A002544 a(n) = binomial(2*n+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

Author

N. J. A. Sloane, Simon Plouffe

Keywords

• nonn
• easy
• nice

Comments

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

References

• 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).

Links

• 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.

Crossrefs

Cf. A085373, A002457, A002674, A202543.

Equals A002736/2.

A diagonal of A331430.

Programs

Maple

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

Mathematica

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

PARI

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

Python

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

Formula

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