A085373 -id:A085373 - cp's OEIS frontend

A002544 a(n) = binomial(2n+1,n)(n+1)^2.

1, 12, 90, 560, 3150, 16632, 84084, 411840, 1969110, 9237800, 42678636, 194699232, 878850700, 3931426800, 17450721000, 76938289920, 337206098790, 1470171918600, 6379820115900, 27569305764000, 118685861314020, 509191949220240, 2177742427450200, 9287309860732800

Offset: 0

N. J. A. Sloane, Simon Plouffe

Coefficients for numerical differentiation.
Take the first n integers 1,2,3..n and find all combinations with repetitions allowed for the first n of them. Find the sum of each of these combinations to get this sequence. Example for 1 and 2: 1,2,1+1,1+2,2+2 gives sum of 12=a(2). - J. M. Bergot, Mar 08 2016
Let cos(x) = 1 -x^2/2 +x^4/4!-x^6/6!.. = Sum_i (-1)^i x^(2i)/(2i)! be the standard power series of the cosine, and y = 2*(1-cos(x)) = 4*sin^2(x/2) = x^2 -x^4/12 +x^6/360 ...= Sum_i 2*(-1)^(i+1) x^(2i)/(2i)! be a closely related series. Then this sequence represents the reversion x^2 = Sum_i 1/a(i) *y^(i+1). - R. J. Mathar, May 03 2022

C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
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).

T. D. Noe, Table of n, a(n) for n=0..200
W. G. Bickley and J. C. P. Miller, Numerical differentiation near the limits of a difference table, Phil. Mag., 33 (1942), 1-12 (plus tables) [Annotated scanned copy]
Ömür Deveci and Anthony G. Shannon, Some aspects of Neyman triangles and Delannoy arrays, Mathematica Montisnigri (2021) Vol. L, 36-43.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
A. Petojevic and N. Dapic, The vAm(a,b,c;z) function, Preprint 2013.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135.
H. E. Salzer, Coefficients for numerical differentiation with central differences, J. Math. Phys., 22 (1943), 115-135. [Annotated scanned copy]
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)
R. Shenton and A. W. Kemp, An S-fraction and ln^2(1+x), Journal of Computational and Applied Mathematics, 26 (1989) 367-370 North-Holland.
T. R. Van Oppolzer, Lehrbuch zur Bahnbestimmung der Kometen und Planeten, Vol. 2, Engelmann, Leipzig, 1880, p. 21.
Mats Vermeeren, A dynamical solution to the Basel problem, arXiv preprint arXiv:1506.05288 [math.CA], 2015.

Cf. A085373, A002457, A002674, A202543.
Equals A002736/2.
A diagonal of A331430.

seq((n+1)^2*(binomial(2*n+2, n+1))/2, n=0..29); # Zerinvary Lajos, May 31 2006

Table[Binomial[2n+1,n](n+1)^2,{n,0,20}] (* Harvey P. Dale, Mar 23 2011 *)

PARI
```
a(n)=binomial(2*n+1,n)*(n+1)^2
```

x='x+O('x^99); Vec((1+2*x)/(1-4*x)^(5/2)) \\ Altug Alkan, Jul 09 2016

from sympy import binomial
def a(n): return binomial(2*n + 1, n)*(n + 1)**2 # Indranil Ghosh, Apr 18 2017

G.f.: (1 + 2x)/(1 - 4x)^(5/2).
a(n-1) = sum(i_1 + i_2 + ... + i_n) where the sum is over 0 <= i_1 <= i_2 <= ... <= i_n <= n; a(n) = (n+1)^2 C(2n+1, n). - David Callan, Nov 20 2003
a(n) = (n+1)^2 * binomial(2*n+2,n+1)/2. - Zerinvary Lajos, May 31 2006
Asymptotics: a(n)-> (1/64) * (128*n^2+176*n+41) * 4^n * n^(-1/2)/(sqrt(Pi)), for n->infinity. - Karol A. Penson, Aug 05 2013
G.f.: 2F1(3/2,2;1;4x). - R. J. Mathar, Aug 09 2015
a(n) = A002457(n)*(n+1). - R. J. Mathar, Aug 09 2015
a(n) = A000217(n)*A000984(n). - J. M. Bergot, Mar 10 2016
a(n-1) = A001791(n)*n*(n+1)/2. - Anton Zakharov, Jul 04 2016
From Ilya Gutkovskiy, Jul 04 2016: (Start)
E.g.f.: ((1 + 2*x)*(1 + 8*x)*BesselI(0,2*x) + 2*x*(3 + 8*x)*BesselI(1,2*x))*exp(2*x).
Sum_{n>=0} 1/a(n) = Pi^2/9 = A100044. (End)
From Peter Bala, Apr 18 2017: (Start)
With x = y^2/(1 + y) we have log^2(1 + y) = Sum_{n >= 0} (-1)^n*x^(n+1)/a(n). See Shenton and Kemp.
Series reversion ( Sum_{n >= 0} (-1)^n*x^(n+1)/a(n) ) = Sum_{n >= 1} 2*x^n/(2*n)! = Sum_{n >= 1} x^n/A002674(n). (End)
D-finite with recurrence n^2*a(n) -2*(n+1)*(2*n+1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)^2 = A202543^2. - Amiram Eldar, May 14 2022

A123160 Triangle read by rows: T(n,k) = n!(n+k-1)!/((n-k)!(n-1)!*(k!)^2) for 0 <= k <= n, with T(0,0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 9, 18, 10, 1, 16, 60, 80, 35, 1, 25, 150, 350, 350, 126, 1, 36, 315, 1120, 1890, 1512, 462, 1, 49, 588, 2940, 7350, 9702, 6468, 1716, 1, 64, 1008, 6720, 23100, 44352, 48048, 27456, 6435, 1, 81, 1620, 13860, 62370, 162162, 252252, 231660, 115830, 24310

Offset: 0

Roger L. Bagula, Oct 02 2006

T(n,k) is also the number of order-preserving partial transformations (of an n-element chain) of width k (width(alpha) = |Dom(alpha)|). - Abdullahi Umar, Aug 25 2008

			Triangle begins:
  1;
  1,  1;
  1,  4,   3;
  1,  9,  18,  10;
  1, 16,  60,  80,  35;
  1, 25, 150, 350, 350, 126;
  ...

Frederick T. Wall, Chemical Thermodynamics, W. H. Freeman, San Francisco, 1965 pages 296 and 305

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.

Cf. A005773, A037965, A059481, A085373, A088218, A088617, A122899, A123164, A178301.

[Binomial(n,k)*Binomial(n+k-1,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jun 19 2022

T:=proc(n,k) if k=0 and n=0 then 1 elif k<=n then n!*(n+k-1)!/(n-k)!/(n-1)!/(k!)^2 else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, m_]= If [n==m==0, 1, n!*(n+m-1)!/((n-m)!*(n-1)!(m!)^2)];
Table[T[n, m], {n,0,10}, {m,0,n}]//Flatten
max = 9; s = (x+1)/(2*Sqrt[(1-x)^2-4*y])+1/2 + O[x]^(max+2) + O[y]^(max+2); T[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {y, 0, k}]; Table[T[n-k, k], {n, 0, max}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 18 2015, after Vladimir Kruchinin *)

def A123160(n,k): return binomial(n, k)*binomial(n+k-1, k)
flatten([[A123160(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jun 19 2022

T(n, m) = n!*(n + m - 1)!/((n - m)!*(n - 1)!(m!)^2), with T(0, 0) = 1.
T(n, k) = binomial(n,k)*binomial(n+k-1,k). The row polynomials (except the first) are (1+x)*P(n,0,1,2x+1), where P(n,a,b,x) denotes the Jacobi polynomial. The columns of this triangle give the diagonals of A122899. - Peter Bala, Jan 24 2008
T(n, k) = binomial(n,k)*(n+k-1)!/((n-1)!*k!).
T(n, k)= binomial(n,k)*binomial(n+k-1,n-1). - Abdullahi Umar, Aug 25 2008
G.f.: (x+1)/(2*sqrt((1-x)^2-4*y)) + 1/2. - Vladimir Kruchinin, Jun 16 2015
From _Peter Bala, Jul 20 2015: (Start)
O.g.f. (1 + x)/( 2*sqrt((1 - x)^2 - 4*x*y) ) + 1/2 = 1 + (1 + y)*x + (1 + 4*y + 3*y^2)*x^2 + ....
For n >= 1, the n-th row polynomial R(n,y) = (1 + y)*r(n-1,y), where r(n,y) is the n-th row polynomial of A178301.
exp( Sum_{n >= 1} R(n,y)*x^n/n ) = 1 + (1 + y)*x + (1 + 3*y + 2*y^2)*x^2 + ... is the o.g.f for A088617. (End)
From G. C. Greubel, Jun 19 2022: (Start)
T(n, n) = A088218(n).
T(n, n-1) = A037965(n).
T(n, n-2) = A085373(n-2).
Sum_{k=0..n} T(n, k) = A123164(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A005773(n). (End)

Edited by N. J. A. Sloane, Oct 26 2006 and Jul 03 2008

A141611 Triangle read by rows: T(n, k) = (n-k+1)(k+1)binomial(n, k).

Original entry on oeis.org

1, 2, 2, 3, 8, 3, 4, 18, 18, 4, 5, 32, 54, 32, 5, 6, 50, 120, 120, 50, 6, 7, 72, 225, 320, 225, 72, 7, 8, 98, 378, 700, 700, 378, 98, 8, 9, 128, 588, 1344, 1750, 1344, 588, 128, 9, 10, 162, 864, 2352, 3780, 3780, 2352, 864, 162, 10, 11, 200, 1215, 3840, 7350, 9072, 7350, 3840, 1215, 200, 11

Offset: 0

Roger L. Bagula and Gary W. Adamson, Aug 22 2008

Read as a square array, this array factorizes as M*transpose(M), where M = ( k*binomial(n, k) )A003506(n,k).%20-%20_Peter%20Bala">{n,k>=1} = A003506(n,k). - _Peter Bala, Mar 06 2017

			Triangle begins as:
   1;
   2,   2;
   3,   8,    3;
   4,  18,   18,    4;
   5,  32,   54,   32,    5;
   6,  50,  120,  120,   50,    6;
   7,  72,  225,  320,  225,   72,    7;
   8,  98,  378,  700,  700,  378,   98,    8;
   9, 128,  588, 1344, 1750, 1344,  588,  128,    9;
  10, 162,  864, 2352, 3780, 3780, 2352,  864,  162,  10;
  11, 200, 1215, 3840, 7350, 9072, 7350, 3840, 1215, 200, 11;
  ...
From _Peter Bala_, Mar 06 2017: (Start)
Factorization as a square array
  /1         \ /1  2  3  4...\ /1  2   3   4...\
  |2  2      | |   2  6 12...| |2  8  12  32...|
  |3  6  3   |*|      3 12...|=|3 18  54 120...|
  |4 12 12 4 | |         4...| |4 32 120 320...|
  |...       | |             | |...            |
(End)

Indranil Ghosh, Table of n, a(n) for n = 1..5151 (Rows 0..100 of triangle, flattened)

Cf. A003506, A007466 (row sums), A037966, A085373.

A141611:= func< n,k | (k+1)*(n-k+1)*Binomial(n,k) >;
[A141611(n,k): k in [0..n], n in [0..14]]; // G. C. Greubel, Sep 22 2024

T[n_, m_]:= (n-m+1)*(m+1)*Binomial[n,m];
Table[T[n, m], {n,0,12}, {m,0,n}]//Flatten

T(n,m)=(n - m + 1)*(m + 1)*binomial(n, m) \\ Charles R Greathouse IV, Feb 15 2017

def A141611(n,k): return (k+1)*(n-k+1)*binomial(n,k)
flatten([[A141611(n,k) for k in range(n+1)] for n in range(15)]) # G. C. Greubel, Sep 22 2024

T(n, k) = (k+1)*(n-k+1)*binomial(n,k).
Sum_{k=0..n} T(n, k) = A007466(n+1) (row sums).
O.g.f.: (1 - (1 + t)*x + 2*t*x^2)/(1 - (1 + t)*x)^3 = 1 + (2 + 2*t)*x + (3 + 8*t + 3*t^2)*x^2 + (4 + 18*t + 18*t^2 + 4*t^3)*x^3 + .... - Peter Bala, Mar 06 2017
From G. C. Greubel, Sep 22 2024: (Start)
T(2*n, n) = A037966(n+1).
T(2*n-1, n) = 2*A085373(n-1), for n >= 1.
Sum_{k=0..n} (-1)^k*T(n, k) = A000007(n) - 2*[n=2]. (End)

Offset corrected by G. C. Greubel, Sep 22 2024

A119578 a(n) = (n + n^2)binomial(2n,n)/2.

Original entry on oeis.org

0, 2, 18, 120, 700, 3780, 19404, 96096, 463320, 2187900, 10161580, 46558512, 210924168, 946454600, 4212243000, 18614102400, 81746933040, 357041751660, 1551848136300, 6715600122000, 28947771052200, 124337568995640, 532337037821160, 2272426880817600, 9674281104930000

Offset: 0

Zerinvary Lajos, May 31 2006

For n > 0, also the number of one-sided prudent walks from (0,0) to (n,n), with n+2 east steps, 2 west steps and n north steps.

Bruno Berselli, Table of n, a(n) for n = 0..500
Shanzhen Gao and Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
S. Gao and H. Niederhausen, Sequences Arising From Prudent Self-Avoiding Walks, 2010.

Cf. A000984, A001622, A085373.

[0] cat [ (n+1)*Factorial(2*n-1)/Factorial(n-1)^2: n in [1..23] ]; // Klaus Brockhaus, Apr 30 2011

[seq ((n+n^2)*(binomial(2*n,n))/2,n=0..29)];

Table[(n+n^2) Binomial[2n,n]/2,{n,0,30}] (* Harvey P. Dale, Jun 02 2016 *)

a(n) = (n+1)*Gamma(2*n)/Gamma(n)^2 for n > 0. - Shanzhen Gao, Apr 26 2011
G.f.: 2 * x * (1 - x) / (1 - 4*x)^(5/2). - Ilya Gutkovskiy, Nov 17 2021
From Amiram Eldar, May 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 2*Pi^2/9 - 2*Pi/sqrt(3) + 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*sqrt(5)*log(phi) - 8*log(phi)^2 - 2, where phi is the golden ratio (A001622). (End)
D-finite with recurrence (-n+1)*a(n) +(5*n-1)*a(n-1) +2*(-2*n+3)*a(n-2)=0. - R. J. Mathar, Jul 08 2022
a(n) = A000217(n)*A000984(n). - R. J. Mathar, Jul 08 2022

A085374 a(n) = binomial(2n+1, n+1)*binomial(n+3, 3).

Original entry on oeis.org

1, 12, 100, 700, 4410, 25872, 144144, 772200, 4011150, 20323160, 100876776, 492156392, 2366136500, 11232648000, 52739956800, 245240799120, 1130632213590, 5172827121000, 23504600427000, 106141827191400, 476627347816620, 2129348151284640, 9468445336740000

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jun 26 2003

Cf. A001700, A002457, A085373.

Cf. A000292, A000984, A001622.

seq(binomial(n+2,3)/2*binomial(2*n,n), n=1..20); # Zerinvary Lajos, Jan 18 2007

Table[Binomial[2 n + 1, n + 1]Binomial[n + 3, 3], {n, 0, 30}]

a(n) = A000292(n+1)*A000984(n+1)/2. - Zerinvary Lajos, Jan 18 2007, corrected Aug 09 2015
From R. J. Mathar, Aug 09 2015: (Start)
D-finite with recurrence n*(n+1)*a(n) - 2*(n+3)*(2*n+1)*a(n-1) = 0.
G.f.: 2F1(3/2,4;2;4x). (End)
a(n) ~ 2^(2*n)*n^(5/2)/(3*sqrt(Pi)). - Stefano Spezia, Aug 31 2025
From Amiram Eldar, Sep 06 2025: (Start)
a(n) = A001700(n) * A000292(n+1).
Sum_{n>=0} 1/a(n) = 10*sqrt(3)*Pi - 8*Pi^2/3 - 27.
Sum_{n>=0} (-1)^n/a(n) = 84*sqrt(5)*log(phi) - 192*log(phi)^2 - 45, where phi is the golden ratio (A001622). (End)

A085375 a(n) = binomial(2n+1, n+1)binomial(n+4, 4).

Original entry on oeis.org

1, 15, 150, 1225, 8820, 58212, 360360, 2123550, 12033450, 66050270, 353068716, 1845586470, 9464546000, 47738754000, 237329805600, 1164893795820, 5653161067950, 27157342385250, 129275302348500, 610315506350550, 2859764086899720, 13308425945529000

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jun 26 2003

Chai Wah Wu, Table of n, a(n) for n = 0..500

Cf. A001700, A002457, A085373, A085374.

Cf. A000332, A000984.

seq(binomial(2*n+1, n+1)*binomial(n+4, 4), n=0..20); # Zerinvary Lajos, Jan 18 2007

Table[Binomial[2*n + 1, n + 1] * Binomial[n + 4, 4], {n, 0, 30}]

from _future_ import division
A085375_list, b = [], 1
for n in range(501):
    A085375_list.append(b)
    b = b*2*(n+5)*(2*n+3)//((n+1)*(n+2)) # Chai Wah Wu, Jan 26 2016

a(n+1) = a(n)*2*(n+5)*(2*n+3)/((n+1)*(n+2)). - Chai Wah Wu, Jan 26 2016

G.f.: (1 - 3*x + 6*x^2 - 5*x^3) / (1 - 4*x)^(9/2). - Ilya Gutkovskiy, Nov 17 2021

cp's OEIS Frontend

A002544 a(n) = binomial(2*n+1,n)*(n+1)^2.

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A123160 Triangle read by rows: T(n,k) = n!*(n+k-1)!/((n-k)!*(n-1)!*(k!)^2) for 0 <= k <= n, with T(0,0) = 1.

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A141611 Triangle read by rows: T(n, k) = (n-k+1)*(k+1)*binomial(n, k).

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A119578 a(n) = (n + n^2)*binomial(2*n,n)/2.

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A085374 a(n) = binomial(2n+1, n+1)*binomial(n+3, 3).

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A085375 a(n) = binomial(2*n+1, n+1)*binomial(n+4, 4).

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A002544 a(n) = binomial(2n+1,n)(n+1)^2.

A123160 Triangle read by rows: T(n,k) = n!(n+k-1)!/((n-k)!(n-1)!*(k!)^2) for 0 <= k <= n, with T(0,0) = 1.

A141611 Triangle read by rows: T(n, k) = (n-k+1)(k+1)binomial(n, k).

A119578 a(n) = (n + n^2)binomial(2n,n)/2.

A085375 a(n) = binomial(2n+1, n+1)binomial(n+4, 4).