) = A001791(n-2)*n + A000984(n-2)*n for n > 1. (End)
From Amiram Eldar, May 16 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)
a(n) = 2^(n-2)*JacobiP(n-1, -1/2, -n+2, 3) for n > 1. - Peter Luschny, Jan 22 2025

Formula, more terms and comments from Vladeta Jovovic, Sep 15 2004

A167859 a(n) = 4^n * Sum_{k=0..n} binomial(2*k, k)^2 / 4^k.

Original entry on oeis.org

1, 8, 68, 672, 7588, 93856, 1229200, 16695424, 232418596, 3293578784, 47309094672, 686870685312, 10059942413584, 148412250014336, 2202990595617344, 32873407393419776, 492791264816231204

Offset: 0

Alexander Adamchuk, Nov 13 2009

Every a(n) from a((p-1)/2) to a(p-1) is divisible by prime p for p = {7, 47, 191, 383, 439, 1151, 1399, 2351, 2879, 3119, 3511, 3559, ...} = A167860, apparently a subset of primes of the form 8n+7 (A007522).
7^3 divides a(13) and 7^2 divides a(10)-a(13).
Every a(n) from a(kp-1 - (p-1)/2) to a(kp-1) is divisible by prime p from A167860.
Every a(n) from a((p^2-1)/2) to a(p^2-1) is divisible by prime p from A167860. For p=7 every a(n) from a((p^3-1)/2) to a(p^3-1) and from a((p^4-1)/2) to a(p^4-1)is divisible by p^2.

Vincenzo Librandi, Table of n, a(n) for n = 0..300

Cf. A000984, A066796, A006134, A082590, A132310, A002457, A144635, A167713, A167860, A007522.

A167859 := proc(n)
    add( (binomial(2*k,k)/2^k)^2,k=0..n) ;
    4^n*% ;
end proc:
seq(A167859(n),n=0..20) ; # R. J. Mathar, Sep 21 2016

Table[4^n*Sum[Binomial[2*k,k]^2/4^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)

a(n) = 4^n*sum(k=0,n, binomial(2*k,k)^2/4^k) \\ Charles R Greathouse IV, Sep 21 2016

Recurrence: n^2*a(n) = 4*(5*n^2 - 4*n + 1)*a(n-1) - 16*(2*n - 1)^2*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(4*n+2)/(3*Pi*n). - Vaclav Kotesovec, Oct 20 2012
G.f.: 2*EllipticK(4*sqrt(x))/(Pi*(1-4*x)), where EllipticK is the complete elliptic integral of the first kind, using the Gradshteyn and Ryzhik convention, also used by Maple.  In the convention of Abramowitz and Stegun, used by Mathematica, this would be written as 2*K(16*x)/(Pi*(1-4*x)). - Robert Israel, Sep 21 2016

More terms from Sean A. Irvine, Apr 14 2010

Further terms from Jon E. Schoenfield, May 09 2010

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)binomial(n,k)binomial(n+k,k) (n >= k >= 0).

Original entry on oeis.org

-1, -1, 2, -1, 6, -6, -1, 12, -30, 20, -1, 20, -90, 140, -70, -1, 30, -210, 560, -630, 252, -1, 42, -420, 1680, -3150, 2772, -924, -1, 56, -756, 4200, -11550, 16632, -12012, 3432, -1, 72, -1260, 9240, -34650, 72072, -84084, 51480, -12870, -1, 90, -1980, 18480, -90090, 252252, -420420, 411840, -218790, 48620, -1, 110, -2970, 34320, -210210, 756756, -1681680, 2333760, -1969110, 923780, -184756

Offset: 0

N. J. A. Sloane, Jan 17 2020

This is Table I of Ser (1933), page 92.
From Petros Hadjicostas, Jul 09 2020: (Start)
Essentially Ser (1933) in his book (and in particular for Tables I-IV) finds triangular arrays that allow him to express the coefficients of various kinds of series in terms of the coefficients of other series.
He uses Newton's series (or some variation of it), factorial series, and inverse factorial series. Unfortunately, he uses unusual notation, and as a result it is difficult to understand what he is actually doing.
Rivoal (2008, 2009) essentially uses factorial series and transformations to other kinds of series to provide new proofs of the irrationality of log(2), zeta(2), and zeta(3). As a result, the triangular array T(n,k) appears in various parts of his papers.
We believe Table I (p. 92) in Ser (1933), regarding the numbers T(n,k), corresponds to four different formulas. We have deciphered the first two of them. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k=0..n) begins:
  -1;
  -1,  2;
  -1,  6,   -6;
  -1, 12,  -30,   20;
  -1, 20,  -90,  140,    -70;
  -1, 30, -210,  560,   -630,   252;
  -1, 42, -420, 1680,  -3150,  2772,   -924;
  -1, 56, -756, 4200, -11550, 16632, -12012, 3432;
  ...
From _Petros Hadjicostas_, Jul 11 2020: (Start)
Its inverse (from Table II, p. 92) is
  -1;
  -1/2, 1/2;
  -1/3, 1/2,   -1/6;
  -1/4, 9/20,  -1/4,  1/20;
  -1/5, 2/5,   -2/7,  1/10, -1/70;
  -1/6, 5/14, -25/84, 5/36, -1/28,  1/252;
  -1/7, 9/28, -25/84, 1/6,  -9/154,  1/84, -1/924;
   ... (End)

J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, pp. 92-93.

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)
Tanguy Rivoal, Applications arithmétiques de l'interpolation lagrangienne, preprint (2008); see pp. 1 and 15.
Tanguy Rivoal, Applications arithmétiques de l'interpolation lagrangienne, Int. J. Number Theory 5.2 (2009), 185-208; see pp. 185 and 199.

A063007 is the same triangle without the minus signs, and has much more information.
Columns 1 and 2 are A002378 and A033487; the last three diagonals are A002544, A002457, A000984.
Cf. A331431, A331432.

/* As triangle: */ [[(-1)^(k+1) * Factorial(n+k) / (Factorial(k) * Factorial(k) * Factorial(n-k)): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jan 19 2020

Table[CoefficientList[-Hypergeometric2F1[-n, n + 1, 1, x], x], {n, 0, 9}] // Flatten (* Georg Fischer, Jan 18 2020 after Peter Luschny in A063007 *)

def T(n,k): return (-1)^(k+1)*falling_factorial(n+k,2*k)/factorial(k)^2
flatten([[T(n,k) for k in (0..n)] for n in (0..10)]) # Peter Luschny, Jul 09 2020

T(n,k) can also be written as (-1)^(k+1)*(n+k)!/(k!*k!*(n-k)!).
From Petros Hadjicostas, Jul 09 2020: (Start)
Ser's first formula from his Table I (p. 92) is the following:
Sum_{k=0..n} T(n,k)*k!/(x*(x+1)*...*(x+k)) = -(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, Sum_{k=0..n} T(n,k)/binomial(m+k, k) = 0 for m = 1..n.
Ser's second formula from his Table I appears also in Rivoal (2008, 2009) in a slightly different form:
Sum_{k=0..n} T(n,k)/(x + k) = (-1)^(n+1)*(x-1)*(x-2)*...*(x-n)/(x*(x+1)*...*(x+n)).
As a result, for m = 1..n, Sum_{k=0..n} T(n,k)/(m + k) = 0. (End)
T(n,k) = (-1)^(k+1)*FallingFactorial(n+k,2*k)/(k!)^2. - Peter Luschny, Jul 09 2020
From Petros Hadjicostas, Jul 10 2020: (Start)
Peter Luschny's formula above is essentially the way the numbers T(n,k) appear in Eq. (7) on p. 86 of Ser's (1933) book. Eq. (7) is essentially equivalent to the first formula above (related to Table I on p. 92).
By inverting that formula (in some way), he gets
n!/(x*(x+1)*...*(x+n)) = Sum_{p=0..n} (-1)^p*(2*p+1)*f_p(n+1)*f_p(x), where f_p(x) = (x-1)*...*(x-p)/(x*(x+1)*...*(x+p)). This is equivalent to Eq. (8) on p. 86 of Ser's book.
The rational coefficients A(n,p) = (2*p+1)*f_p(n+1) = (2*p+1)*(n*(n-1)*...*(n+1-p))/((n+1)*...*(n+1+p)) appear in Table II on p. 92 of Ser's book.
If we consider the coefficients T(n,k) and (-1)^(p+1)*A(n,p) as infinite lower triangular matrices, then they are inverses of one another (see the example below). This means that, for m >= s,
Sum_{k=s..m} T(m,k)*(-1)^(s+1)*A(k,s) = I(s=m) = Sum_{k=s..m} (-1)^(k+1)*A(m,k)*T(k,s), where I(s=m) = 1, if s = m, and = 0, otherwise.
Without the (-1)^p, we get the formula
1/(x+n) = Sum_{p=0..n} (2*p+1)*f_p(n+1)*f_p(x),
which apparently is the inversion of the second of Ser's formulas (related to Table I on p. 92).
In all of the above formulas, an empty product is by definition 1, so f_0(x) = 1/x. (End)

Thanks to Bob Selcoe, who noticed a typo in one of the entries, which, when corrected, led to an explicit formula for the whole of Ser's Table I.

cp's OEIS Frontend

A090816 a(n) = (3*n+1)!/((2*n)! * n!).

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A092443 Sequence arising from enumeration of domino tilings of Aztec Pillow-like regions.

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A167713 a(n) = 16^n * Sum_{k=0..n} binomial(2*k, k) / 16^k.

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A331511 Square array T(n,k), n >= 0, k >= 0, read by descending antidiagonals, where column k is the expansion of (1 - (k-3)*x)/(1 - 2*(k-1)*x + ((k-3)*x)^2)^(3/2).

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A337464 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1+2*(k-4)*x+((k+4)*x)^2) * (1-(k+4)*x+sqrt(1+2*(k-4)*x+((k+4)*x)^2)) )).

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A062344 Triangle of binomial(2*n, k) with n >= k.

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A093375 Array T(m,n) read by ascending antidiagonals: T(m,n) = m*binomial(n+m-2, n-1) for m, n >= 1.

A090816 a(n) = (3n+1)!/((2n)! * n!).

A331511 Square array T(n,k), n >= 0, k >= 0, read by descending antidiagonals, where column k is the expansion of (1 - (k-3)x)/(1 - 2(k-1)x + ((k-3)x)^2)^(3/2).

A337464 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt(2 / ( (1+2(k-4)x+((k+4)x)^2) (1-(k+4)x+sqrt(1+2(k-4)x+((k+4)x)^2)) )).

A331430 Triangle read by rows: T(n, k) = (-1)^(k+1)binomial(n,k)binomial(n+k,k) (n >= k >= 0).